Problem DefinitionBuild a Date class and a main function to test i.pdf
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Problem Definition Build a Date class and a main function to test it. Specifications Below is the interface for the Date class: it is our \"contract\" with you: you have to implement everything it describes, and show us that it works with a test harness that puts it through its paces. The comments in the interface below should be sufficient for you to understand the project (use these comments in your Date declaration), without the need of any further documentation. But of course, as always, you can ask us any questions you may have on Piazza. Note: Placing the error messages into the constructors like is not necessarily a good way to handle constructor errors, but until you learn about exceptions in CS 14, it\'s the best we can do. Private Member Functions The functions declared private above, isLeap, daysPerMonth, name, number, are helper functions - member functions that will never be needed by a user of the class, and so do not belong to the public interface (which is why they are \"private\"). They are, however, needed by the interface functions (public member functions), which use them to test the validity of arguments and construct valid dates. For example, the constructor that passes in the month as a string will call the number function to assign a value to the unsigned member variable month. isLeap: The rule for whether a year is a leap year is: (year % 4 == 0) implies leap year except (year % 100 == 0) implies NOT leap year except (year % 400 == 0) implies leap year So, for instance, year 2000 is a leap year, but 1900 is NOT a leap year. Years 2004, 2008, 2012, 2016, etc. are all leap years. Years 2005, 2006, 2007, 2009, 2010, etc. are NOT leap years. Output Specifications Read the specifications for the print function carefully. The only cout statements within your Date member functions should be: the \"Invalid Date\" warnings in the constructors in your two print functions Required Main Function You must use this main function and global function getDate as they are here. You may not change these functions at all. Copy-and-paste these into your main.cpp file and then add the Date class. Solution // Date.h #include #include #include #include using namespace std; class Date { private: unsigned day; unsigned month; string monthName; unsigned year; public: Date(); Date(unsigned m, unsigned d, unsigned y); Date(const string &mn, unsigned d, unsigned y); void printNumeric() const; void printAlpha() const; private: bool isLeap(unsigned y) const; unsigned daysPerMonth(unsigned m, unsigned y) const; string name(unsigned m) const; unsigned number(const string &mn) const; }; //Date.cpp #include #include #include #include using namespace std; #include \"Date.h\" // creates the date January 1st, 2000. Date::Date() { day = 1; month = 1; monthName = \"January\"; year = 2000; } /* parameterized constructor: month number, day, year - e.g. (3, 1, 2010) will construct the date March 1st, 2010 If any of the arguments are invalid (e.g. 15 for month o.
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CounterTest.javaimport static org.junit.Assert.;import org.jun.pdf
CounterTest.java:
import static org.junit.Assert.*;
import org.junit.Test;
public class CounterTest {
// FIRST
// Write a class Counter such that the following test works
@Test
public void testZero() {
Counter cnt = new Counter(0);
assertEquals(\"initial value of (0) failed\", 0, cnt.getCount());
cnt.increase();
assertEquals(\"increased value of (0) failed\", 0, cnt.getCount());
cnt.decrease();
assertEquals(\"decreased value of (0) failed\", 0, cnt.getCount());
}
// SECOND
// Uncomment the following method,
// modify your Counter class so that this test works.
/*
@Test
public void testIncrease() {
Counter cnt = new Counter(7);
assertEquals(\"initial value of (7) failed\", 0, cnt.getCount());
cnt.increase();
assertEquals(\"increased once value of (7) failed\", 7, cnt.getCount());
cnt.increase();
assertEquals(\"twice increased once value of (7) failed\", 14,
cnt.getCount());
}
*/
// THIRD
// Uncomment the following method,
// modify your Counter class so that this test works.
/*
@Test
public void testDecrease() {
Counter cnt = new Counter(11);
cnt.decrease();
assertEquals(\"decreased value of (11) failed\", -11, cnt.getCount());
cnt.decrease();
assertEquals(\"twice decreased value of (11) failed\", -22,
cnt.getCount());
}
*/
// FOURTH
// Uncomment the following method,
// modify your Counter class so that this test works.
/*
@Test
public void testNegative() {
Counter cnt = new Counter(-1);
cnt.decrease();
assertEquals(\"decreased value of (-1) failed\", 1, cnt.getCount());
cnt.increase();
assertEquals(\"decreased/increased value of (-1) failed\", 0,
cnt.getCount());
cnt.increase();
assertEquals(\"decreased/increased/increased value of (-1) failed\", -1,
cnt.getCount());
}
*/
}
Date.java:
/**
*
* This class represents a date given the month and the day of the month. For the
* given date, this class provides a method for determining the season in the northern hemisphere
for the date.
*
* For example, the given code fragment the output to the console should be WINTER.
*
* Date jan1 = new Date(1, 1);
* String season = jan1.getSeason();
* System.out.println(season);
*
* @author parks
*
*/
public class Date
{
private static final Month year[] = {
new Month(\"January\", 1, 31),
new Month(\"February\", 2, 29),
new Month(\"March\", 3, 31),
new Month(\"April\", 4, 30),
new Month(\"May\", 5, 31),
new Month(\"June\", 6, 30),
new Month(\"July\", 7, 31),
new Month(\"August\", 8, 31),
new Month(\"September\", 9, 30),
new Month(\"October\", 10, 31),
new Month(\"November\", 11, 30),
new Month(\"December\", 12, 31),
};
private int month;
private int day;
/**
* Constructs a new Date object. The month should be a value
* from 1 -12 and the day from 1 - 31.
*
* @param theMonth the month
* @param theDay the day
*/
public Date(int theMonth, int theDay)
{
month = theMonth;
day = theDay;
}
/**
* This method returns the String representation for the month. For example:
* \"January\" for month == 1, \"February\" for month == 2, etc
*
* @return string representation of th.
Hi,I have implemented increment() method. Please find the below up.pdf
Hi,
I have implemented increment() method. Please find the below updated code.
DaysBetween Class:
import java.text.DateFormatSymbols;
import java.util.Calendar;
import java.util.Scanner;
class DateClass {
protected int year;
protected int month;
protected int day;
public static final int MINYEAR = 1583;
// Constructor
public DateClass(int newMonth, int newDay, int newYear)
{
month = newMonth;
day = newDay;
year = newYear;
}
// Observers
public int getYear()
{
return year;
}
public int getMonth()
{
return month;
}
public int getDay()
{
return day;
}
public int lilian()
{
// Returns the Lilian Day Number of this date.
// Precondition: This Date is a valid date after 10/14/1582.
//
// Computes the number of days between 1/1/0 and this date as if no calendar
// reforms took place, then subtracts 578,100 so that October 15, 1582 is day 1.
final int subDays = 578100; // number of calculated days from 1/1/0 to 10/14/1582 November
17, 1858
int numDays;
// Add days in years.
numDays = year * 365;
// Add days in the months.
if (month <= 2)
numDays = numDays + (month - 1) * 31;
else
numDays = numDays + ((month - 1) * 31) - ((4 * (month-1) + 27) / 10);
// Add days in the days.
numDays = numDays + day;
// Take care of leap years.
numDays = numDays + (year / 4) - (year / 100) + (year / 400);
// Handle special case of leap year but not yet leap day.
if (month < 3)
{
if ((year % 4) == 0) numDays = numDays - 1;
if ((year % 100) == 0) numDays = numDays + 1;
if ((year % 400) == 0) numDays = numDays - 1;
}
// Subtract extra days up to 10/14/1582.
numDays = numDays - subDays;
return numDays;
}
Override
public String toString()
// Returns this date as a String.
{
String monthString = new DateFormatSymbols().getMonths()[month-1];
return(monthString + \"/\" + day + \"/\" + year);
}
public class mjd
{
public int mjd()
{
final int subDays = 678941;
int numDays;
numDays = year * 365;
if (month <= 2)
numDays = numDays + (month - 1) * 31;
else
numDays = numDays + ((month -1) * 31) - ((4 * (month-1) + 27)/10);
numDays = numDays + day;
numDays = numDays + (year / 4) - (year / 100) + (year / 400);
if (month < 3)
{
if ((year % 4) == 0) numDays = numDays -1;
if ((year % 100) == 0) numDays = numDays + 1;
if ((year % 400) == 0) numDays -= numDays -1;
}
// Days subtracted up to 10/14/1582
numDays = numDays - subDays;
return numDays;
}
}
public class djd
{
public int djd()
{
final int subDays = 693961; // number of calculated days from 1/1/0 to January 1,1900
int numDays;
// Add days in years.
numDays = year * 365;
// Add days in the months.
if (month <= 2)
numDays = numDays + (month - 1) * 31;
else
numDays = numDays + ((month - 1) * 31) - ((4 * (month-1) + 27) / 10);
// Add days in the days.
numDays = numDays + day;
// Take care of leap years.
numDays = numDays + (year / 4) - (year / 100) + (year / 400);
// Handle special case of leap year but not yet leap day.
if (month < 3)
{
if ((year % 4) == 0) numDays = numDays - 1;
if ((year % 100) == 0) numDays = numDays + 1;
if ((year % .
Modify the Date class that was covered in the lecture which overload.pdf
Modify the Date class that was covered in the lecture which overloaded the increment and stream
insertion operators.
The new version of the class should have the following overloaded operators:
(=) subtraction assignment operator :: subtracts the right operand from the left operand and
assigns the result to the left operand.
This operator should cause appropriate number of decrements to the object's 'day' member. It
also checks appropriate decrements to the 'month' and 'year' data members, if necessary.
(>>) cin 's stream extraction operator :: This operator should prompt the user for a date to be
stored in a Date object (you can specify your own format and prompt the user for the same).
Write a driver program and test the operators with the Date objects.
The program should have the following additional requirements:
The operator overloaded functions should be non-member functions.
Input validation (day, month) for the Date object.
The subtraction assignment operator (=) should work with end of month, end of year, and leap
year conditions as shown in the example output below.
Minimum three files (main.cpp, Date.h, Date.cpp). Submit your code as:
LastName_FirstName_Q2.zip (containing all files).
Example output:
Enter a date in format mm-dd-yyyy
01-01-2023
Date d1 is: January 1, 2023 // Check a valid date
Date (d1 -= 3) is: December 29, 2022 // Decrement by 3 days
Date d2 is: March 2, 2008
Date (d2-=2) is: February 29, 2008 // leap year has 29 days
Date.h
// Fig. 10.6: Date.h
// Date class definition with overloaded increment operators.
#ifndef DATE_H
#define DATE_H
#include
#include
class Date {
friend std::ostream& operator<<(std::ostream&, const Date&);
public:
Date(int m = 1, int d = 1, int y = 1900); // default constructor
void setDate(int, int, int); // set month, day, year
Date& operator++(); // prefix increment operator
Date operator++(int); // postfix increment operator
Date& operator+=(unsigned int); // add days, modify object
static bool leapYear(int); // is year a leap year?
bool endOfMonth(int) const; // is day at the end of month?
private:
unsigned int month;
unsigned int day;
unsigned int year;
static const std::array days; // days per month
void helpIncrement(); // utility function for incrementing date
};
#endif
Date.cpp
#include
#include
#include "Date.h"
using namespace std;
// initialize static member; one classwide copy
const array Date::days{
0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
// Date constructor
Date::Date(int month, int day, int year) {
setDate(month, day, year);
}
// set month, day and year
void Date::setDate(int mm, int dd, int yy) {
if (mm >= 1 && mm <= 12) {
month = mm;
}
else {
throw invalid_argument{"Month must be 1-12"};
}
if (yy >= 1900 && yy <= 2100) {
year = yy;
}
else {
throw invalid_argument{"Year must be >= 1900 and <= 2100"};
}
// test for a leap year
if ((month == 2 && leapYear(year) && dd >= 1 && dd <= 29) ||
(dd >= 1 && dd <= days[month])) {
day = dd;
}
else {
throw invalid_argument{
"Da.
Date class that represents a date consisting of a year, month, and a.pdf
Date class that represents a date consisting of a year, month, and a day
import java.util.TimeZone;
public class Date {
// constants
private static final int JANUARY = 1;
private static final int FEBRUARY = 2;
private static final int DECEMBER = 12;
private static final int DAYS_PER_WEEK = 7;
private static final int DAYS_PER_YEAR = 365;
private static final int DAYS_PER_LEAP_YEAR = 366;
private static final String[] DAY_NAMES = {\"SUNDAY\", \"MONDAY\", \"TUESDAY\",
\"WEDNESDAY\", \"THURSDAY\", \"FRIDAY\", \"SATRDAY\"};
private static final int[] DAYS_PER_MONTH = { -1,
// 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12
31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31,
};
// fields
private int year;
private int month;
private int day;
/**
Constructs a new object representing the given year, month, and day.
@throws IllegalArgumentException if month is not between 1 and 12, or if day is not between 1
and the number of days in that month.
*/
public Date(int year, int month, int day) {
this.year = year;
this.month = month;
this.day = day;
if (month < 1 || month > 12 || day < 1 || day > getDaysInMonth()) {
throw new IllegalArgumentException(\"Invalid day or month: \" + toString());
}
}
/** Constructs a new object representing today\'s date. */
public Date() {
this(2006, july
, 22);
int daysSinceEpoch = (int) ((System.currentTimeMillis() +
TimeZone.getDefault().getRawOffset()) / 1000 / 60 / 60 / 24);
for (int i = 0; i < daysSinceEpoch; i++) {
nextDay();
}
}
/**
Adds the given number of days to this Date.
@throws IllegalArgumentException if days < 0.
*/
public void addDays(int days) {
while (days > getDaysInYear()) {
days -= getDaysInYear();
year++;
}
for (int i = 1; i <= days; i++) {
nextDay();
}
}
/**
Returns whether o refers to a Date object representing the same
year, month, and day as this one.
*/
public boolean equals(Object o) {
Date other = (Date) o;
return day == other.day &&
month == other.month &&
year == other.year;
}
/**
Returns the day represented by this Date object.
@return the day, between 1 and the number of days in this Date\'s month
(which varies from 28 to 31)
*/
public int getDay() {
return day;
}
/** Returns the number of days in the month represented by the current Date. */
public int getDaysInMonth() {
int result = DAYS_PER_MONTH[month];
if (month == FEBRUARY && isLeapYear()) {
result++;
}
return result;
}
public int getDaysInYear() {
if (isLeapYear()) {
return 366;
} else {
return 365;
}
}
/**
Returns a string such as \"Monday\" representing what day of the week
this Date fell on.
@return the day as either \"Sunday\", \"Monday\", \"Tuesday, \"Wednesday\", \"Thursday\",
\"Friday\", or \"Satrday\"
*/
public String getDayOfWeek() {
int index = 1;
Date temp = new Date(1753, JANUARY, 1);
while (temp != this) {
temp.nextDay();
index = (index + 1) % DAYS_PER_WEEK;
}
return DAY_NAMES[index];
}
/**
Returns the month represented by this Date object.
@return the month, between 1 and 12
*/
public int getMonth() {
return month;
}
/**
Returns the yea.
Excel DATEDIFF Function
This presentation is developed to impart understanding of Microsoft Excel's one of the most unique and powerful function DATEDIF. Which helps in calculating difference between dates.
