The Java Program for the above given isimport java.io.File;impo.pdf
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The Java Program for the above given is: import java.io.File; import java.io.FileNotFoundException; import java.io.IOException; import java.util.Scanner; import java.util.NoSuchElementException; import java.io.PrintWriter; public class InputPrototype { public static void main(String[] args) throws FileNotFoundException { Scanner in = new Scanner(System.in); System.out.print(\"Input File Name: \"); String inputFileName = in.next(); System.out.print(\"Output File Name: \"); String outputFileName = in.next(); File inputFile = new File(inputFileName); Scanner input = new Scanner(inputFile); PrintWriter out = new PrintWriter(outputFileName); double total= 0; while (in.hasNextDouble()) { double value = in.nextDouble(); out.printf(\"%15.2f\ \", value); total = total + value; } out.printf(\"Total: %8.2f\ \", total); in.close(); out.close(); } } } Solution The Java Program for the above given is: import java.io.File; import java.io.FileNotFoundException; import java.io.IOException; import java.util.Scanner; import java.util.NoSuchElementException; import java.io.PrintWriter; public class InputPrototype { public static void main(String[] args) throws FileNotFoundException { Scanner in = new Scanner(System.in); System.out.print(\"Input File Name: \"); String inputFileName = in.next(); System.out.print(\"Output File Name: \"); String outputFileName = in.next(); File inputFile = new File(inputFileName); Scanner input = new Scanner(inputFile); PrintWriter out = new PrintWriter(outputFileName); double total= 0; while (in.hasNextDouble()) { double value = in.nextDouble(); out.printf(\"%15.2f\ \", value); total = total + value; } out.printf(\"Total: %8.2f\ \", total); in.close(); out.close(); } } }.
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![The Java Program for the above given is:
import java.io.File;
import java.io.FileNotFoundException;
import java.io.IOException;
import java.util.Scanner;
import java.util.NoSuchElementException;
import java.io.PrintWriter;
public class InputPrototype
{
public static void main(String[] args)
throws FileNotFoundException
{
Scanner in = new Scanner(System.in);
System.out.print("Input File Name: ");
String inputFileName = in.next();
System.out.print("Output File Name: ");
String outputFileName = in.next();
File inputFile = new File(inputFileName);
Scanner input = new Scanner(inputFile);
PrintWriter out = new PrintWriter(outputFileName);
double total= 0;
while (in.hasNextDouble())
{
double value = in.nextDouble();
out.printf("%15.2f ", value);
total = total + value;
}
out.printf("Total: %8.2f ", total);
in.close();
out.close();
}
}
}
Solution](https://image.slidesharecdn.com/thejavaprogramfortheabovegivenisimportjava-230409001940-6eac8543/85/The-Java-Program-for-the-above-given-isimport-java-io-File-impo-pdf-1-320.jpg)
![The Java Program for the above given is:
import java.io.File;
import java.io.FileNotFoundException;
import java.io.IOException;
import java.util.Scanner;
import java.util.NoSuchElementException;
import java.io.PrintWriter;
public class InputPrototype
{
public static void main(String[] args)
throws FileNotFoundException
{
Scanner in = new Scanner(System.in);
System.out.print("Input File Name: ");
String inputFileName = in.next();
System.out.print("Output File Name: ");
String outputFileName = in.next();
File inputFile = new File(inputFileName);
Scanner input = new Scanner(inputFile);
PrintWriter out = new PrintWriter(outputFileName);
double total= 0;
while (in.hasNextDouble())
{
double value = in.nextDouble();
out.printf("%15.2f ", value);
total = total + value;
}
out.printf("Total: %8.2f ", total);
in.close();
out.close();
}
}
}](data:image/gif;base64,R0lGODlhAQABAIAAAAAAAP///yH5BAEAAAAALAAAAAABAAEAAAIBRAA7)
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Write a java program that allows the user to input the name of a fil.docx
The document describes a Java program that takes in a file containing positive integers, reads the integers, separates the even and odd numbers, and writes them to separate text files. The program prompts the user to enter the file name and number of integers, then uses loops and conditionals to read each integer, check if it is even or odd, and write it to the corresponding file. On completion, it notifies the user that the files were successfully written.