Answer main.cpp.pdf
Answer
****************************************main.cpp******************************
****************************
#include
#include
#include
#include \"Month.h\"
int main()
{
//testing the constructors here
Month myMonth1;
Month myMonth12(12);
Month myMonth3(\"March\");
string newMonth = \"May\";
int newNum = 4;
//testing the get and set functions
cout << \"\ ---------------------------------------------\" << endl;
cout << \"Testing Constructors, Set() and Get()\" << endl;
cout << \"---------------------------------------------\" << endl;
cout << \"myMonth1.getMonthName() gives: \" << myMonth1.getMonthName() << endl;
cout << \"myMonth1.getMonthNumber() gives: \" << myMonth1.getMonthNumber() << endl;
cout << \"myMonth1.getMonthName() gives: \" << myMonth12.getMonthName() << endl;
cout << \"myMonth12.getMonthNumber() gives: \" << myMonth12.getMonthNumber() <<
endl;
cout << \"myMonth3.getMonthName() gives: \" << myMonth3.getMonthName() << endl;
cout << \"myMonth3.getMonthNumber() gives: \" << myMonth3.getMonthNumber() << endl;
cout << endl << \"Setting number to \" << newNum << endl;
myMonth1.setMonthNumber(newNum);
cout << \"Now myMonth1.getMonthName() gives: \" << myMonth1.getMonthName() << endl;
cout << \"And myMonth1.getMonthNumber() gives: \" << myMonth1.getMonthNumber() <<
endl;
cout << endl << \"Setting number to 13 (invalid) \" << endl;
myMonth1.setMonthNumber(13);
cout << \"Now myMonth1.getMonthName() gives: \" << myMonth1.getMonthName() << endl;
cout << \"And myMonth1.getMonthNumber() gives: \" << myMonth1.getMonthNumber() <<
endl;
cout << endl << \"Setting name to \" << newMonth << endl;
myMonth1.setMonthName(newMonth);
cout << \"Now myMonth1.getMonthName() gives: \" << myMonth1.getMonthName() << endl;
cout << \"And myMonth1.getMonthNumber() gives: \" << myMonth1.getMonthNumber() <<
endl;
cout << endl << \"Setting name to XYZ (invalid) \"<< endl;
myMonth1.setMonthName(\"XYZ\");
cout << \"Now myMonth1.getMonthName() gives: \" << myMonth1.getMonthName() << endl;
cout << \"And myMonth1.getMonthNumber() gives: \" << myMonth1.getMonthNumber() <<
endl;
//testing inputs, outputs, increments and decrements
cout << \"\ ---------------------------------------------\" << endl;
cout << \"Testing cin, cout, increments and decrements\" << endl;
cout << \"---------------------------------------------\" << endl;
cout << \"myMonth12 is \" << myMonth12<< endl;
cout << \"myMonth12++ gives \"<< myMonth12++ << endl;
cout << \"But myMonth12 is now \" << myMonth12 << endl;
cout << \"++myMonth12 gives \"<< ++myMonth12 << endl;
cout << \"myMonth12-- gives \"<< myMonth12-- << endl;
cout << \"But myMonth2 is now \" << myMonth12 << endl;
cout << \"--myMonth12 gives \"<< --myMonth12 << endl;
Month myMonth5;
cin >> myMonth5;
cout << \"myMonth5 is set to \"<< myMonth5 << endl;
cout << \"Press Enter to quit.\" << endl;
cin.ignore();
cin.get();
return 0;
}
***********************************************month.h************************
****************************
#if.
Recommended
CounterTest.javaimport static org.junit.Assert.;import org.jun.pdf
CounterTest.java:
import static org.junit.Assert.*;
import org.junit.Test;
public class CounterTest {
// FIRST
// Write a class Counter such that the following test works
@Test
public void testZero() {
Counter cnt = new Counter(0);
assertEquals(\"initial value of (0) failed\", 0, cnt.getCount());
cnt.increase();
assertEquals(\"increased value of (0) failed\", 0, cnt.getCount());
cnt.decrease();
assertEquals(\"decreased value of (0) failed\", 0, cnt.getCount());
}
// SECOND
// Uncomment the following method,
// modify your Counter class so that this test works.
/*
@Test
public void testIncrease() {
Counter cnt = new Counter(7);
assertEquals(\"initial value of (7) failed\", 0, cnt.getCount());
cnt.increase();
assertEquals(\"increased once value of (7) failed\", 7, cnt.getCount());
cnt.increase();
assertEquals(\"twice increased once value of (7) failed\", 14,
cnt.getCount());
}
*/
// THIRD
// Uncomment the following method,
// modify your Counter class so that this test works.
/*
@Test
public void testDecrease() {
Counter cnt = new Counter(11);
cnt.decrease();
assertEquals(\"decreased value of (11) failed\", -11, cnt.getCount());
cnt.decrease();
assertEquals(\"twice decreased value of (11) failed\", -22,
cnt.getCount());
}
*/
// FOURTH
// Uncomment the following method,
// modify your Counter class so that this test works.
/*
@Test
public void testNegative() {
Counter cnt = new Counter(-1);
cnt.decrease();
assertEquals(\"decreased value of (-1) failed\", 1, cnt.getCount());
cnt.increase();
assertEquals(\"decreased/increased value of (-1) failed\", 0,
cnt.getCount());
cnt.increase();
assertEquals(\"decreased/increased/increased value of (-1) failed\", -1,
cnt.getCount());
}
*/
}
Date.java:
/**
*
* This class represents a date given the month and the day of the month. For the
* given date, this class provides a method for determining the season in the northern hemisphere
for the date.
*
* For example, the given code fragment the output to the console should be WINTER.
*
* Date jan1 = new Date(1, 1);
* String season = jan1.getSeason();
* System.out.println(season);
*
* @author parks
*
*/
public class Date
{
private static final Month year[] = {
new Month(\"January\", 1, 31),
new Month(\"February\", 2, 29),
new Month(\"March\", 3, 31),
new Month(\"April\", 4, 30),
new Month(\"May\", 5, 31),
new Month(\"June\", 6, 30),
new Month(\"July\", 7, 31),
new Month(\"August\", 8, 31),
new Month(\"September\", 9, 30),
new Month(\"October\", 10, 31),
new Month(\"November\", 11, 30),
new Month(\"December\", 12, 31),
};
private int month;
private int day;
/**
* Constructs a new Date object. The month should be a value
* from 1 -12 and the day from 1 - 31.
*
* @param theMonth the month
* @param theDay the day
*/
public Date(int theMonth, int theDay)
{
month = theMonth;
day = theDay;
}
/**
* This method returns the String representation for the month. For example:
* \"January\" for month == 1, \"February\" for month == 2, etc
*
* @return string representation of th.
Hi,I have implemented increment() method. Please find the below up.pdf
Hi,
I have implemented increment() method. Please find the below updated code.
DaysBetween Class:
import java.text.DateFormatSymbols;
import java.util.Calendar;
import java.util.Scanner;
class DateClass {
protected int year;
protected int month;
protected int day;
public static final int MINYEAR = 1583;
// Constructor
public DateClass(int newMonth, int newDay, int newYear)
{
month = newMonth;
day = newDay;
year = newYear;
}
// Observers
public int getYear()
{
return year;
}
public int getMonth()
{
return month;
}
public int getDay()
{
return day;
}
public int lilian()
{
// Returns the Lilian Day Number of this date.
// Precondition: This Date is a valid date after 10/14/1582.
//
// Computes the number of days between 1/1/0 and this date as if no calendar
// reforms took place, then subtracts 578,100 so that October 15, 1582 is day 1.
final int subDays = 578100; // number of calculated days from 1/1/0 to 10/14/1582 November
17, 1858
int numDays;
// Add days in years.
numDays = year * 365;
// Add days in the months.
if (month <= 2)
numDays = numDays + (month - 1) * 31;
else
numDays = numDays + ((month - 1) * 31) - ((4 * (month-1) + 27) / 10);
// Add days in the days.
numDays = numDays + day;
// Take care of leap years.
numDays = numDays + (year / 4) - (year / 100) + (year / 400);
// Handle special case of leap year but not yet leap day.
if (month < 3)
{
if ((year % 4) == 0) numDays = numDays - 1;
if ((year % 100) == 0) numDays = numDays + 1;
if ((year % 400) == 0) numDays = numDays - 1;
}
// Subtract extra days up to 10/14/1582.
numDays = numDays - subDays;
return numDays;
}
Override
public String toString()
// Returns this date as a String.
{
String monthString = new DateFormatSymbols().getMonths()[month-1];
return(monthString + \"/\" + day + \"/\" + year);
}
public class mjd
{
public int mjd()
{
final int subDays = 678941;
int numDays;
numDays = year * 365;
if (month <= 2)
numDays = numDays + (month - 1) * 31;
else
numDays = numDays + ((month -1) * 31) - ((4 * (month-1) + 27)/10);
numDays = numDays + day;
numDays = numDays + (year / 4) - (year / 100) + (year / 400);
if (month < 3)
{
if ((year % 4) == 0) numDays = numDays -1;
if ((year % 100) == 0) numDays = numDays + 1;
if ((year % 400) == 0) numDays -= numDays -1;
}
// Days subtracted up to 10/14/1582
numDays = numDays - subDays;
return numDays;
}
}
public class djd
{
public int djd()
{
final int subDays = 693961; // number of calculated days from 1/1/0 to January 1,1900
int numDays;
// Add days in years.
numDays = year * 365;
// Add days in the months.
if (month <= 2)
numDays = numDays + (month - 1) * 31;
else
numDays = numDays + ((month - 1) * 31) - ((4 * (month-1) + 27) / 10);
// Add days in the days.
numDays = numDays + day;
// Take care of leap years.
numDays = numDays + (year / 4) - (year / 100) + (year / 400);
// Handle special case of leap year but not yet leap day.
if (month < 3)
{
if ((year % 4) == 0) numDays = numDays - 1;
if ((year % 100) == 0) numDays = numDays + 1;
if ((year % .
Modify the Date class that was covered in the lecture which overload.pdf
Modify the Date class that was covered in the lecture which overloaded the increment and stream
insertion operators.
The new version of the class should have the following overloaded operators:
(=) subtraction assignment operator :: subtracts the right operand from the left operand and
assigns the result to the left operand.
This operator should cause appropriate number of decrements to the object's 'day' member. It
also checks appropriate decrements to the 'month' and 'year' data members, if necessary.
(>>) cin 's stream extraction operator :: This operator should prompt the user for a date to be
stored in a Date object (you can specify your own format and prompt the user for the same).
Write a driver program and test the operators with the Date objects.
The program should have the following additional requirements:
The operator overloaded functions should be non-member functions.
Input validation (day, month) for the Date object.
The subtraction assignment operator (=) should work with end of month, end of year, and leap
year conditions as shown in the example output below.
Minimum three files (main.cpp, Date.h, Date.cpp). Submit your code as:
LastName_FirstName_Q2.zip (containing all files).
Example output:
Enter a date in format mm-dd-yyyy
01-01-2023
Date d1 is: January 1, 2023 // Check a valid date
Date (d1 -= 3) is: December 29, 2022 // Decrement by 3 days
Date d2 is: March 2, 2008
Date (d2-=2) is: February 29, 2008 // leap year has 29 days
Date.h
// Fig. 10.6: Date.h
// Date class definition with overloaded increment operators.
#ifndef DATE_H
#define DATE_H
#include
#include
class Date {
friend std::ostream& operator<<(std::ostream&, const Date&);
public:
Date(int m = 1, int d = 1, int y = 1900); // default constructor
void setDate(int, int, int); // set month, day, year
Date& operator++(); // prefix increment operator
Date operator++(int); // postfix increment operator
Date& operator+=(unsigned int); // add days, modify object
static bool leapYear(int); // is year a leap year?
bool endOfMonth(int) const; // is day at the end of month?
private:
unsigned int month;
unsigned int day;
unsigned int year;
static const std::array days; // days per month
void helpIncrement(); // utility function for incrementing date
};
#endif
Date.cpp
#include
#include
#include "Date.h"
using namespace std;
// initialize static member; one classwide copy
const array Date::days{
0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
// Date constructor
Date::Date(int month, int day, int year) {
setDate(month, day, year);
}
// set month, day and year
void Date::setDate(int mm, int dd, int yy) {
if (mm >= 1 && mm <= 12) {
month = mm;
}
else {
throw invalid_argument{"Month must be 1-12"};
}
if (yy >= 1900 && yy <= 2100) {
year = yy;
}
else {
throw invalid_argument{"Year must be >= 1900 and <= 2100"};
}
// test for a leap year
if ((month == 2 && leapYear(year) && dd >= 1 && dd <= 29) ||
(dd >= 1 && dd <= days[month])) {
day = dd;
}
else {
throw invalid_argument{
"Da.
Date class that represents a date consisting of a year, month, and a.pdf
Date class that represents a date consisting of a year, month, and a day
import java.util.TimeZone;
public class Date {
// constants
private static final int JANUARY = 1;
private static final int FEBRUARY = 2;
private static final int DECEMBER = 12;
private static final int DAYS_PER_WEEK = 7;
private static final int DAYS_PER_YEAR = 365;
private static final int DAYS_PER_LEAP_YEAR = 366;
private static final String[] DAY_NAMES = {\"SUNDAY\", \"MONDAY\", \"TUESDAY\",
\"WEDNESDAY\", \"THURSDAY\", \"FRIDAY\", \"SATRDAY\"};
private static final int[] DAYS_PER_MONTH = { -1,
// 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12
31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31,
};
// fields
private int year;
private int month;
private int day;
/**
Constructs a new object representing the given year, month, and day.
@throws IllegalArgumentException if month is not between 1 and 12, or if day is not between 1
and the number of days in that month.
*/
public Date(int year, int month, int day) {
this.year = year;
this.month = month;
this.day = day;
if (month < 1 || month > 12 || day < 1 || day > getDaysInMonth()) {
throw new IllegalArgumentException(\"Invalid day or month: \" + toString());
}
}
/** Constructs a new object representing today\'s date. */
public Date() {
this(2006, july
, 22);
int daysSinceEpoch = (int) ((System.currentTimeMillis() +
TimeZone.getDefault().getRawOffset()) / 1000 / 60 / 60 / 24);
for (int i = 0; i < daysSinceEpoch; i++) {
nextDay();
}
}
/**
Adds the given number of days to this Date.
@throws IllegalArgumentException if days < 0.
*/
public void addDays(int days) {
while (days > getDaysInYear()) {
days -= getDaysInYear();
year++;
}
for (int i = 1; i <= days; i++) {
nextDay();
}
}
/**
Returns whether o refers to a Date object representing the same
year, month, and day as this one.
*/
public boolean equals(Object o) {
Date other = (Date) o;
return day == other.day &&
month == other.month &&
year == other.year;
}
/**
Returns the day represented by this Date object.
@return the day, between 1 and the number of days in this Date\'s month
(which varies from 28 to 31)
*/
public int getDay() {
return day;
}
/** Returns the number of days in the month represented by the current Date. */
public int getDaysInMonth() {
int result = DAYS_PER_MONTH[month];
if (month == FEBRUARY && isLeapYear()) {
result++;
}
return result;
}
public int getDaysInYear() {
if (isLeapYear()) {
return 366;
} else {
return 365;
}
}
/**
Returns a string such as \"Monday\" representing what day of the week
this Date fell on.
@return the day as either \"Sunday\", \"Monday\", \"Tuesday, \"Wednesday\", \"Thursday\",
\"Friday\", or \"Satrday\"
*/
public String getDayOfWeek() {
int index = 1;
Date temp = new Date(1753, JANUARY, 1);
while (temp != this) {
temp.nextDay();
index = (index + 1) % DAYS_PER_WEEK;
}
return DAY_NAMES[index];
}
/**
Returns the month represented by this Date object.
@return the month, between 1 and 12
*/
public int getMonth() {
return month;
}
/**
Returns the yea.
Excel DATEDIFF Function
This presentation is developed to impart understanding of Microsoft Excel's one of the most unique and powerful function DATEDIF. Which helps in calculating difference between dates.
Answer main.cpp.pdf
Answer
****************************************main.cpp******************************
****************************
#include
#include
#include
#include \"Month.h\"
int main()
{
//testing the constructors here
Month myMonth1;
Month myMonth12(12);
Month myMonth3(\"March\");
string newMonth = \"May\";
int newNum = 4;
//testing the get and set functions
cout << \"\ ---------------------------------------------\" << endl;
cout << \"Testing Constructors, Set() and Get()\" << endl;
cout << \"---------------------------------------------\" << endl;
cout << \"myMonth1.getMonthName() gives: \" << myMonth1.getMonthName() << endl;
cout << \"myMonth1.getMonthNumber() gives: \" << myMonth1.getMonthNumber() << endl;
cout << \"myMonth1.getMonthName() gives: \" << myMonth12.getMonthName() << endl;
cout << \"myMonth12.getMonthNumber() gives: \" << myMonth12.getMonthNumber() <<
endl;
cout << \"myMonth3.getMonthName() gives: \" << myMonth3.getMonthName() << endl;
cout << \"myMonth3.getMonthNumber() gives: \" << myMonth3.getMonthNumber() << endl;
cout << endl << \"Setting number to \" << newNum << endl;
myMonth1.setMonthNumber(newNum);
cout << \"Now myMonth1.getMonthName() gives: \" << myMonth1.getMonthName() << endl;
cout << \"And myMonth1.getMonthNumber() gives: \" << myMonth1.getMonthNumber() <<
endl;
cout << endl << \"Setting number to 13 (invalid) \" << endl;
myMonth1.setMonthNumber(13);
cout << \"Now myMonth1.getMonthName() gives: \" << myMonth1.getMonthName() << endl;
cout << \"And myMonth1.getMonthNumber() gives: \" << myMonth1.getMonthNumber() <<
endl;
cout << endl << \"Setting name to \" << newMonth << endl;
myMonth1.setMonthName(newMonth);
cout << \"Now myMonth1.getMonthName() gives: \" << myMonth1.getMonthName() << endl;
cout << \"And myMonth1.getMonthNumber() gives: \" << myMonth1.getMonthNumber() <<
endl;
cout << endl << \"Setting name to XYZ (invalid) \"<< endl;
myMonth1.setMonthName(\"XYZ\");
cout << \"Now myMonth1.getMonthName() gives: \" << myMonth1.getMonthName() << endl;
cout << \"And myMonth1.getMonthNumber() gives: \" << myMonth1.getMonthNumber() <<
endl;
//testing inputs, outputs, increments and decrements
cout << \"\ ---------------------------------------------\" << endl;
cout << \"Testing cin, cout, increments and decrements\" << endl;
cout << \"---------------------------------------------\" << endl;
cout << \"myMonth12 is \" << myMonth12<< endl;
cout << \"myMonth12++ gives \"<< myMonth12++ << endl;
cout << \"But myMonth12 is now \" << myMonth12 << endl;
cout << \"++myMonth12 gives \"<< ++myMonth12 << endl;
cout << \"myMonth12-- gives \"<< myMonth12-- << endl;
cout << \"But myMonth2 is now \" << myMonth12 << endl;
cout << \"--myMonth12 gives \"<< --myMonth12 << endl;
Month myMonth5;
cin >> myMonth5;
cout << \"myMonth5 is set to \"<< myMonth5 << endl;
cout << \"Press Enter to quit.\" << endl;
cin.ignore();
cin.get();
return 0;
}
***********************************************month.h************************
****************************
#if.
Write a program to generate the entire calendar for one year. The pr.pdf
Write a program to generate the entire calendar for one year. The program must get two values
from the user: (1) the year and (2) the day of the week for January 1st of that year. The year,
which should be positive, is needed to check for and handle leap years1. The day of the week for
January 1st is needed so that you know where to start the calendar. The user should enter 0 for
Sunday, 1 for Monday, ... or 6 for Saturday. As always, you need to validate the user\'s input. To
actually print the calendar, you must use a single method that prints out the calendar for one
month and then call this function 12 times from main(), once for each month in the year. To
check for a leap year you will need to write another method that takes the year as a parameter
and returns true if it’s a leap year, or false otherwise. Stubs (i.e. method signatures without any
code) for both of these methods have been provided for you.
The calendar should be printed in the form below. In this example, January starts on a Saturday
(day 6). Note that February starts on a Tuesday in this example because January ended on a
Monday.
January
1
2 3 4 5 6 7 8
9 10 11 12 13 14 15
16 17 18 19 20 21 22
23 24 25 26 27 2829
30 31
February
1 2 3 4 5
6 7 8 9 10 11 12 ...
The format he gave us is as follows:
public class PA6a {
/**
* Error to output if year is not positive
*/
static final String E_YEAR = \"The year must be positive!\";
/**
* Error to output if the day is not between 0 and 6
*/
static final String E_DAY = \"The day of January 1st must be between 0 and 6!\";
/**
* Determines if an input is a leap year
*
* @param year year in question
* @ereturn true if a leap year
*/
public static boolean isLeapYear(int year) {
return false; // TODO: replace with your code
}
/**
* Outputs a month to the console
*
* @param month title
* @param startDay 0=Sunday ... 6=Saturday
* @param numDays number of days in the month
* @return day of the week of the last day of the month
*/
public static int printMonth(String month, int startDay, int numDays) {
return 0; // TODO: replace with your code
}
/**
* Program execution point:
* input year, day of the week (0-6) of january 1
* output calendar for that year
*
* @param args command-line arguments (ignored)
*/
public static void main(String[] args) {
// TODO: write your code here
}
I am confused especially with the print month method, help is greatly appreciated.
Solution
package org.students;
import java.util.Scanner;
import com.sun.org.apache.bcel.internal.generic.IUSHR;
public class Calander {
//declaring static variables
static final String E_YEAR = \"The year must be positive!\";
static final String E_DAY = \"The day of January 1st must be between 0 and 6!\";
public static void main(String[] args) {
//Declaring variables
int getFirstDay;
int year;
Scanner scanner = new Scanner(System.in);
//This while loop continue to execute until user enters valid year
while (true) {
//Getting the year entered by the user
System.out.print(\"Enter the year:.
publicclass Date {privatestatic String DATE_SEPARATOR = ;pr.pdf
publicclass Date {
privatestatic String DATE_SEPARATOR = \"/\";
privatestaticintDAYS_PER_WEEK = 7;
//Attributes
privateint day;
privateint month;
privateint year;
/**
* Default Constructor
* Instantiates an object of type Date to 1/1/2000
*/
public Date() {
this.day = 1;
this.month = 1;
this.year = 2000;
}
/**
* Constructs a new date object to represent the given date.
* @param day
* @param month
* @param year
*/
public Date(int day, int month, int year) {
if(isValid(day, month, year)) {
this.day = day;
this.month = month;
this.year = year;
} else
System.out.println(\"Invalid Date.\");
}
/**
* Returns the day value of this date for example, for the date 2006/07/22, returns 22.
* @return
*/
publicint getDay() {
return day;
}
/**
* Returns the month value of this date ,for example, for the date 2006/07/22, returns 7.
* @return
*/
publicint getMonth() {
return month;
}
/**
* Returns the year value of this date, for example , the date 2006/07/22, returns 2006.
* @return
*/
publicint getYear() {
return year;
}
/**
* @param day the day to set
*/
publicvoid setDay(int day) {
this.day = day;
}
/**
* @param month the month to set
*/
publicvoid setMonth(int month) {
this.month = month;
}
/**
* @param year the year to set
*/
publicvoid setYear(int year) {
this.year = year;
}
/**
* Returns true if the year of this date is a leap year.
* A leap year occurs every 4 years , except for multiples of 100 that are not multiples of 400.
* For example, 1956,1844,1600,and 2000 are leap years, but 1983,2002,1700,and 1900 are not.
* @return
*/
publicboolean isLeapYear(){
if(((this.year % 400) == 0) || (((this.year % 4) == 0) && ((this.year % 100) != 0)))
returntrue;
else
returnfalse;
}
/**
* Checks if the date is valid
* @param day
* @param month
* @param year
* @return
*/
publicboolean isValid(int day, int month, int year) {
if((month < 1) || (12 < month))
returnfalse;
else {
if(((month == 1) || (month == 3) || (month == 5) || (month == 7) ||
(month == 8) || (month == 10) || (month == 12)) && ((day < 1) || (31 < day)))
returnfalse;
elseif(((month == 4) || (month == 6) || (month == 9) || (month == 11)) && ((day < 1) || (30 <
day)))
returnfalse;
elseif(month == 2) {
if(isLeapYear() && ((day < 1) || (29 < day)))
returnfalse;
elseif((day < 1) || (28 < day))
returnfalse;
}
}
returntrue;
}
/**
* Returns the maximum number of days in a month
* @return
*/
publicint maxMonthDays() {
if(this.month == 2) {
if(isLeapYear())
return 29;
else
return 28;
} elseif((this.month == 1) || (this.month == 3) || (this.month == 5) || (this.month == 7) ||
(this.month == 8) || (this.month == 10) || (this.month == 12))
return 31;
else
return 30;
}
/**
* Checks if this dat is same as other date
* @param other
* @return
*/
publicboolean isEqual(Date other) {
if((this.day == other.day) && (this.month == other.month) && (this.year == other.year))
returntrue;
else
returnfalse;
}
/**
* Moves this Date object forward in time by the given number of days .
* @param days
*/
publicvoid addDays(int days).
Les03 (Using Single Row Functions To Customize Output)
Using Single Row FUnction (uploaded by achmatim.net)
Hello I need some help on this question this is for C++ PEARSON REVE.pdf
Hello I need some help on this question this is for C++ PEARSON REVEL
I have a code that does not seem to be correct and I want to know to the correct way.
There is also the reason for the error and the question itself.
here is the code:
#include
#include
using namespace std;
class Month {
private:
int num;
static string monthName[12];
public:
Month() {
num = 0;
}
Month(int n) {
num = n;
}
void setNum(int n) {
num = n;
}
void setName(string name) {
for (int i = 0; i < 12; i++) {
if (name == monthName[i]) {
num = i + 1;
return;
}
}
num = 0;
}
int getNum() const {
return num;
}
string getName() const {
if (num < 1 || num > 12) {
return "Unknown";
}
return monthName[num - 1];
}
Month& operator++() {
num++;
if (num > 12) {
num = 1;
}
return *this;
}
Month operator++(int) {
Month temp = *this;
++(*this);
return temp;
}
Month& operator--() {
num--;
if (num < 1) {
num = 12;
}
return *this;
}
Month operator--(int) {
Month temp = *this;
--(*this);
return temp;
}
friend ostream& operator<<(ostream& os, const Month& month) {
os << month.getName();
return os;
}
friend istream& operator>>(istream& is, Month& month) {
int n;
is >> n;
month.setNum(n);
return is;
}
};
string Month::monthName[12] = {"January,", "February,", "March,", "April,", "May,", "June,",
"July,",
"August,", "September,", "October,", "November,", "December,"};
int main() {
int n;
cout << "Enter: month number (1-12):";
cin >> n;
Month month(n);
cout << month << endl;
for (int i = 0; i < 7; i++) {
++month;
cout << month << endl;
}
cout << "Enter: another month number (1 - 12): ";
cin >> n;
month.setNum(n);
cout << month << endl;
for (int i = 0; i < 7; i++) {
--month;
cout << month << endl;
}
return 0;
}
and here is the errors that it is giving me:
Logic errors.
And here is the question for the answer that i am trying to do:
Design a class named Month. The class should have the following members:
Data Members:
An int variable named num to hold the month number, which is an integer in the range of 1
through 12.
A static array of twelve strings named monthName, initialized with the names of the months.
For example, monthName[0] should be set to "January", monthName[1] should be set to
"February", etc. The static array will be used as a lookup table to look up the name of a month.
Constructors:
The class should have a default constructor that sets the num member variable to 0 as well as a
constructor that accepts an int argument for the num member variable.
Mutators:
A setNum function that accepts an int argument for the month number. The setNum function
should assign this value to the num member variable.
A setName function that accepts a string argument for the name of the month. The setName
function should find the name in the monthName array, and then update the num member
variable with the correct number. For example, if the argument is "January", the num variable
will be set to 1. If the argument is "April", the num variable will be set to 4. And so on. If the
argument is not found in th.
I am having some trouble getting this to compile, I have made seve.docx
I am having some trouble getting this to compile, I have made several changes and still not having any luck. Can someone please give me some guidance on how to correct this. thank you in advance.
Chad
#include <stdio.h>
#include <stdlib.h>
int main()
{//Created by Chad Hartley
//October 30, 2015
char Month;//Variable for the character the user will input.
//Ask which month needs to be converted.
//I used the third character in the months that had the same first letter, except for June I had to use the second character.
//I included a Key for the user to reference to know which character is expected.
printf("Please use the following letter to represent the month you want converted: \n");
printf("J=January,F=February,M=March,A=April,Y=May,U=June,L=July,G=August,S=September,O=October,N=November,D=December \n");
scanf("%c",&Month);
switch(Month)//switching statement used to look for the character inputed
{
case J:
printf("01");
printf("\n");//used to print a new line
break;//I used break to not have a continuous output
case F:
printf("02");
printf("\n");
break;
case M:
printf("03");
printf("\n");
break;
case A:
printf("04");
printf("\n");
break;
case Y:
printf(“05");
printf("\n");
break;
case U:
printf("06");
printf("\n");
break;
case L:
printf("07");
printf("\n");
break;
case G:
printf("08");
printf("\n");
break;
case S:
printf("09");
printf("\n");
break;
case O:
printf("10");
printf("\n");
break;
case N:
printf("11");
printf("\n");
break;
case D:
printf("12");
printf("\n");
break;
default:
printf("Please enter a valid character.");//This is a default statement to cover any other character the user may input.
printf("\n");
}
return 0;
}
PROVIDE COMMENT HERE FOR ABOVE ANSWER
PROVIDE COMMENT FOR THIS TOO
This program will allow a character value for a month to be entered and an output a number corresponding to the month. A switch statement will be used to convert from a character to a number the print out the result.
#include <stdio.h>
int main(void) {
/* variable definition */
int J = 1; 6; 7;
int F = 2;
int M = 3; 5;
int A = 4; 8;
int S = 9;
int O = 10;
int N = 11;
int D = 12;
/* the switch statement */
switch (1) <-------- The switch statement is where the number of the character will be placed to convert.
{
case 1: <-----The case scenario execute the conversion
case 6:
case 7:
printf ("J is equal to 1, 6 or 7\n"); <--------The switch is then printed out.
break; <------ This break ends case 1 etc.
case 2:
printf ("F is equal 2\n");
break;
case 3:
.
I got the codes written down below- Basically- I am trying to implemen.pdf
I got the codes written down below. Basically, I am trying to implement OOP of employees and
payments with LinkedList, but when I test it, I keep getting errors like "uncleared Identifier."
Please help me with my programs.(Language: C)
The above images are the expected output and graphic illustrations of the linked lists.
Please help me debug the program and provide the correct version of the codes. DO NOT COPY
others' answers or chatGPT, and then, I'll give you an upvote.
Main:
///////////////////////////////////////////////////////////////////////////////////////////////////
#include "employee.h"
int main(){
//-----------------------------------------------------
//testCase -1
/*
EMPLOYEE e;
e.name="james";
e.id=123456;
e.next=NULL;
e.payments=NULL;
printEmployee(e);
*/
//-----------------------------------------------------
//testCase -2
/*
PAYMENT p1;
p1.date="01/01/2010";
p1.amount=150;
p1.next=NULL;
PAYMENT p2;
p2.date="01/05/2010";
p2.amount=200;
p2.next=&p1;
EMPLOYEE e;
e.name="sarah";
e.id=123456;
e.payments=&p2;
e.next=NULL;
printEmployee(e);
*/
//-----------------------------------------------------
//testCase -3
/*
addEmployee(123,"alma");
addEmployee(456,"mike");
printEmployees();
*/
//-----------------------------------------------------
//testCase -4
/*
addEmployee(123,"alma");
addPayment(123,"01/02/2020",3000);
addPayment(123,"10/07/2021",5000);
printEmployees();
*/
//-----------------------------------------------------
//testCase -5
/*
addEmployee(123,"alma");
addPayment(123,"01/02/2020",3000);
addPayment(123,"10/07/2021",5000);
addEmployee(456,"mike");
addPayment(456,"06/03/2020",700);
addPayment(456,"07/02/2020",200);
addPayment(456,"06/04/2021",600);
printEmployees();
*/
//-----------------------------------------------------
//testCase -6
/*
addEmployee(123,"alma");
addPayment(123,"01/02/2020",3000);
addPayment(123,"10/07/2021",5000);
addEmployee(456,"mike");
addPayment(456,"06/03/2020",700);
addPayment(456,"07/02/2020",200);
addPayment(456,"06/04/2021",600);
deleteEmployee(123,"alma");
printEmployees();
*/
//-----------------------------------------------------
//testCase -7
/*
addEmployee(123,"alma");
addPayment(123,"01/02/2020",3000);
addPayment(123,"10/07/2021",5000);
addEmployee(456,"mike");
addPayment(456,"06/03/2020",700);
addPayment(456,"07/02/2020",200);
addPayment(456,"06/04/2021",600);
deleteEmployee(456,"mike");
deleteEmployee(123,"alma");
printEmployees();
*/
}
///////////////////////////////////////////////////////////////////////////////////////////////////
Constructor:
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
typedef struct payment{
char* date; // MM/DD/YY as text
float amount;
struct payment* next;
}PAYMENT;
typedef struct employee{
int id;
char* name;
PAYMENT* payments;
struct employee* next;
}EMPLOYEE;
void printEmployee(EMPLOYEE e);
void printEmployees();
void addEmployee(int e_id,char* e_name);
int addPayment(int e_id, char* p_date, float p_amount);
int deleteEmployee(int e_id, char.