Write a java program that allows the user to input the name of a file.docx
Write a java program that allows the user to input the name of a file that contains positive integers (one integer per line) The number of integers in the file The program should find all odd and even numbers and write the odd numbers to a file \"odd.txt\" and write the even numbers to another file \"even.txt\"
Solution
package expertques;
import java.io.File;
import java.io.PrintWriter;
import java.util.Scanner;
public class evenoddwithfile {
public static void main(String[] args) throws Exception {
File in;//file input
File even = new File(\"f://even.txt\");//creates file even
File odd = new File(\"f://odd.txt\");//creates file odd
Scanner s = new Scanner(System.in);//reads input from console or user
System.out.println(\"Enter the file name with path: \");
in = new File(s.nextLine());
System.out.println(\"Enter the number of integers n the file: \");
int n = s.nextInt();
if (in.exists()) {//checks if input file exist or not
Scanner ins = new Scanner(in);//reads input file
PrintWriter pe = new PrintWriter(even);//write to even file
PrintWriter po = new PrintWriter(odd);//writes to odd file
for (int i = 0; i < n; i++) {//loop till the numbers exists
int x = ins.nextInt();//reads from input file
if (x % 2 == 0) {//checks for even or odd
pe.println(x);//if even writes in even file
} else {
po.println(x);//if odd writes in odd file
}
}
pe.close();//closes even file very important else nothing will be written in the file
po.close();//closes odd file very important else nothing will be written in the file
System.out.println(\"File successfully written with odds and evens\");
} else {
System.out.println(\"Error file not found at specified path\");//if file not exist
}
}
}
output
run:
Enter the file name with path:
f://integers.txt
Enter the number of integers n the file:
8
File successfully written with odds and evens
BUILD SUCCESSFUL (total time: 6 seconds)
.
PAGE 1Input output for a file tutorialStreams and File IOI.docx
PAGE
1Input output for a file tutorialStreams and File I/O
In this tutorial we study the following concepts:
Text File I/O
Techniques for a file: Class File
Binary File I/O
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1 An Overview of stream and file I/O
A stream is an object that either outputs data or inputs data. The Java System.out is a stream object that outputs data on the screen.
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2 Text-File I/O
In this section we learn how to input from a text file and how to output to a text file. We first look at output instructions for a text file.
2.1 Text-File Output with PrintWriter class.
To be able to write to a file we need the Java class PrintWriter. This class is in JDK java.io, therefore we need to import it . Another class that we need is : FileOutputStream. This class also is in java.io. A file output stream is an output stream for writing data to a File.
Example 1. The following program to write an integer, a floating point number, a Boolean value and a string into a text file.
import java. io. PrintWriter;
import java. io. FileNotFoundException;
publicclass Main {
publicstaticvoid main( String[] args) {
PrintWriter out = null;
try {
out = new PrintWriter("out.txt");
} catch( FileNotFoundException e) {
System. out. println("Error: opening the file out.txt");
System. exit( 0);
}
out.println(-12);
out.println(25.5);
out.println(true);
out.println("John Doe.");
//Note that we also can write:
out.println(-12+"\n"+ 25.5 + "\n" + true + "\n" + "John Doe.");
out.close();
} }
The content of the file out.txt would be:
-12
25.5
true
John Doe.
-12
25.5
true
John Doe.
Description:
First we need to declare a variable that refers to and object of class PrintWriter. This class has several constructors. One of them needs the name of the output file on the hard drive (or other storages): out = new PrintWriter("out.txt")
This instruction co.
Hi, I need help with a java programming project. specifically practi.pdf
Given 8 distinct integers from the set {1, 2, . . . , 16}, prove that there exists k such that ai aj = k
has at least 3 distinct solutions (ai , aj ) ?
Solution
For any k (unless the K is zero), at least 3 distinct solutions need not be existing.
Let\'s look at the next most favourable case
If k= 2;
We have to consider the case where the number of solutions of ai-aj=2 are minimum since we
have to show that at least 3 distinct solutions exist (which means even in the most unfavourable
case for a chosen k there will be 3 distinct solutions satisfying the equation ai-aj=2)
If k = 2; we start by taking numbers that are not at all 2 units apart from each other;
So we start with 1 and move 3 units successively ahead
1, 4, 7, 10,13, 16; Now we notice that these 6 numbers are of 3k+1 form; remaining numbers
can be any 2 ranomdly selected say 2 and 11; we get 4-2 = 2 and 13-11=2; Hence only 2 distinct
solutions are possible..
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Write a java program that allows the user to input the name of a file.docx
Write a java program that allows the user to input the name of a file that contains positive integers (one integer per line) The number of integers in the file The program should find all odd and even numbers and write the odd numbers to a file \"odd.txt\" and write the even numbers to another file \"even.txt\"
Solution
package expertques;
import java.io.File;
import java.io.PrintWriter;
import java.util.Scanner;
public class evenoddwithfile {
public static void main(String[] args) throws Exception {
File in;//file input
File even = new File(\"f://even.txt\");//creates file even
File odd = new File(\"f://odd.txt\");//creates file odd
Scanner s = new Scanner(System.in);//reads input from console or user
System.out.println(\"Enter the file name with path: \");
in = new File(s.nextLine());
System.out.println(\"Enter the number of integers n the file: \");
int n = s.nextInt();
if (in.exists()) {//checks if input file exist or not
Scanner ins = new Scanner(in);//reads input file
PrintWriter pe = new PrintWriter(even);//write to even file
PrintWriter po = new PrintWriter(odd);//writes to odd file
for (int i = 0; i < n; i++) {//loop till the numbers exists
int x = ins.nextInt();//reads from input file
if (x % 2 == 0) {//checks for even or odd
pe.println(x);//if even writes in even file
} else {
po.println(x);//if odd writes in odd file
}
}
pe.close();//closes even file very important else nothing will be written in the file
po.close();//closes odd file very important else nothing will be written in the file
System.out.println(\"File successfully written with odds and evens\");
} else {
System.out.println(\"Error file not found at specified path\");//if file not exist
}
}
}
output
run:
Enter the file name with path:
f://integers.txt
Enter the number of integers n the file:
8
File successfully written with odds and evens
BUILD SUCCESSFUL (total time: 6 seconds)
.