I Have the following Java program in which converts Date to Words an.pdf
I Have the following Java program in which converts Date to Words and the only thing I have to
do now is add a boolean for a leap year. ( I am using Apache NetBeans) please explain where the
boolean goes.
Like this:
I need the code working with the boolean added, Thank you!!!
package a1_demo;
import java.util.Scanner;
public class MainApp {
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
System.out.print("Welcome to the Date to Words Converter!\n\n");
while ( true ) {
// get user input
String str_MM_DD_YYYY = Validator.getString(
Validator.sc,
"Enter date in MM/DD/YYYY format: "
);
// convert text to ints:
// "11/22/3333" - > int, int int
String[] dateComponents = str_MM_DD_YYYY.split( "/" );
if ( dateComponents.length != 3 ) {
System.out.println("\t\t Bad input, please retry...");
continue;
}
int month = 0;
int day = 0;
int year = 0;
try {
month = Integer.parseInt(dateComponents[ 0 ]);
day = Integer.parseInt(dateComponents[ 1 ]);
year = Integer.parseInt(dateComponents[ 2 ]);
} catch ( java.lang.NumberFormatException ex ) {
System.out.println("\t\t numeric values are expected, please retry...");
continue;
}
Date2TextConverter converter = new Date2TextConverter( month, day, year );
System.out.print( converter.month2text() );
System.out.print( " " );
System.out.print( converter.day2text() );
System.out.print( " " );
System.out.print( converter.year2text() );
System.out.println();
}// endless while loop
}//main
}//class MainApp
package a1_demo;
public class Date2TextConverter {
//---------------------------------------
// data
//---------------------------------------
int month = 0;
int day = 0;
int year = 0;
//---------------------------------------
// constructors
//---------------------------------------
public Date2TextConverter(int month, int day, int year )
{
this.month = month;
this.day = day;
this.year = year;
}//Date2TextConverter
//---------------------------------------
// operations
//---------------------------------------
public String month2text()
{
switch( month )
{
case 1: return "Jan";
case 2: return "Feb";
case 3: return "Mar";
case 4: return"April";
case 5: return"May";
case 6: return"June";
case 7: return"July";
case 8: return"Aug";
case 9: return"Sep";
case 10: return"Oct";
case 11: return"Nov";
case 12: return"Dec";
//...
default:
return "Invalid Month";
}
}//month2text
public String day2text()
{
//return Integer.toString( day );
return number2words (day);
}//day2text
public String year2text()
{
int century = year / 100;
int yearInCentury = year % 100;
String centuryText = number2words( century );
String yearText = number2words( yearInCentury );
if ( yearInCentury < 10 ) {
return centuryText + " zero " + yearText;
} else {
return centuryText + " " + yearText;
}
}//year2text
private String number2words( int number ) {
// convert to text a number 0 through 99
if ( number < 20 ) {
// 0, 1, 2, 3, ..., 19
return zero2nineteen[ number ];
} else if ( number % 10 == 0 ) {
// 20, 30, 40, ... 90
return .
Operator Overloading
How to use to operator overloading in C++ to improve readability and re-usability of your source code.
struct procedure { Date dateOfProcedure; int procedureID.pdf
struct procedure
{
Date dateOfProcedure;
int procedureID;
int procedureProviderID;
};
Data.h
#include
#include
class Date
{
friend ostream& operator<<( ostream &, const Date & );
// allows easy output to a ostream
public:
Date( int m = 1, int d = 1, int y = 1900 ); // constructor, note the default values
void setDate( int, int, int ); // set the date const
Date &operator+=( int ); // add days, modify object
bool leapYear( int) const; // is this a leap year?
bool endOfMonth( int ) const; // is this end of month?
int getMonth ( ) const;
int getDay ( ) const;
int getYear ( ) const;
string getMonthString( ) const;
private:
int month;
int day;
int year;
static const int days[]; // array of days per month static const string monthName[]; // array of
month names
void helpIncrement();
// utility function
}; #endif Data.cpp // Member function definitions for Date
class in separate
date.cpp
#include
#include \"date.h\"
#include // Initialize static members at file scope; // one class-wide copy.
const int Date::days[] = { 0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31 }; const string
Date::monthName[] = { \"January\", \"February\", \"March\", \"April\", \"May\", \"June\",
\"July\", \"August\", \"September\", \"October\", \"November\", \"December\" };
// Date constructor
Date::Date(int m, int d, int y)
{
setDate(m, d, y); } // Set the date
void Date::setDate(int mm, int dd, int yy)
{
month = (mm >= 1 && mm <= 12) ? mm : 1; year = (yy >=1900&& yy <= 2100) ? yy
: 1900; // test for a leap year
if (month == 2 && leapYear(year)) day = (dd >= 1 && dd <= 29) ? dd : 1;
else
day =(dd >= 1 && dd <= days[month]) ? dd : 1;
} // Add a specific number of days to a date
const Date &Date::operator+=(int additionalDays)
{
for (int i = 0; i < additionalDays; i++) helpIncrement();
return *this; // enables cascading
} // If the year is leap year, return true; //
otherwise, return false
bool Date::leapYear(int testYear)
const
{
if (testYear % 400 == 0 || (testYear %100 != 0 && testYear % 4 == 0)) return true; // a
leap year
else
return false; // not a leap year
} // Determine if the day is the end of the month
bool Date::endOfMonth(int testDay)
const
{
if (month == 2 && leapYear(year))
return (testDay == 29); // last day of Feb. in leap year
else
return (testDay == days[month]);
} // Function to help increment the date
void Date::helpIncrement()
{
if (!endOfMonth(day))
{
// date is not at the end of the month
day++;
}
else if (month < 12)
{
// date is at the end of the month, but month < 12 day = 1; ++month;
}
else
{
// end of month and year: last day of the year
day = 1;
month = 1;
++year;
} }
// Overloaded output operator
ostream &operator<<(ostream& output, const Date &d)
{
output << d.monthName[d.month] << \' \' << d.day << \", \"<< d.year;
return output; // enables cascading
}
int Date::getMonth() const //
{
Public:
int month[20];
If(month[]!=”/0”)
Cout<<”Enter Month :-<>month;
Elseif (month[]<12)
{
Cout<>month[];
else
return month;
}
int Date::getDay() const //
{
Pu.
(----IN C) Incorrect output on some test cases- I'm using a linked lis.pdf
(****IN C) Incorrect output on some test cases.
I'm using a linked list to take in information about a person and using structs to find the oldest
and youngest person in the list; however, I am having trouble with some test cases. For the most
part, The code seems to work the way I want it to, but certain test cases do have wrong outputs.
For example:
Input File Has Four Entries(2 Youngest are in the Same Month)
Here is the text file:
Smith,Bob,July,31,2002,CS/BS,Junior
Joe,Average,February,3,2002,CS/BS,Freshman
Doe,Jane,July,27,2003,CS/BS,Sophomore
Doe,John,July,12,2003,CS/BS,Sophomore:
Output:
Oldest - John Doe & Youngest Jane Doe.
John is clearly not the oldest in the list so please help, below is my code.
//add birthday struct here
struct birthday
{
int day, year;
char month[50];
}; typedef struct birthday bday;
//add node_t struct here (this struct will represent a node in the linked list)
struct node
{
char firstName[50], lastName[50], major[50], classStanding[50];
bday bd;
struct node* next;
};
typedef struct node node_t;
struct stats
{
// node pointer pointing at the oldest and youngest person in the list
node_t* oldest;
node_t* youngest;
// holding the number of birthdays in each month (0 = Jan. & 11 = Dec.)
int numBD[12];
}; typedef struct stats list_stats_t;
void add(node_t **node, node_t **head)
{
(*node)->next = *head;
*head = *node;
}
node_t* readNodeInfo(FILE* input)
{
node_t *node;
node = (node_t*)malloc(sizeof(node_t)); // allocates memory
char newLine[1]; // reads new line character
// uses scanset to read information between commas
fscanf(input, "%[^,], %[^,], %[^,], %d, %d, %[^,], %[^\n] %[\n]",
node->lastName, node->firstName, node->bd.month,
&node->bd.day, &node->bd.year, node->major,
node->classStanding, newLine);
return node;
}
node_t* createList(FILE* in, node_t** head)
{
node_t *temp;
while(!feof(in)) // loops until end of file in
{
temp = readNodeInfo(in); // calls function and assigns it to temp
add(&temp, head); // adds to linked list
}
return *head;
}
void printList(FILE* out, node_t* head) // prints data
{
// if there is no data in the linked list, print error code to terminal
if (head == NULL)
{
fprintf(stderr, "LIST IS EMPTY\n");
return;
}
node_t *temp = head;
printBorder(out);
fprintf(out, "\nList Info:\n");
while (temp != NULL)
{
fprintf(out, "Name:\t%s %s\n", temp->firstName, temp->lastName);
fprintf(out, "Date of Birth:\t%s %d, %d\n", temp->bd.month,
temp->bd.day,
temp->bd.year);
fprintf(out, "Major:\t%s\n", temp->major);
fprintf(out, "Year:\t%s\n\n", temp->classStanding);
temp = temp->next; // next element
}
printBorder(out);
}
void printBorder(FILE* out)
{
int i = 0;
for(i = 0; i < 80; i++)
fprintf(out,"*");
fprintf(out, "\n");
}
void deleteList(node_t** head) // deletes all nodes within the linked list
{
// deref head to get the real head
node_t* current = *head;
node_t* temp = NULL;
while (current != NULL)
{
temp = current->next;
free(current);
current = temp;
}
// deref head reference to affect the re.
A husband and wife, Ted and Suzanne, share a digital music player th.pdf
A husband and wife, Ted and Suzanne, share a digital music player that has a feature that
randomly selects which song to play. A total of 3476 songs have been loaded into the player,
some by Ted and the rest by Suzanne. They are interested in determining whether they have each
loaded a different proportion of songs into the player. Suppose that when the player was in the
random-selection mode, 22 of the first 29 songs selected were songs loaded by Suzanne. Let p
denote the proportion of songs that were loaded by Suzanne. State the null and alternative
hypotheses to be tested. How strong is the evidence that Ted and Suzanne have each loaded a
different proportion of songs into the player? (Round your test statistic to two decimal places and
your P-value to four decimal places. Assume a 95% confidence level.)
Hypotheses:
H0: p = 0.5
Ha: p < 0.5H0: p = 0.5
Ha: p > 0.5 H0: p = 0.5
Ha: p 0.5
Conclusion:
There is strong evidence that the proportion of songs downloaded by Ted and Suzanne differs
from 0.5.
There is not enough evidence to conclude that the proportion of songs downloaded by Ted and
Suzanne differs from 0.5. z=P-value=
Solution
Formulating the null and alternatuve hypotheses,
Ho: p = 0.5
Ha: p =/= 0.5
*****************************************
As we see, the hypothesized po = 0.5
Getting the point estimate of p, p^,
p^ = x / n = 0.75862069
Getting the standard error of p^, sp,
sp = sqrt[po (1 - po)/n] = 0.092847669
Getting the z statistic,
z = (p^ - po)/sp = 2.785430073
***********************************************
As this is a 2 tailed test, then, getting the p value,
p = 0.002672838
***********************************************
significance level = 0.05
Comparing p and the significance value, we REJECT THE NULL HYPOTHESIS.
Conclusion:
There is strong evidence that the proportion of songs downloaded by Ted and Suzanne differs
from 0.5.
***********************************************.
Why the average of two extreme observations is the best predictor of.pdf
Why is a constant in matrices called a scalar? Don\'t assume I know anything.
Solution
solution:
a matrix consists of variable values as well as constant values (which are nothing but the real
numbers) and in linear algebra reals numbers are considered as scalers hence constant values in
the matrix are called as scalers.
Why is a constant in matrices called a scalar Dont assume I know .pdf
Why are animals like a snail, a clam and a squid, which all look very different, considered to be
in the same phylum?
Solution
Animals of same phyla share maximum common characteristics of their reproductive, digestive,
or nutrition, locomotion, habitat, morphology, body symmetry and other characters. Those
characteristics evaluate with score on matching and animals of maximum resemblance score take
under the same phyla. A new criterion is also use to set phyla, on the basis of genetic equality..
What is the mission andor vision of BMO Harriis bankSolution.pdf
What is the difference between a secured bond and a debenture bond? Respond using at least one
internet resource. Explain why a bond may sell at a premium.
Solution
The bonds are secured by the collateral assets i.e. the assets are pledged as security that if the
company fails to pay the principle and the interest on the bond within stipulated time, the holders
could discharge their debts by seizing and selling the company\'s pledged assets.
The debentures are the debt security issued by a company which is not secured by any specified
assets, but it denotes a general credit of the company.
A bond can be sold at a premium when it offers a coupon rate that is higher than prevailing
market interest rates. Therefore, the investors would pay more for the bonds if they are provided
with a higher yield on the said bond.
===================================.
Using the Thin-lens equation A 1f = 1p + 1q. with equation B L =.pdf
Using the Thin-lens equation: A 1/f = 1/p + 1/q. with equation: B L = p + q and q = p + d p = L -
d/2 and q = L + d/2. Derive equation: C f = L^2 - d^2/4L. Explain mathematically why no image
is formed if the object and image are separated by a distance that is less than 4f.
Solution
A. given:
1/f = 1/p + 1/q
which gives
f = pq/(p+q)
p =(L - d)/2 and q = (L+d)/2
f = [(L - d)/2]*[(L + d)/2]/[(L - d)/2 + (L + d)/2]
f = (L^2 - d^2)/4L
B. u = object distance,
v = image distance
s = separation of object and image,
f = focal length of the lens.
1/u + 1/v = 1/f
1/u + 1/(s - u) = 1/f
(s-u+u)/(u*(s-u)) = 1/f
fs = su - u^2
u^2 = (u - f)s
s = u^2/(u - f)
By differentiating it
ds/du = ((u - f)(2u) - u^2)/(u - f)^2
= (u^2 - 2uf)/(u - f)^2
= u(u - 2f)/(u - f)^2
ds / du = 0 when:
u(u - 2f) = 0
u = 0 or, for a real image, u = 2f.
When u < 2f, ds / du < 0.
When u > 2f, ds / du > 0.
u = 2f is therefore a minimum.
By same method v = 2f is therefore a minimum image distance.
which means total separation 4f..
Using C language to write an algorithm that accepts a list implement.pdf
Using C language to write an algorithm that accepts a list implemented using a linked list
traverses it, and returns the data in the node with the smallest key value.
Thank you very much
Solution
#include
#include
#include
struct node
{
int data;
struct node *link;
};
struct node *head,*x,*y,*z;
void main()
{
void create();
void insbeg();
void insmid();
void insend();
void delbeg();
void delmid();
void delend();
void display();
int ch;
clrscr();
while(1)
{
printf(\"\ 1.Creation\");
printf(\"\ 2.Insertion at beginning\");
printf(\"\ 3.Insertion at middle\");
printf(\"\ 4.Insertion at end\");
printf(\"\ 5.Deletion at beginning\");
printf(\"\ 6.Deletion at middle\");
printf(\"\ 7.Deletion at end\");
printf(\"\ 8.Display\");
printf(\"\ 9.Exit\");
printf(\"\ Enter ur choice:\");
scanf(\"%d\",&ch);
switch(ch)
{
case 1:
create();
break;
case 2:
insbeg();
break;
case 3:
insmid();
break;
case 4:
insend();
break;
case 5:
delbeg();
break;
case 6:
delmid();
break;
case 7:
delend();
break;
case 8:
display();
break;
default:
exit(0);
}
}
}
void create()
{
int c;
head=NULL;
x=(struct node*)malloc(sizeof(struct node));
printf(\"\ Enter the data:\");
scanf(\"%d\",&x->data);
x->link=NULL;
head=x;
printf(\"\ Do u wish to continue press 1 otherwise 0:\");
scanf(\"%d\",&c);
while(c!=0)
{
y=(struct node*)malloc(sizeof(struct node));
printf(\"\ Enter the data:\");
scanf(\"%d\",&y->data);
y->link=NULL;
x->link=y;
x=y;
printf(\"\ Do u wish to continue press 1 otherwise 0:\");
scanf(\"%d\",&c);
}
}
void insbeg()
{
y=(struct node*)malloc(sizeof(struct node));
printf(\"\ Enter the data:\");
scanf(\"%d\",&y->data);
y->link=head;
head=y;
}
void insmid()
{
int pos,c=1;
y=(struct node*)malloc(sizeof(struct node));
printf(\"\ Enter the data:\");
scanf(\"%d\",&y->data);
printf(\"\ Enter the position to be inserted:\");
scanf(\"%d\",&pos);
x=head;
while(clink;
c++;
}
y->link=x;
z->link=y;
}
void insend()
{
y=(struct node*)malloc(sizeof(struct node));
printf(\"\ Enter the data:\");
scanf(\"%d\",&y->data);
y->link=NULL;
x=head;
while(x->link!=NULL)
{
x=x->link;
}
x->link=y;
}
void delbeg()
{
if(head==NULL)
printf(\"\ List is empty\");
else
{
x=head;
head=x->link;
free(x);
}
}
void delmid()
{
int pos,c=1;
if(head==NULL)
printf(\"\ List is empty\");
else
{
printf(\"\ Enter the position to be deleted:\");
scanf(\"%d\",&pos);
x=head;
while(clink;
c++;
}
z->link=x->link;
free(x);
}
}
void delend()
{
if(head==NULL)
printf(\"\ List is empty\");
else
{
x=head;
while(x->link!=NULL)
{
y=x;
x=x->link;
}
y->link=NULL;
free(x);
}
}
void display()
{
if(head==NULL)
printf(\"\ List is empty\");
else
{
x=head;
printf(\"\ List elements are \ \");
while(x->link!=NULL)
{
printf(\"%d->\",x->data);
x=x->link;
}
printf(\"%d\",x->data);
}
}.
More Related Content
Similar to Problem DefinitionBuild a Date class and a main function to test i.pdf
Write a program to generate the entire calendar for one year. The pr.pdf
Write a program to generate the entire calendar for one year. The program must get two values
from the user: (1) the year and (2) the day of the week for January 1st of that year. The year,
which should be positive, is needed to check for and handle leap years1. The day of the week for
January 1st is needed so that you know where to start the calendar. The user should enter 0 for
Sunday, 1 for Monday, ... or 6 for Saturday. As always, you need to validate the user\'s input. To
actually print the calendar, you must use a single method that prints out the calendar for one
month and then call this function 12 times from main(), once for each month in the year. To
check for a leap year you will need to write another method that takes the year as a parameter
and returns true if it’s a leap year, or false otherwise. Stubs (i.e. method signatures without any
code) for both of these methods have been provided for you.
The calendar should be printed in the form below. In this example, January starts on a Saturday
(day 6). Note that February starts on a Tuesday in this example because January ended on a
Monday.
January
1
2 3 4 5 6 7 8
9 10 11 12 13 14 15
16 17 18 19 20 21 22
23 24 25 26 27 2829
30 31
February
1 2 3 4 5
6 7 8 9 10 11 12 ...
The format he gave us is as follows:
public class PA6a {
/**
* Error to output if year is not positive
*/
static final String E_YEAR = \"The year must be positive!\";
/**
* Error to output if the day is not between 0 and 6
*/
static final String E_DAY = \"The day of January 1st must be between 0 and 6!\";
/**
* Determines if an input is a leap year
*
* @param year year in question
* @ereturn true if a leap year
*/
public static boolean isLeapYear(int year) {
return false; // TODO: replace with your code
}
/**
* Outputs a month to the console
*
* @param month title
* @param startDay 0=Sunday ... 6=Saturday
* @param numDays number of days in the month
* @return day of the week of the last day of the month
*/
public static int printMonth(String month, int startDay, int numDays) {
return 0; // TODO: replace with your code
}
/**
* Program execution point:
* input year, day of the week (0-6) of january 1
* output calendar for that year
*
* @param args command-line arguments (ignored)
*/
public static void main(String[] args) {
// TODO: write your code here
}
I am confused especially with the print month method, help is greatly appreciated.
Solution
package org.students;
import java.util.Scanner;
import com.sun.org.apache.bcel.internal.generic.IUSHR;
public class Calander {
//declaring static variables
static final String E_YEAR = \"The year must be positive!\";
static final String E_DAY = \"The day of January 1st must be between 0 and 6!\";
public static void main(String[] args) {
//Declaring variables
int getFirstDay;
int year;
Scanner scanner = new Scanner(System.in);
//This while loop continue to execute until user enters valid year
while (true) {
//Getting the year entered by the user
System.out.print(\"Enter the year:.
publicclass Date {privatestatic String DATE_SEPARATOR = ;pr.pdf
publicclass Date {
privatestatic String DATE_SEPARATOR = \"/\";
privatestaticintDAYS_PER_WEEK = 7;
//Attributes
privateint day;
privateint month;
privateint year;
/**
* Default Constructor
* Instantiates an object of type Date to 1/1/2000
*/
public Date() {
this.day = 1;
this.month = 1;
this.year = 2000;
}
/**
* Constructs a new date object to represent the given date.
* @param day
* @param month
* @param year
*/
public Date(int day, int month, int year) {
if(isValid(day, month, year)) {
this.day = day;
this.month = month;
this.year = year;
} else
System.out.println(\"Invalid Date.\");
}
/**
* Returns the day value of this date for example, for the date 2006/07/22, returns 22.
* @return
*/
publicint getDay() {
return day;
}
/**
* Returns the month value of this date ,for example, for the date 2006/07/22, returns 7.
* @return
*/
publicint getMonth() {
return month;
}
/**
* Returns the year value of this date, for example , the date 2006/07/22, returns 2006.
* @return
*/
publicint getYear() {
return year;
}
/**
* @param day the day to set
*/
publicvoid setDay(int day) {
this.day = day;
}
/**
* @param month the month to set
*/
publicvoid setMonth(int month) {
this.month = month;
}
/**
* @param year the year to set
*/
publicvoid setYear(int year) {
this.year = year;
}
/**
* Returns true if the year of this date is a leap year.
* A leap year occurs every 4 years , except for multiples of 100 that are not multiples of 400.
* For example, 1956,1844,1600,and 2000 are leap years, but 1983,2002,1700,and 1900 are not.
* @return
*/
publicboolean isLeapYear(){
if(((this.year % 400) == 0) || (((this.year % 4) == 0) && ((this.year % 100) != 0)))
returntrue;
else
returnfalse;
}
/**
* Checks if the date is valid
* @param day
* @param month
* @param year
* @return
*/
publicboolean isValid(int day, int month, int year) {
if((month < 1) || (12 < month))
returnfalse;
else {
if(((month == 1) || (month == 3) || (month == 5) || (month == 7) ||
(month == 8) || (month == 10) || (month == 12)) && ((day < 1) || (31 < day)))
returnfalse;
elseif(((month == 4) || (month == 6) || (month == 9) || (month == 11)) && ((day < 1) || (30 <
day)))
returnfalse;
elseif(month == 2) {
if(isLeapYear() && ((day < 1) || (29 < day)))
returnfalse;
elseif((day < 1) || (28 < day))
returnfalse;
}
}
returntrue;
}
/**
* Returns the maximum number of days in a month
* @return
*/
publicint maxMonthDays() {
if(this.month == 2) {
if(isLeapYear())
return 29;
else
return 28;
} elseif((this.month == 1) || (this.month == 3) || (this.month == 5) || (this.month == 7) ||
(this.month == 8) || (this.month == 10) || (this.month == 12))
return 31;
else
return 30;
}
/**
* Checks if this dat is same as other date
* @param other
* @return
*/
publicboolean isEqual(Date other) {
if((this.day == other.day) && (this.month == other.month) && (this.year == other.year))
returntrue;
else
returnfalse;
}
/**
* Moves this Date object forward in time by the given number of days .
* @param days
*/
publicvoid addDays(int days).
Les03 (Using Single Row Functions To Customize Output)
Using Single Row FUnction (uploaded by achmatim.net)
Hello I need some help on this question this is for C++ PEARSON REVE.pdf
Hello I need some help on this question this is for C++ PEARSON REVEL
I have a code that does not seem to be correct and I want to know to the correct way.
There is also the reason for the error and the question itself.
here is the code:
#include
#include
using namespace std;
class Month {
private:
int num;
static string monthName[12];
public:
Month() {
num = 0;
}
Month(int n) {
num = n;
}
void setNum(int n) {
num = n;
}
void setName(string name) {
for (int i = 0; i < 12; i++) {
if (name == monthName[i]) {
num = i + 1;
return;
}
}
num = 0;
}
int getNum() const {
return num;
}
string getName() const {
if (num < 1 || num > 12) {
return "Unknown";
}
return monthName[num - 1];
}
Month& operator++() {
num++;
if (num > 12) {
num = 1;
}
return *this;
}
Month operator++(int) {
Month temp = *this;
++(*this);
return temp;
}
Month& operator--() {
num--;
if (num < 1) {
num = 12;
}
return *this;
}
Month operator--(int) {
Month temp = *this;
--(*this);
return temp;
}
friend ostream& operator<<(ostream& os, const Month& month) {
os << month.getName();
return os;
}
friend istream& operator>>(istream& is, Month& month) {
int n;
is >> n;
month.setNum(n);
return is;
}
};
string Month::monthName[12] = {"January,", "February,", "March,", "April,", "May,", "June,",
"July,",
"August,", "September,", "October,", "November,", "December,"};
int main() {
int n;
cout << "Enter: month number (1-12):";
cin >> n;
Month month(n);
cout << month << endl;
for (int i = 0; i < 7; i++) {
++month;
cout << month << endl;
}
cout << "Enter: another month number (1 - 12): ";
cin >> n;
month.setNum(n);
cout << month << endl;
for (int i = 0; i < 7; i++) {
--month;
cout << month << endl;
}
return 0;
}
and here is the errors that it is giving me:
Logic errors.
And here is the question for the answer that i am trying to do:
Design a class named Month. The class should have the following members:
Data Members:
An int variable named num to hold the month number, which is an integer in the range of 1
through 12.
A static array of twelve strings named monthName, initialized with the names of the months.
For example, monthName[0] should be set to "January", monthName[1] should be set to
"February", etc. The static array will be used as a lookup table to look up the name of a month.
Constructors:
The class should have a default constructor that sets the num member variable to 0 as well as a
constructor that accepts an int argument for the num member variable.
Mutators:
A setNum function that accepts an int argument for the month number. The setNum function
should assign this value to the num member variable.
A setName function that accepts a string argument for the name of the month. The setName
function should find the name in the monthName array, and then update the num member
variable with the correct number. For example, if the argument is "January", the num variable
will be set to 1. If the argument is "April", the num variable will be set to 4. And so on. If the
argument is not found in th.
I am having some trouble getting this to compile, I have made seve.docx
I am having some trouble getting this to compile, I have made several changes and still not having any luck. Can someone please give me some guidance on how to correct this. thank you in advance.
Chad
#include <stdio.h>
#include <stdlib.h>
int main()
{//Created by Chad Hartley
//October 30, 2015
char Month;//Variable for the character the user will input.
//Ask which month needs to be converted.
//I used the third character in the months that had the same first letter, except for June I had to use the second character.
//I included a Key for the user to reference to know which character is expected.
printf("Please use the following letter to represent the month you want converted: \n");
printf("J=January,F=February,M=March,A=April,Y=May,U=June,L=July,G=August,S=September,O=October,N=November,D=December \n");
scanf("%c",&Month);
switch(Month)//switching statement used to look for the character inputed
{
case J:
printf("01");
printf("\n");//used to print a new line
break;//I used break to not have a continuous output
case F:
printf("02");
printf("\n");
break;
case M:
printf("03");
printf("\n");
break;
case A:
printf("04");
printf("\n");
break;
case Y:
printf(“05");
printf("\n");
break;
case U:
printf("06");
printf("\n");
break;
case L:
printf("07");
printf("\n");
break;
case G:
printf("08");
printf("\n");
break;
case S:
printf("09");
printf("\n");
break;
case O:
printf("10");
printf("\n");
break;
case N:
printf("11");
printf("\n");
break;
case D:
printf("12");
printf("\n");
break;
default:
printf("Please enter a valid character.");//This is a default statement to cover any other character the user may input.
printf("\n");
}
return 0;
}
PROVIDE COMMENT HERE FOR ABOVE ANSWER
PROVIDE COMMENT FOR THIS TOO
This program will allow a character value for a month to be entered and an output a number corresponding to the month. A switch statement will be used to convert from a character to a number the print out the result.
#include <stdio.h>
int main(void) {
/* variable definition */
int J = 1; 6; 7;
int F = 2;
int M = 3; 5;
int A = 4; 8;
int S = 9;
int O = 10;
int N = 11;
int D = 12;
/* the switch statement */
switch (1) <-------- The switch statement is where the number of the character will be placed to convert.
{
case 1: <-----The case scenario execute the conversion
case 6:
case 7:
printf ("J is equal to 1, 6 or 7\n"); <--------The switch is then printed out.
break; <------ This break ends case 1 etc.
case 2:
printf ("F is equal 2\n");
break;
case 3:
.
I got the codes written down below- Basically- I am trying to implemen.pdf
I got the codes written down below. Basically, I am trying to implement OOP of employees and
payments with LinkedList, but when I test it, I keep getting errors like "uncleared Identifier."
Please help me with my programs.(Language: C)
The above images are the expected output and graphic illustrations of the linked lists.
Please help me debug the program and provide the correct version of the codes. DO NOT COPY
others' answers or chatGPT, and then, I'll give you an upvote.
Main:
///////////////////////////////////////////////////////////////////////////////////////////////////
#include "employee.h"
int main(){
//-----------------------------------------------------
//testCase -1
/*
EMPLOYEE e;
e.name="james";
e.id=123456;
e.next=NULL;
e.payments=NULL;
printEmployee(e);
*/
//-----------------------------------------------------
//testCase -2
/*
PAYMENT p1;
p1.date="01/01/2010";
p1.amount=150;
p1.next=NULL;
PAYMENT p2;
p2.date="01/05/2010";
p2.amount=200;
p2.next=&p1;
EMPLOYEE e;
e.name="sarah";
e.id=123456;
e.payments=&p2;
e.next=NULL;
printEmployee(e);
*/
//-----------------------------------------------------
//testCase -3
/*
addEmployee(123,"alma");
addEmployee(456,"mike");
printEmployees();
*/
//-----------------------------------------------------
//testCase -4
/*
addEmployee(123,"alma");
addPayment(123,"01/02/2020",3000);
addPayment(123,"10/07/2021",5000);
printEmployees();
*/
//-----------------------------------------------------
//testCase -5
/*
addEmployee(123,"alma");
addPayment(123,"01/02/2020",3000);
addPayment(123,"10/07/2021",5000);
addEmployee(456,"mike");
addPayment(456,"06/03/2020",700);
addPayment(456,"07/02/2020",200);
addPayment(456,"06/04/2021",600);
printEmployees();
*/
//-----------------------------------------------------
//testCase -6
/*
addEmployee(123,"alma");
addPayment(123,"01/02/2020",3000);
addPayment(123,"10/07/2021",5000);
addEmployee(456,"mike");
addPayment(456,"06/03/2020",700);
addPayment(456,"07/02/2020",200);
addPayment(456,"06/04/2021",600);
deleteEmployee(123,"alma");
printEmployees();
*/
//-----------------------------------------------------
//testCase -7
/*
addEmployee(123,"alma");
addPayment(123,"01/02/2020",3000);
addPayment(123,"10/07/2021",5000);
addEmployee(456,"mike");
addPayment(456,"06/03/2020",700);
addPayment(456,"07/02/2020",200);
addPayment(456,"06/04/2021",600);
deleteEmployee(456,"mike");
deleteEmployee(123,"alma");
printEmployees();
*/
}
///////////////////////////////////////////////////////////////////////////////////////////////////
Constructor:
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
typedef struct payment{
char* date; // MM/DD/YY as text
float amount;
struct payment* next;
}PAYMENT;
typedef struct employee{
int id;
char* name;
PAYMENT* payments;
struct employee* next;
}EMPLOYEE;
void printEmployee(EMPLOYEE e);
void printEmployees();
void addEmployee(int e_id,char* e_name);
int addPayment(int e_id, char* p_date, float p_amount);
int deleteEmployee(int e_id, char.