PAGE 1Input output for a file tutorialStreams and File IOI.docx
PAGE
1Input output for a file tutorialStreams and File I/O
In this tutorial we study the following concepts:
Text File I/O
Techniques for a file: Class File
Binary File I/O
File Objects and File Name
Note: For question 4 of the midterm you only need to look at the sections 2.2 and 2.3. I do not think you need java class: StringTokenizer. But I just added a small section about it to the end of this tutorial.
So far we learned keyboard/screen I/O. It is time to learn File I/O.
1 An Overview of stream and file I/O
A stream is an object that either outputs data or inputs data. The Java System.out is a stream object that outputs data on the screen.
As our program gets bigger and bigger we may need to enter a lot of data every time we run our program. We also may have a lot of data as the output of the program. It is not practical to enter a sizable input from the keyboard. Therefore, we save the data in a file and let our program to input from this file. Similarly, it is not practical to look at a sizable output. Moreover, we may need to save the output for further study. Therefore, we let our program to output to a file.
We have two kinds of I/O files: Text files and binary files. A text file is readable and understandable by human, but a binary file is not. The size of a text file is bigger than the size of a binary file. We can write arrays without using a loop into a binary file. We can also write objects to a binary file.
2 Text-File I/O
In this section we learn how to input from a text file and how to output to a text file. We first look at output instructions for a text file.
2.1 Text-File Output with PrintWriter class.
To be able to write to a file we need the Java class PrintWriter. This class is in JDK java.io, therefore we need to import it . Another class that we need is : FileOutputStream. This class also is in java.io. A file output stream is an output stream for writing data to a File.
Example 1. The following program to write an integer, a floating point number, a Boolean value and a string into a text file.
import java. io. PrintWriter;
import java. io. FileNotFoundException;
publicclass Main {
publicstaticvoid main( String[] args) {
PrintWriter out = null;
try {
out = new PrintWriter("out.txt");
} catch( FileNotFoundException e) {
System. out. println("Error: opening the file out.txt");
System. exit( 0);
}
out.println(-12);
out.println(25.5);
out.println(true);
out.println("John Doe.");
//Note that we also can write:
out.println(-12+"\n"+ 25.5 + "\n" + true + "\n" + "John Doe.");
out.close();
} }
The content of the file out.txt would be:
-12
25.5
true
John Doe.
-12
25.5
true
John Doe.
Description:
First we need to declare a variable that refers to and object of class PrintWriter. This class has several constructors. One of them needs the name of the output file on the hard drive (or other storages): out = new PrintWriter("out.txt")
This instruction co.
Hi, I need help with a java programming project. specifically practi.pdf
Given 8 distinct integers from the set {1, 2, . . . , 16}, prove that there exists k such that ai aj = k
has at least 3 distinct solutions (ai , aj ) ?
Solution
For any k (unless the K is zero), at least 3 distinct solutions need not be existing.
Let\'s look at the next most favourable case
If k= 2;
We have to consider the case where the number of solutions of ai-aj=2 are minimum since we
have to show that at least 3 distinct solutions exist (which means even in the most unfavourable
case for a chosen k there will be 3 distinct solutions satisfying the equation ai-aj=2)
If k = 2; we start by taking numbers that are not at all 2 units apart from each other;
So we start with 1 and move 3 units successively ahead
1, 4, 7, 10,13, 16; Now we notice that these 6 numbers are of 3k+1 form; remaining numbers
can be any 2 ranomdly selected say 2 and 11; we get 4-2 = 2 and 13-11=2; Hence only 2 distinct
solutions are possible..
Understanding java streams
The document discusses Java streams and I/O. It defines streams as abstract representations of input/output devices that are sources or destinations of data. It describes byte and character streams, the core stream classes in java.io, predefined System streams, common stream subclasses, reading/writing files and binary data with byte streams, and reading/writing characters with character streams. It also covers object serialization/deserialization and compressing files with GZIP.