I Have the following Java program in which converts Date to Words an.pdf
I Have the following Java program in which converts Date to Words and the only thing I have to
do now is add a boolean for a leap year. ( I am using Apache NetBeans) please explain where the
boolean goes.
Like this:
I need the code working with the boolean added, Thank you!!!
package a1_demo;
import java.util.Scanner;
public class MainApp {
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
System.out.print("Welcome to the Date to Words Converter!\n\n");
while ( true ) {
// get user input
String str_MM_DD_YYYY = Validator.getString(
Validator.sc,
"Enter date in MM/DD/YYYY format: "
);
// convert text to ints:
// "11/22/3333" - > int, int int
String[] dateComponents = str_MM_DD_YYYY.split( "/" );
if ( dateComponents.length != 3 ) {
System.out.println("\t\t Bad input, please retry...");
continue;
}
int month = 0;
int day = 0;
int year = 0;
try {
month = Integer.parseInt(dateComponents[ 0 ]);
day = Integer.parseInt(dateComponents[ 1 ]);
year = Integer.parseInt(dateComponents[ 2 ]);
} catch ( java.lang.NumberFormatException ex ) {
System.out.println("\t\t numeric values are expected, please retry...");
continue;
}
Date2TextConverter converter = new Date2TextConverter( month, day, year );
System.out.print( converter.month2text() );
System.out.print( " " );
System.out.print( converter.day2text() );
System.out.print( " " );
System.out.print( converter.year2text() );
System.out.println();
}// endless while loop
}//main
}//class MainApp
package a1_demo;
public class Date2TextConverter {
//---------------------------------------
// data
//---------------------------------------
int month = 0;
int day = 0;
int year = 0;
//---------------------------------------
// constructors
//---------------------------------------
public Date2TextConverter(int month, int day, int year )
{
this.month = month;
this.day = day;
this.year = year;
}//Date2TextConverter
//---------------------------------------
// operations
//---------------------------------------
public String month2text()
{
switch( month )
{
case 1: return "Jan";
case 2: return "Feb";
case 3: return "Mar";
case 4: return"April";
case 5: return"May";
case 6: return"June";
case 7: return"July";
case 8: return"Aug";
case 9: return"Sep";
case 10: return"Oct";
case 11: return"Nov";
case 12: return"Dec";
//...
default:
return "Invalid Month";
}
}//month2text
public String day2text()
{
//return Integer.toString( day );
return number2words (day);
}//day2text
public String year2text()
{
int century = year / 100;
int yearInCentury = year % 100;
String centuryText = number2words( century );
String yearText = number2words( yearInCentury );
if ( yearInCentury < 10 ) {
return centuryText + " zero " + yearText;
} else {
return centuryText + " " + yearText;
}
}//year2text
private String number2words( int number ) {
// convert to text a number 0 through 99
if ( number < 20 ) {
// 0, 1, 2, 3, ..., 19
return zero2nineteen[ number ];
} else if ( number % 10 == 0 ) {
// 20, 30, 40, ... 90
return .
Operator Overloading
How to use to operator overloading in C++ to improve readability and re-usability of your source code.
struct procedure { Date dateOfProcedure; int procedureID.pdf
struct procedure
{
Date dateOfProcedure;
int procedureID;
int procedureProviderID;
};
Data.h
#include
#include
class Date
{
friend ostream& operator<<( ostream &, const Date & );
// allows easy output to a ostream
public:
Date( int m = 1, int d = 1, int y = 1900 ); // constructor, note the default values
void setDate( int, int, int ); // set the date const
Date &operator+=( int ); // add days, modify object
bool leapYear( int) const; // is this a leap year?
bool endOfMonth( int ) const; // is this end of month?
int getMonth ( ) const;
int getDay ( ) const;
int getYear ( ) const;
string getMonthString( ) const;
private:
int month;
int day;
int year;
static const int days[]; // array of days per month static const string monthName[]; // array of
month names
void helpIncrement();
// utility function
}; #endif Data.cpp // Member function definitions for Date
class in separate
date.cpp
#include
#include \"date.h\"
#include // Initialize static members at file scope; // one class-wide copy.
const int Date::days[] = { 0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31 }; const string
Date::monthName[] = { \"January\", \"February\", \"March\", \"April\", \"May\", \"June\",
\"July\", \"August\", \"September\", \"October\", \"November\", \"December\" };
// Date constructor
Date::Date(int m, int d, int y)
{
setDate(m, d, y); } // Set the date
void Date::setDate(int mm, int dd, int yy)
{
month = (mm >= 1 && mm <= 12) ? mm : 1; year = (yy >=1900&& yy <= 2100) ? yy
: 1900; // test for a leap year
if (month == 2 && leapYear(year)) day = (dd >= 1 && dd <= 29) ? dd : 1;
else
day =(dd >= 1 && dd <= days[month]) ? dd : 1;
} // Add a specific number of days to a date
const Date &Date::operator+=(int additionalDays)
{
for (int i = 0; i < additionalDays; i++) helpIncrement();
return *this; // enables cascading
} // If the year is leap year, return true; //
otherwise, return false
bool Date::leapYear(int testYear)
const
{
if (testYear % 400 == 0 || (testYear %100 != 0 && testYear % 4 == 0)) return true; // a
leap year
else
return false; // not a leap year
} // Determine if the day is the end of the month
bool Date::endOfMonth(int testDay)
const
{
if (month == 2 && leapYear(year))
return (testDay == 29); // last day of Feb. in leap year
else
return (testDay == days[month]);
} // Function to help increment the date
void Date::helpIncrement()
{
if (!endOfMonth(day))
{
// date is not at the end of the month
day++;
}
else if (month < 12)
{
// date is at the end of the month, but month < 12 day = 1; ++month;
}
else
{
// end of month and year: last day of the year
day = 1;
month = 1;
++year;
} }
// Overloaded output operator
ostream &operator<<(ostream& output, const Date &d)
{
output << d.monthName[d.month] << \' \' << d.day << \", \"<< d.year;
return output; // enables cascading
}
int Date::getMonth() const //
{
Public:
int month[20];
If(month[]!=”/0”)
Cout<<”Enter Month :-<>month;
Elseif (month[]<12)
{
Cout<>month[];
else
return month;
}
int Date::getDay() const //
{
Pu.
(----IN C) Incorrect output on some test cases- I'm using a linked lis.pdf
(****IN C) Incorrect output on some test cases.
I'm using a linked list to take in information about a person and using structs to find the oldest
and youngest person in the list; however, I am having trouble with some test cases. For the most
part, The code seems to work the way I want it to, but certain test cases do have wrong outputs.
For example:
Input File Has Four Entries(2 Youngest are in the Same Month)
Here is the text file:
Smith,Bob,July,31,2002,CS/BS,Junior
Joe,Average,February,3,2002,CS/BS,Freshman
Doe,Jane,July,27,2003,CS/BS,Sophomore
Doe,John,July,12,2003,CS/BS,Sophomore:
Output:
Oldest - John Doe & Youngest Jane Doe.
John is clearly not the oldest in the list so please help, below is my code.
//add birthday struct here
struct birthday
{
int day, year;
char month[50];
}; typedef struct birthday bday;
//add node_t struct here (this struct will represent a node in the linked list)
struct node
{
char firstName[50], lastName[50], major[50], classStanding[50];
bday bd;
struct node* next;
};
typedef struct node node_t;
struct stats
{
// node pointer pointing at the oldest and youngest person in the list
node_t* oldest;
node_t* youngest;
// holding the number of birthdays in each month (0 = Jan. & 11 = Dec.)
int numBD[12];
}; typedef struct stats list_stats_t;
void add(node_t **node, node_t **head)
{
(*node)->next = *head;
*head = *node;
}
node_t* readNodeInfo(FILE* input)
{
node_t *node;
node = (node_t*)malloc(sizeof(node_t)); // allocates memory
char newLine[1]; // reads new line character
// uses scanset to read information between commas
fscanf(input, "%[^,], %[^,], %[^,], %d, %d, %[^,], %[^\n] %[\n]",
node->lastName, node->firstName, node->bd.month,
&node->bd.day, &node->bd.year, node->major,
node->classStanding, newLine);
return node;
}
node_t* createList(FILE* in, node_t** head)
{
node_t *temp;
while(!feof(in)) // loops until end of file in
{
temp = readNodeInfo(in); // calls function and assigns it to temp
add(&temp, head); // adds to linked list
}
return *head;
}
void printList(FILE* out, node_t* head) // prints data
{
// if there is no data in the linked list, print error code to terminal
if (head == NULL)
{
fprintf(stderr, "LIST IS EMPTY\n");
return;
}
node_t *temp = head;
printBorder(out);
fprintf(out, "\nList Info:\n");
while (temp != NULL)
{
fprintf(out, "Name:\t%s %s\n", temp->firstName, temp->lastName);
fprintf(out, "Date of Birth:\t%s %d, %d\n", temp->bd.month,
temp->bd.day,
temp->bd.year);
fprintf(out, "Major:\t%s\n", temp->major);
fprintf(out, "Year:\t%s\n\n", temp->classStanding);
temp = temp->next; // next element
}
printBorder(out);
}
void printBorder(FILE* out)
{
int i = 0;
for(i = 0; i < 80; i++)
fprintf(out,"*");
fprintf(out, "\n");
}
void deleteList(node_t** head) // deletes all nodes within the linked list
{
// deref head to get the real head
node_t* current = *head;
node_t* temp = NULL;
while (current != NULL)
{
temp = current->next;
free(current);
current = temp;
}
// deref head reference to affect the re.
Similar to Problem DefinitionBuild a Date class and a main function to test i.pdf (16)
Write a program to generate the entire calendar for one year. The pr.pdf
Write a program to generate the entire calendar for one year. The pr.pdf
publicclass Date {privatestatic String DATE_SEPARATOR = ;pr.pdf
publicclass Date {privatestatic String DATE_SEPARATOR = ;pr.pdf
Les03 (Using Single Row Functions To Customize Output)
Les03 (Using Single Row Functions To Customize Output)
Hello I need some help on this question this is for C++ PEARSON REVE.pdf
Hello I need some help on this question this is for C++ PEARSON REVE.pdf
I am having some trouble getting this to compile, I have made seve.docx
I am having some trouble getting this to compile, I have made seve.docx
I got the codes written down below- Basically- I am trying to implemen.pdf
I got the codes written down below- Basically- I am trying to implemen.pdf
I Have the following Java program in which converts Date to Words an.pdf
I Have the following Java program in which converts Date to Words an.pdf
struct procedure { Date dateOfProcedure; int procedureID.pdf
struct procedure { Date dateOfProcedure; int procedureID.pdf
C programming & data structure [arrays & pointers]
C programming & data structure [arrays & pointers]
(----IN C) Incorrect output on some test cases- I'm using a linked lis.pdf
(----IN C) Incorrect output on some test cases- I'm using a linked lis.pdf
More from smitaguptabootique
A husband and wife, Ted and Suzanne, share a digital music player th.pdf
A husband and wife, Ted and Suzanne, share a digital music player that has a feature that
randomly selects which song to play. A total of 3476 songs have been loaded into the player,
some by Ted and the rest by Suzanne. They are interested in determining whether they have each
loaded a different proportion of songs into the player. Suppose that when the player was in the
random-selection mode, 22 of the first 29 songs selected were songs loaded by Suzanne. Let p
denote the proportion of songs that were loaded by Suzanne. State the null and alternative
hypotheses to be tested. How strong is the evidence that Ted and Suzanne have each loaded a
different proportion of songs into the player? (Round your test statistic to two decimal places and
your P-value to four decimal places. Assume a 95% confidence level.)
Hypotheses:
H0: p = 0.5
Ha: p < 0.5H0: p = 0.5
Ha: p > 0.5 H0: p = 0.5
Ha: p 0.5
Conclusion:
There is strong evidence that the proportion of songs downloaded by Ted and Suzanne differs
from 0.5.
There is not enough evidence to conclude that the proportion of songs downloaded by Ted and
Suzanne differs from 0.5. z=P-value=
Solution
Formulating the null and alternatuve hypotheses,
Ho: p = 0.5
Ha: p =/= 0.5
*****************************************
As we see, the hypothesized po = 0.5
Getting the point estimate of p, p^,
p^ = x / n = 0.75862069
Getting the standard error of p^, sp,
sp = sqrt[po (1 - po)/n] = 0.092847669
Getting the z statistic,
z = (p^ - po)/sp = 2.785430073
***********************************************
As this is a 2 tailed test, then, getting the p value,
p = 0.002672838
***********************************************
significance level = 0.05
Comparing p and the significance value, we REJECT THE NULL HYPOTHESIS.
Conclusion:
There is strong evidence that the proportion of songs downloaded by Ted and Suzanne differs
from 0.5.
***********************************************.
Why the average of two extreme observations is the best predictor of.pdf
Why is a constant in matrices called a scalar? Don\'t assume I know anything.
Solution
solution:
a matrix consists of variable values as well as constant values (which are nothing but the real
numbers) and in linear algebra reals numbers are considered as scalers hence constant values in
the matrix are called as scalers.
Why is a constant in matrices called a scalar Dont assume I know .pdf
Why are animals like a snail, a clam and a squid, which all look very different, considered to be
in the same phylum?
Solution
Animals of same phyla share maximum common characteristics of their reproductive, digestive,
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What is the mission andor vision of BMO Harriis bankSolution.pdf
What is the difference between a secured bond and a debenture bond? Respond using at least one
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The bonds are secured by the collateral assets i.e. the assets are pledged as security that if the
company fails to pay the principle and the interest on the bond within stipulated time, the holders
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===================================.
Using the Thin-lens equation A 1f = 1p + 1q. with equation B L =.pdf
Using the Thin-lens equation: A 1/f = 1/p + 1/q. with equation: B L = p + q and q = p + d p = L -
d/2 and q = L + d/2. Derive equation: C f = L^2 - d^2/4L. Explain mathematically why no image
is formed if the object and image are separated by a distance that is less than 4f.
Solution
A. given:
1/f = 1/p + 1/q
which gives
f = pq/(p+q)
p =(L - d)/2 and q = (L+d)/2
f = [(L - d)/2]*[(L + d)/2]/[(L - d)/2 + (L + d)/2]
f = (L^2 - d^2)/4L
B. u = object distance,
v = image distance
s = separation of object and image,
f = focal length of the lens.
1/u + 1/v = 1/f
1/u + 1/(s - u) = 1/f
(s-u+u)/(u*(s-u)) = 1/f
fs = su - u^2
u^2 = (u - f)s
s = u^2/(u - f)
By differentiating it
ds/du = ((u - f)(2u) - u^2)/(u - f)^2
= (u^2 - 2uf)/(u - f)^2
= u(u - 2f)/(u - f)^2
ds / du = 0 when:
u(u - 2f) = 0
u = 0 or, for a real image, u = 2f.
When u < 2f, ds / du < 0.
When u > 2f, ds / du > 0.
u = 2f is therefore a minimum.
By same method v = 2f is therefore a minimum image distance.
which means total separation 4f..
Using C language to write an algorithm that accepts a list implement.pdf
Using C language to write an algorithm that accepts a list implemented using a linked list
traverses it, and returns the data in the node with the smallest key value.
Thank you very much
Solution
#include
#include
#include
struct node
{
int data;
struct node *link;
};
struct node *head,*x,*y,*z;
void main()
{
void create();
void insbeg();
void insmid();
void insend();
void delbeg();
void delmid();
void delend();
void display();
int ch;
clrscr();
while(1)
{
printf(\"\ 1.Creation\");
printf(\"\ 2.Insertion at beginning\");
printf(\"\ 3.Insertion at middle\");
printf(\"\ 4.Insertion at end\");
printf(\"\ 5.Deletion at beginning\");
printf(\"\ 6.Deletion at middle\");
printf(\"\ 7.Deletion at end\");
printf(\"\ 8.Display\");
printf(\"\ 9.Exit\");
printf(\"\ Enter ur choice:\");
scanf(\"%d\",&ch);
switch(ch)
{
case 1:
create();
break;
case 2:
insbeg();
break;
case 3:
insmid();
break;
case 4:
insend();
break;
case 5:
delbeg();
break;
case 6:
delmid();
break;
case 7:
delend();
break;
case 8:
display();
break;
default:
exit(0);
}
}
}
void create()
{
int c;
head=NULL;
x=(struct node*)malloc(sizeof(struct node));
printf(\"\ Enter the data:\");
scanf(\"%d\",&x->data);
x->link=NULL;
head=x;
printf(\"\ Do u wish to continue press 1 otherwise 0:\");
scanf(\"%d\",&c);
while(c!=0)
{
y=(struct node*)malloc(sizeof(struct node));
printf(\"\ Enter the data:\");
scanf(\"%d\",&y->data);
y->link=NULL;
x->link=y;
x=y;
printf(\"\ Do u wish to continue press 1 otherwise 0:\");
scanf(\"%d\",&c);
}
}
void insbeg()
{
y=(struct node*)malloc(sizeof(struct node));
printf(\"\ Enter the data:\");
scanf(\"%d\",&y->data);
y->link=head;
head=y;
}
void insmid()
{
int pos,c=1;
y=(struct node*)malloc(sizeof(struct node));
printf(\"\ Enter the data:\");
scanf(\"%d\",&y->data);
printf(\"\ Enter the position to be inserted:\");
scanf(\"%d\",&pos);
x=head;
while(clink;
c++;
}
y->link=x;
z->link=y;
}
void insend()
{
y=(struct node*)malloc(sizeof(struct node));
printf(\"\ Enter the data:\");
scanf(\"%d\",&y->data);
y->link=NULL;
x=head;
while(x->link!=NULL)
{
x=x->link;
}
x->link=y;
}
void delbeg()
{
if(head==NULL)
printf(\"\ List is empty\");
else
{
x=head;
head=x->link;
free(x);
}
}
void delmid()
{
int pos,c=1;
if(head==NULL)
printf(\"\ List is empty\");
else
{
printf(\"\ Enter the position to be deleted:\");
scanf(\"%d\",&pos);
x=head;
while(clink;
c++;
}
z->link=x->link;
free(x);
}
}
void delend()
{
if(head==NULL)
printf(\"\ List is empty\");
else
{
x=head;
while(x->link!=NULL)
{
y=x;
x=x->link;
}
y->link=NULL;
free(x);
}
}
void display()
{
if(head==NULL)
printf(\"\ List is empty\");
else
{
x=head;
printf(\"\ List elements are \ \");
while(x->link!=NULL)
{
printf(\"%d->\",x->data);
x=x->link;
}
printf(\"%d\",x->data);
}
}.
type 1-2 page (typewritten) research paper on the issue of voting.pdf
type 1-2 page (typewritten) research paper on the issue of voting rights for convicted felons in
Texas. Your paper must meet the following requirements:
1) Identify and explain the current TEXAS law regarding the voting rights of the following types
of Texas citizens:
a. Those currently incarcerated;
b. Those released from prison but who are still on parole;
c. Those who are ex-convicts and have completed their parole.
2) Explain as to whether you think which, if any or all, of these laws should be changed. Give
specific reasons for your position.
3) Describe who may or may not benefit from the position you take in 2) above. Consider the
impact on political parties, or on candidates for various Texas State offices and also consider the
impact in other areas, e.g., economic, political, social.
4) List at least two things you could do in order to engage the political process in Texas in
pursuit of your position in 2) above.
5) Cite at least three (3) reliable, quality references from the media and/or the internet within the
body of the paper using the parenthetical citation style (see below for link). One of the citations
MUST come from the class textbook (-10 point penalty points if this is not done), and NONE
may be from ANY encyclopedia, including Wikipedia (-10 point penalty if this is done). For full
credit, you must have three sources and at least three citations.
You must also have a properly formatted reference section at the end of the paper. The paper’s
in-text citations AND works cited section must conform to the format found in the examples
from this website: http://www.chicagomanualofstyle.org/tools_citationguide.html (be sure to
click on the Author-Date tab!).
you must be very careful when they examine internet websites and provide at minimum the
author/publisher, date published or accessed, and the title of the article or website. The URL
should also be available.
All written assignments must be typed, in 12-point font, and double-spaced.
Solution
1.a,Those currently incarcerated:
Incarceration in the United States is one of the main forms of punishment, rehabilitation, or both
for the commission of felony and other offenses. The United States has the largest prison
population in the world.
1.The number of people incarcerated in America quadrupled-from roughly 500,000 to 2.3
million people.
2.Today, the US is 5% of the World population and has 25% of world prisoners.
3.Combining the number of people in prison and jail with those under parole or probation
supervision, 1 in ever y 31 adults, or 3.2 percent of the population is under some form of
correctional control.
b).Parole is the provisional release of a prisoner who agrees to certain conditions prior to the
completion of the maximum sentence period.Most states have the ability to release prisoners on
parole after they almost surely pose no threat to the public.
c).I believe that convicted felons should be allowed to vote, indeed I believe they should actually
be allowed to vot.
Two figures have a ration if similarity of 34. If the are of the la.pdf
Two figures have a ration if similarity of 3:4. If the are of the larger figure is 100 square units,
what is the area of the smaller figure?
Solution
Smaller figure area/ larger figure area =3/4
Given larger figure area = 100 square units
Plugging in into the above relation
Smaller figure area/100 =3/4
Smaller figure area = 3/4 * 100 = 75 square units.
There is an area that has been classified, according to the IEC requ.pdf
There is an area that has been classified, according to the IEC requirements, a being ‘Group IIA’.
Compare this to something that has been classified, by the Americans or Canadians, as ‘Class I,
Group A’. Which would be the most dangerous are of the two?
Solution
Among Group IIA and clasS I the GrouP IIA is dangerous.
Group IIA contains flammable liquid ,combustible liquid and flammable gases.
These are caused at the time of explosion which are dangerous.
This produces high temperature which causes harmful to humans ,animals ,environment.
It may releases some different gases in to earth which causes effect to environment and it
may causes new health problems.
This group IIA can mix with oxygen and pollute the air.
Class I: It contains liquid ,vapour and flammable gas and acetylene.
Which does not contains more pollution and effect when compared to Group II A.
compare class I and Group II A Group IIA is dangerous and high tendency to mix with air..
The game of Pig is a simple two-player dice game in which the first .pdf
The game of Pig is a simple two-player dice game in which the first player to reach 100 or more
points wins. Players take turns. On each turn, a player rolls a six-sided die:
If the player rolls a 1, then the player gets no new points and it becomes the other player’s turn.
If the player rolls 2 through 6, then he or she can either
ROLL AGAIN
or
HOLD:
At this point, the sum of all rolls is added to the player’s score and it becomes the other
player’s turn.
Write a program that plays the game of Pig, where one player is a human and the other is the
computer. When it is the human’s turn, the program should show the score of both players and
the previous roll. Allow the human to input whether to roll again or hold.
The computer program should play according to the following rules:
Keep rolling when it is the computer’s turn until it has accumulated 20 or more points, then hold.
If the computer wins or rolls a 1, then the turn ends immediately.
Allow the human to roll first.
Notes
To make your task easier, you can modularize your codes as the following methods:
static int usersTurn(int oldTotal): user rolls until they either:
hold: the turn total is added to the oldTotal and the sum is returned
roll a 1: the oldTotal is returned
static int computersTurn(int oldTotal): computer rolls until it either:
holds (once it has accumulated 20 or more points for this turn): the turn total is added to the
oldTotal and the sum is returned
rolls a 1: the oldTotal is returned
static int rollDie(): simulates rolling a die (random number from 1 to 6)
Sample Output (user input shown in red)
Welcome to the game of Pig!
You rolled: 4
Your turn score is 4 and your total score is 0
If you hold, you will have 4 points.
Enter \'r\' to roll again, \'s\' to stop.
r
You rolled: 2
Your turn score is 6 and your total score is 0
If you hold, you will have 6 points.
Enter \'r\' to roll again, \'s\' to stop.
r
You rolled: 4
Your turn score is 10 and your total score is 0
If you hold, you will have 10 points.
Enter \'r\' to roll again, \'s\' to stop.
r
You rolled: 4
Your turn score is 14 and your total score is 0
If you hold, you will have 14 points.
Enter \'r\' to roll again, \'s\' to stop.
r
You rolled: 3
Your turn score is 17 and your total score is 0
If you hold, you will have 17 points.
Enter \'r\' to roll again, \'s\' to stop.
r
You rolled: 5
Your turn score is 22 and your total score is 0
If you hold, you will have 22 points.
Enter \'r\' to roll again, \'s\' to stop.
s
Your score is 22
It is the computer\'s turn.
The computer rolled: 2
The computer rolled: 5
The computer rolled: 5
The computer rolled: 3
The computer rolled: 2
The computer rolled: 4
The computer holds.
The computer\'s score is 21
You rolled: 1
You lose your turn! Your total is 22
Your score is 22
It is the computer\'s turn.
The computer rolled: 3
The computer rolled: 6
The computer rolled: 3
The computer rolled: 2
The computer rolled: 3
The computer rolled: 1
The computer lost its turn! Computer total is 2.
Problem 2 DSB-SC, Matched Filtering Consider the Double-Sideband Sup.pdf
Problem 2: DSB-SC, Matched Filtering Consider the Double-Sideband Suppressed Carrier
(DSB-SC) system depicted in Fig. 1 which is present in many modern digital communication
systems p(t) Bandpass Transmit Signal T\' r1 [n] Pulse cos(2m fet) sin(2 fct) Pulse Tsampling
BandpasS Received Signal Pulse g(t) ci(t) co(t) Tsampling Pulse g(t) Fig. 1: DSB-SC scheme for
the Problem 2.
Solution
trying to upload image
but could nt.
One of the popular tourist attractions in Alaska is watching black b.pdf
One of the popular tourist attractions in Alaska is watching black bears catch salmon swimming
upstream to spawn. Not all “black” bears are black, though— some are tan-colored. Suppose that
six black bears and three tan-colored bears are working the rapids of a salmon stream. Over the
course of an hour, six different bears are sighted. What is the probability that those six include at
least twice as many black bears as tan-colored bears?
Solution.
Im doing a marketing plan on how Apple can improve their products .pdf
I\'m doing a marketing plan on how Apple can improve their products below is the outlined
structure pls help xhibit 21-4 Summary Outline of Different Sections of Marketing Plan
Company Analysis Company objectives and overall marketing objectives Company resources
(marketing, production, financial, human, etc.) Other marketing plans (marketing program)
Previous marketing strategy Major screening criteria relevant to product-market opportunity
selected Quantitative (ROI, profitability, risk level, etc.) Qualitative (nature of business
preferred, social responsibility environment, etc,) Major constraints Marketing collaborators
(current and potential) Customer Analysis (organizational customers and/or final consumers)
Product-market Possible segmenting dimensions (customer needs, other characteristics)
Qualifying dimensions and determining dimensions Identify target market(s) (one or more
specific segments) Operational characteristics (demographics, geographic locations, etc.)
Potential size (number of people, dollar purchase potential, etc.) and likely growth Key
economic, psychological, and social influences on buying Type of buying situation Nature of
relationship with customers Competitor Analysis Nature of current/likely competition Current
and prospective competitors (or rivals) Current strategies and likely responses to plan
Competitive barriers to overcome and sources of potential competitive advantage Analysis of the
Market Context-External Market Environment Economic environment Technological
environment Political and legal environment Cultural and social environment Key Factors from
Situation Analysis W.O.T: Strengths, weaknesses, opportunities, and threats from situation
analysis Specific objectives to be achieved with the marketing strategy How will marketing mix
be differentiated from the competition? How will the market offering be positioned? Positioning
statement Marketing Strategy Overview of the Marketing Strategy General direction for the
marketing strategy Description of how the four Ps fit together Target Market(s) Summary of
characteristics of the target market(s) to be approached Product class (type of consumer or
business product) Current product life-cycle stage New-product development requirements
(people, dollars, time, etc.) Product liability safety, and social responsibility considerations
Specification of core physical good or service Product Features, quality etc. Supporting
service(s) needed Warranty (what is covered, timing, who will support, etc Branding
(manufacturer versus dealer, family versus individual, etc)
Solution
Apple iPhone Marketing Plan
•Executive Summery
•Situational Analysis
•SWOT Analysis
•Marketing Objectives
•Marketing Strategy
•Implementation
•Budget
•Control
Executive Summery
The Past– Steve Jobs, Steve Wozniak and Ronald Wayne established Appleon April 1, 1976 in
order to sell the Apple 1 Computer Kit that was handbuilt by Steve Wozniak. The Apple 1 was
sold as a motherboard (with CPU,RAM a.
If we were to implement a better budgeting process in which lower-le.pdf
If we were to implement a better budgeting process in which lower-level managers and/or
employees are involved, who would be the various players, specifically?
What would be their individual and collective roles and responsibilities?
Solution
If the lower-level managers are involved in implementing a better budgeting process, the players
in the budgeting process would be top management, middle management and lower level
management. They would be the people who look after the budgeting process.
The collective role of the people involved is to acheive the budgeted results to the organisation as
a whole and their respective roles. There responsibilities would include individually, they need to
be aware of the results of their respective departments individually..
I can change a link layer protocol and the application layer protoco.pdf
I can change a link layer protocol and the application layer protocol does not need to know or be
made aware of the change. Explain why?
Solution
sol:
yes its need not to be know if there is any change because
the internet uses one of the standard method for communication that is nothing but tcp/ip
,generally higher layers
does not identify the lower layer.when we see the lower layer from higher layer there are
abstracted,so they dont know what is internally happening in the lower layers.
Could a poly-Ala sequence form a beta sheet Why or why notSolu.pdf
Could a poly-Ala sequence form a beta sheet? Why or why not?
Solution
Protein secondary structures are formed due to hydrogen interactions between amino acids
present in linear polypeptide. Alfa helix and beta sheets are the typical secondary structures. Alfa
helix formed due to intra chain hydrogen interactions and beta chains formed due to inter chain
hydrogen interactions. In both cases hydrogen interactions play crucial role. The hydrogen
interactions formed between partial charged atoms. Partial charged atoms exists due to dipole
moment. Dipole moment observed in molecule with different electro negative atoms. Here
alanine contain, a non ionic, hydrophobic side chain, that is Alanine. So it cannot form hydrogen
interactions. So cannot form beta sheets.
Cell Physiology Discuss the ramifications of the Human Genome Proje.pdf
Cell Physiology: Discuss the ramifications of the Human Genome Project and the evolving
definition of a gene based on expanding knowledge of the complexities of genome transcription.
Solution
Human genome project: Human genome project initially stated to examine the single nucleotide
polymorphism among the human populations. The main goals are the type of genetic methods
used predominantly to produce genetic maps of all human chromosomes finally sequencing the
approximately three billion nitrogen base pairs within human DNA. Human genome project is
useful to analyze the existence of genetic variants in dissimilar progeny.
According to human genome project, NHGRI, \"trait is defined as the phenotypic feature i.e.
mainly representing a specific encoding character of a gene in any organism. The ramifications
of the Human Genome Project involve the gene transcription based on the \"genome annotation,
genome evolution followed by segmental duplication\". These genomic alterations are going to
trigger evolutionary aspects of changes in gene expression so that genomics and proteomics have
meticulous importance in explaining complexities of genome transcription. This is because a
majority of human genome is transposbale elements and only 1.5% of genome is going to code
for proteins
So that Genome is defined as the total genetic content of the organism including organelles,
inside the chromosomes that represent that organism\'s genetic character. Intron is the noncoding
nucleotide codons (A, G, C, U) on messenger RNA for any amino acid. It is produced through
gene expression or transcription in the presence of RNA polymerase. It is removed through
splicing by spliceosomes. Exon is the coding nucletode codons (A, G, C, U) on messenger RNA
for any amino acid. It is produced through gene expression or transcription in the presence of
RNA polymerase. It is not removed through splicing by spliceosomes.
Haplotype: Accoridng to human genome project, NHGRI, a haplotype is the inheriting genetic
material of DNA sequences (combinations of alleles) i.e. situated on the adjacent location of
chromosomes of the organism.
Epigenetics- inheritance phenomenon of human genome project: This explained by the following
examples including genomic imprinting by which certain genes expressed in a parent-of-origin-
specific manner. If the allele inherited from the father imprinted, thereby gene silenced, and only
the allele from the mother expressed. A gene, which needed efficient placental growth to supply
nutrients and its expression, is active from maternal allele and inactive of paternal allele.
Proteomics methods are predominantly have been using by the researchers to analyze the
evolving definition of a gene based on expanding knowledge of the complexities of genome
transcription. This is done via the genomic wide sequencing analysis of horizontal gene transfer
of bacterial evolution & intragenomic alterations in eukaryotes finally it is possible to deduce an
evolutiona.
Early on the morning of April 15, 1912, the RMS Titanic collided wit.pdf
Early on the morning of April 15, 1912, the RMS Titanic collided with an iceberg as she made
her maiden voyage from Southampton UK to New York City USA. Although this British
passenger liner was designed with advanced safety features including watertight compartments
with remotely activated watertight doors, she only carried enough lifeboats for 1,178 people due
to outdated maritime safety regulations. The ship\'s hull plates buckled inwards along her
starboard side due to the collision, flooding five of her sixteen watertight compartments. 1,514 of
the 2,224 aboard perished, making this sinking one of the deadliest commercial peacetime
maritime disasters
1.Suppose that 30% of those onboard came onboard at Cherbourg. What is the probability
(correct to three decimal places) that at most 4 of 10 randomly selected people onboard came
onboard at Cherbourg?
Solution
We use binomial probability for this one.
Using a cumulative binomial distribution table or technology, matching
n = number of trials = 10
p = the probability of a success = 0.3
x = the maximum number of successes = 4
Then the cumulative probability is
P(x <= 4 ) = 0.849731667 [ANSWER].
Discuss why ethnoarchaeology is so important in helping us understan.pdf
Discuss why ethnoarchaeology is so important in helping us understand different cultures.
Solution
Study of social organisation and other features helps to understand different cultures, We, all
know that culture is , learned behaviour which has been socially acquired. Culture is the product
of human societies, and man is largely product of his cultural environment. Culture is transmitted
from one generation to another through learning process. Role of culture: It provides
mechanisms and beheaviour which secure for an individual his personal and social survival.
Culture stands for customs, beliefs,laws, religion, arts and other capabilities and skills acquired
by man as a member of society. Cultural factors in health and disease play important role. Every
culture has its own customs, some of which has very high influence on incidence of disease For
example, cancer of lungs from smoking and cirrhosis of liver from drinking are results of abuse
of widely claimed social habits. In India, chewing pan is associated with oral cancer So, that
cultural factors are deeply involved in matters of personal hygiene, nutrition, immunisation,
seeking early medical care , family planning, child rearing, disposal of refuse and excreta,
outlook or health and disease. In short, the whole way of life..
Briefly discuss five issues which crawling insects face and how they .pdf
Briefly discuss five issues which crawling insects face and how they overcome them in the
environment?
Solution
Briefly discuss five issues which crawling insects face and how they overcome them in the
environment?
issues which crawling insects face
how they overcome them In the environment
1
Escaping enemy
Multiple and jointed legs for fast walking Ex: Centipedes and millipedes
2
Maintaining Grip with the surface
Legs have special gripping modified parts with secretions, sometimes they also cause itching or
acts as allergins.
Ex: Head louce in humans or animals
3
Protecting themselves by producing chemicals through special organs like stings etc.,
Some insects like ants secretes formic acid, alkaloids and piperidines, Venoms- scorpions
4
Food availability and its location identification
Modified mouthparts, sucking type, piercing type of mouth parts were developed Ex: Ants,
mites, ticks
Sensory adaptations were seen to identify the food sources ex: Ants
5
Area for habitation, within very small area large number of insects stay
Jointed body segments in order to move into any small areas or holes to escape from enemies Ex;
Cockroaches
6.
Modified digestive or excretory system parts for using the food which is available. Sometimes
water availability will be very less.
Termites have microbes for digestion of cellulose in their guts. Ex: silverfish, Termites etc
The excretory system consists of branching tubules that are extending from the alimentary canal
that absorbs solutes, water, and wastes from the surrounding hemolymph. Most of them have
Malphigian tubule system
7
Body is generally hard in order to escape from enemies- Exoskeleton generally called as shells
It protect animal body In general Chitin forms the exoskeleton in most of the insects
Ex: arthropods including insects,
And also in arachnids such as spiders
issues which crawling insects face
how they overcome them In the environment
1
Escaping enemy
Multiple and jointed legs for fast walking Ex: Centipedes and millipedes
2
Maintaining Grip with the surface
Legs have special gripping modified parts with secretions, sometimes they also cause itching or
acts as allergins.
Ex: Head louce in humans or animals
3
Protecting themselves by producing chemicals through special organs like stings etc.,
Some insects like ants secretes formic acid, alkaloids and piperidines, Venoms- scorpions
4
Food availability and its location identification
Modified mouthparts, sucking type, piercing type of mouth parts were developed Ex: Ants,
mites, ticks
Sensory adaptations were seen to identify the food sources ex: Ants
5
Area for habitation, within very small area large number of insects stay
Jointed body segments in order to move into any small areas or holes to escape from enemies
Ex; Cockroaches
6.
Modified digestive or excretory system parts for using the food which is available. Sometimes
water availability will be very less.
Termites have microbes for digestion of cellulose in their guts. Ex: silverfi.
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YouTube:-
https://youtube.com/@special_education_needs
Teligram :- https://t.me/special_education_needs
Slide Shere :-
https://www.slideshare.net/shabnambano20?utm_campaign=profiletracking&utm_medium=sssite&utm_source=ssslideview
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Objective:
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Mule 4.6 & Java 17 Upgrade | MuleSoft Mysore Meetup #46
Mule 4.6 & Java 17 Upgrade | MuleSoft Mysore Meetup #46
Event Link:-
https://meetups.mulesoft.com/events/details/mulesoft-mysore-presents-exploring-gemini-ai-and-integration-with-mulesoft/
Agenda
● Java 17 Upgrade Overview
● Why and by when do customers need to upgrade to Java 17?
● Is there any immediate impact to upgrading to Mule Runtime 4.6 and beyond?
● Which MuleSoft products are in scope?
For Upcoming Meetups Join Mysore Meetup Group - https://meetups.mulesoft.com/mysore/
YouTube:- youtube.com/@mulesoftmysore
Mysore WhatsApp group:- https://chat.whatsapp.com/EhqtHtCC75vCAX7gaO842N
Speaker:-
Shubham Chaurasia - https://www.linkedin.com/in/shubhamchaurasia1/
Priya Shaw - https://www.linkedin.com/in/priya-shaw
Organizers:-
Shubham Chaurasia - https://www.linkedin.com/in/shubhamchaurasia1/
Giridhar Meka - https://www.linkedin.com/in/giridharmeka
Priya Shaw - https://www.linkedin.com/in/priya-shaw
Shyam Raj Prasad-
https://www.linkedin.com/in/shyam-raj-prasad/
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Problem DefinitionBuild a Date class and a main function to test i.pdf
- 1. Problem Definition Build a Date class and a main function to test it. Specifications Below is the interface for the Date class: it is our "contract" with you: you have to implement everything it describes, and show us that it works with a test harness that puts it through its paces. The comments in the interface below should be sufficient for you to understand the project (use these comments in your Date declaration), without the need of any further documentation. But of course, as always, you can ask us any questions you may have on Piazza. Note: Placing the error messages into the constructors like is not necessarily a good way to handle constructor errors, but until you learn about exceptions in CS 14, it's the best we can do. Private Member Functions The functions declared private above, isLeap, daysPerMonth, name, number, are helper functions - member functions that will never be needed by a user of the class, and so do not belong to the public interface (which is why they are "private"). They are, however, needed by the interface functions (public member functions), which use them to test the validity of arguments and construct valid dates. For example, the constructor that passes in the month as a string will call the number function to assign a value to the unsigned member variable month. isLeap: The rule for whether a year is a leap year is: (year % 4 == 0) implies leap year except (year % 100 == 0) implies NOT leap year except (year % 400 == 0) implies leap year So, for instance, year 2000 is a leap year, but 1900 is NOT a leap year. Years 2004, 2008, 2012, 2016, etc. are all leap years. Years 2005, 2006, 2007, 2009, 2010, etc. are NOT leap years. Output Specifications Read the specifications for the print function carefully. The only cout statements within your Date member functions should be: the "Invalid Date" warnings in the constructors in your two print functions Required Main Function You must use this main function and global function getDate as they are here. You may not change these functions at all. Copy-and-paste these into your main.cpp file and then add the Date class. Solution
- 2. // Date.h #include #include #include #include using namespace std; class Date { private: unsigned day; unsigned month; string monthName; unsigned year; public: Date(); Date(unsigned m, unsigned d, unsigned y); Date(const string &mn, unsigned d, unsigned y); void printNumeric() const; void printAlpha() const; private: bool isLeap(unsigned y) const; unsigned daysPerMonth(unsigned m, unsigned y) const; string name(unsigned m) const; unsigned number(const string &mn) const; }; //Date.cpp #include #include #include #include using namespace std; #include "Date.h" // creates the date January 1st, 2000. Date::Date()
- 3. { day = 1; month = 1; monthName = "January"; year = 2000; } /* parameterized constructor: month number, day, year - e.g. (3, 1, 2010) will construct the date March 1st, 2010 If any of the arguments are invalid (e.g. 15 for month or 32 for day) then the constructor will construct instead a valid Date as close as possible to the arguments provided - e.g. in above example, Date(15, 32, 2010), the Date would be corrected to Dec 31st, 2010. In case of such invalid input, the constructor will issue a console error message: Invalid date values: Date corrected to 12/31/2010. (with a newline at the end). */ Date::Date (unsigned m, unsigned d, unsigned y) { bool invalid = false; //if invalid month input, change to closest month if (m < 1) { m = 1; invalid = true; } else if (m > 12) { m = 12; invalid = true; } //invalid day if (d > daysPerMonth(m, y)) { d = daysPerMonth(m,y);
- 4. invalid = true; } day = d; month = m; monthName = name(m); year = y; if (invalid) { cout << "Invalid date values: Date corrected to "; cout << month << "/" << day << "/" << year << "." << endl; } } /* parameterized constructor: month name, day, year - - e.g. (December, 15, 2012) will construct the date December 15th, 2012 If the constructor is unable to recognize the string argument as a valid month name, then it will issue a console error message: Invalid month name: the Date was set to 1/1/2000. (with a newline at the end). If the day argument is invalid for the given month (but the month name was valid), then the constructor will handle this error in the same manner as the other parameterized constructor. This constructor will recognize both "december" and "December" as month name. */ Date::Date (const string &mName, unsigned d, unsigned y) { //will change to true if invalid day bool invalidDay = false; //will change to true if invalid month bool invalidMonth = false; if (mName == "January" || mName == "january") { month = number("January");
- 5. monthName = "January"; } else if (mName == "February" || mName == "february") { month = number("February"); monthName = "February"; } else if (mName == "March" || mName == "march") { month = number("March"); monthName = "March"; } else if (mName == "April" || mName == "april") { month = number("April"); monthName = "April"; } else if (mName == "May" || mName == "may") { month = number("May"); monthName = "May"; } else if (mName == "June" || mName == "june") { month = number("June"); monthName = "June"; } else if (mName == "July" || mName == "july") { month = number("July"); monthName = "July"; } else if (mName == "August" || mName == "august") { month = number("August"); monthName = "August";
- 6. } else if (mName == "September" || mName == "september") { month = number("September"); monthName = "September"; } else if (mName == "October" || mName == "october") { month = number("October"); monthName = "October"; } else if (mName == "November" || mName == "november") { month = number("November"); monthName = "November"; } else if (mName == "December" || mName == "december") { month = number("December"); monthName = "December"; } //If invalid month name, change date to 1/1/2000 else { day = 1; month = number("January"); monthName = "January"; year = 2000; invalidMonth = true; } //if monthName is correct but day number is invalid, change to closest day if ((!invalidMonth) && (d > daysPerMonth(month, y))) { day = daysPerMonth(month,y); invalidDay = true;
- 7. } //outputs message if input was invalid if (invalidDay || invalidMonth) { cout << "Invalid date values: Date corrected to "; if (invalidDay) { year = y; } cout << month << "/" << day << "/" << year << "." << endl; } else { day = d; year = y; } } /* Outputs to the console (cout) a Date exactly in the format "3/1/2012". Does not output a newline at the end. */ void Date::printNumeric () const { cout << month << "/" << day << "/" << year; } /* Outputs to the console (cout) a Date exactly in the format "March 1, 2012". The first letter of the month name is upper case, and the month name is printed in full - January, not Jan, jan, or january. Does not output a newline at the end. */ void Date::printAlpha () const { cout << monthName << " " << day << ", " << year; } /* Returns true if the year passed in is a leap year, otherwise returns false.
- 8. */ bool Date::isLeap(unsigned y) const { //implies leap year if (y % 4 == 0) { //does not imply leap year if (y % 100 == 0) { //unless its a multiple of 400 if (y % 400 == 0) { return true; } return false; } return true; } return false; } /* Returns number of days allowed in a given month - e.g. daysPerMonth(9, 2000) returns 30. Calculates February's days for leap and non--leap years, thus, the reason year is also a parameter. */ unsigned Date::daysPerMonth(unsigned m, unsigned y) const { if (m == 1 || m == 3 || m == 5 || m == 7 || m == 8 || m == 10 || m == 12) { return 31; } else if (m == 4 || m == 6 || m == 9 || m == 11) { return 30; } else if (isLeap(y))
- 9. { return 29; } return 28; } /* Returns the name of a given month - e.g. name(12) returns the string "December" */ string Date::name(unsigned m) const { if (m <= 1) { return "January"; } else if (m >= 12) { return "December"; } if (m == 2) { return "February"; } else if (m == 3) { return "March"; } else if (m == 4) { return "April"; } else if (m == 5) { return "May"; } else if (m == 6)
- 10. { return "June"; } else if (m == 7) { return "July"; } else if (m == 8) { return "August"; } else if (m == 9) { return "September"; } else if (m == 10) { return "October"; } else if (m == 11) { return "November"; } } /* Returns the number of a given named month - e.g. number("March") returns 3 */ unsigned Date::number(const string &mName) const { if (mName == "January") { return 1; } else if (mName == "February") {
- 11. return 2; } else if (mName == "March") { return 3; } else if (mName == "April") { return 4; } else if (mName == "May") { return 5; } else if (mName == "June") { return 6; } else if (mName == "July") { return 7; } else if (mName == "August") { return 8; } else if (mName == "September") { return 9; } else if (mName == "October") { return 10; } else if (mName == "November") {
- 12. return 11; } else if (mName == "December") { return 12; } } // main.cpp #include #include #include #include #include "Date.h" using namespace std; Date getDate(); int main() { Date testDate; testDate = getDate(); cout << endl; cout << "Numeric: "; testDate.printNumeric(); cout << endl; cout << "Alpha: "; testDate.printAlpha(); cout << endl; return 0; } Date getDate() { int choice; unsigned monthNumber, day, year; string monthName; cout << "Which Date constructor? (Enter 1, 2, or 3)" << endl
- 13. << "1 - Month Number" << endl << "2 - Month Name" << endl << "3 - default" << endl; cin >> choice; cout << endl; if (choice == 1) { cout << "month number? "; cin >> monthNumber; cout << endl; cout << "day? "; cin >> day; cout << endl; cout << "year? "; cin >> year; cout << endl; return Date(monthNumber, day, year); } else if (choice == 2) { cout << "month name? "; cin >> monthName; cout << endl; cout << "day? "; cin >> day; cout << endl; cout << "year? "; cin >> year; cout << endl; return Date(monthName, day, year); } else { return Date(); } } /* output: Which Date constructor? (Enter 1, 2, or 3) 1 - Month Number
- 14. 2 - Month Name 3 - default 3 Numeric: 1/1/2000 Alpha: January 1, 2000 Which Date constructor? (Enter 1, 2, or 3) 1 - Month Number 2 - Month Name 3 - default 2 month name? march day? 32 year? 1994 Invalid date values: Date corrected to 3/31/1994. Numeric: 3/31/1994 Alpha: March 31, 1994 */