Modify the Total program so that it writes the values in two columns.pdf
Modify the Total program so that it writes the values in two columns, like this:
Previous answer Chegg provided displayed the correct output for in.txt but now, how do I make
the empty blank as shown?
Previous solution provided by Chegg:
import java.io.File;
import java.io.FileNotFoundException;
import java.io.PrintWriter;
import java.util.Scanner;
public class Total {
public static void main(String[] args) throws FileNotFoundException {
Scanner console = new Scanner(System.in);
System.out.print("Input file: ");
String inputFileName = console.next();
String outputFileName = "out.txt";
File inputFile = new File(inputFileName);
Scanner in = new Scanner(inputFile);
PrintWriter out = new PrintWriter(outputFileName);
double total = 0;
while (in.hasNext()) {
double value1 = in.hasNextDouble() ? in.nextDouble() : 0;
double value2 = in.hasNextDouble() ? in.nextDouble() : 0;
out.printf("%8.2f %7.2f%n", value1, value2);
total += (value1 + value2);
}
out.printf("Total: %10.2f%n", total);
in.close();
out.close();
}
} Your file content Your program produced no output Expected file content
1: Compare output Output differs. See highlights below. Special character legend 2: Compare
output A Input in2.txt.
import java.io.BufferedReader;import java.io.File;import java.io.pdf
import java.io.BufferedReader;
import java.io.File;
import java.io.FileReader;
import java.io.IOException;
import java.util.ArrayList;
import java.util.List;
import java.util.Scanner;
public class FileEdit {
public static List lines = new ArrayList();
public static void main(String[] args) throws IOException
{
Scanner s = new Scanner(System.in);
System.out.print("EDIT: ");
String fileName = s.nextLine();
BufferedReader br = new BufferedReader(new FileReader(new File(fileName)));
String str="";
int count=0;
while((str=br.readLine())!=null)
{
lines.add(count, str);
System.out.println((++count) + "> "+ str);
}
String text="";
String command = "";
System.out.print((++count)+"> ");
command=s.nextLine();
String commandArr[];
int flag;
while(!command.equals("E"))
{
flag=0;
commandArr = command.split("\\s");
if(commandArr[0].equals("I")) //Insertion
{
while(true)
{
if(commandArr.length==1&&flag!=2)
{
System.out.print((count)+"> ");
text = s.nextLine();
insertLine(text,count-1);
}
else if(commandArr.length==2&&flag!=2)
{
System.out.print((commandArr[1])+"> ");
count=Integer.parseInt(commandArr[1]);
text = s.nextLine();
insertLine(text,Integer.parseInt(commandArr[1])-1);
}
else
insertLine(text,count-1);
System.out.print((++count)+"> ");
command=s.nextLine();
commandArr = command.split("\\s");
if(commandArr[0].equals("I")||commandArr[0].equals("L")||commandArr[0].equals("D")||comm
andArr[0].equals("E"))
{
flag=1;
break;
}
else
{
flag=2;
text=command;
}
}
}
else if(commandArr[0].equals("L")) //Listing
{
printList();
}
else if(commandArr[0].equals("D")) //Deletion
{
if(commandArr.length==1)
{
deleteLine(count-1,count-1);
}
else if(commandArr.length==2)
{
deleteLine(Integer.parseInt(commandArr[1])-1,Integer.parseInt(commandArr[1])-1);
}
else if(commandArr.length==3)
{
deleteLine(Integer.parseInt(commandArr[1])-1,Integer.parseInt(commandArr[2])-1);
}
count=lines.size();
}
if(flag!=1)
{
System.out.print((count++)+"> ");
command = s.nextLine();
}
}
}
public static void insertLine(String text,int count)
{
if (count >= lines.size()) {
for(int i = lines.size(); i <= count; i++) {
lines.add(null);
}
}
lines.set(count,text);
}
public static void deleteLine(int start,int end)
{
for(int i=start;i<=end;i++)
lines.remove(i);
}
public static void printList()
{
for(int i=0;i "+lines.get(i));
}
}
Here is my output:
EDIT: textin.txt
1> The First Line
1>
2> And another line
3> I 3
3> The second line
4> One more line
5> L
1> null
2> null
3> The second line
4> One more line
5> D 2
3> L
1> null
2> The second line
3> One more line
4> E
How can I keep it from printing "null" after ">" and it just be blank. I need the output to look like
this:
1> The first line
2>
3> And another line
4> I 3
3> The second line
4> One more line
5> L
1> The first line
2>
3> The second line
4> One more line
5> And another line
5> D 2
4> L
1> The first line
2> The second line
3> One more line
4> And another line
5> E.
JAVA Q2- Write a program that reads strings from the user and writes t.docx
JAVA
Q2: Write a program that reads strings from the user and writes them to an output file called userStrings.txt. Stop processing when the user enters the string \"DONE\". Do not write the sentinel string (\"DONE\") to the output file. [15 points] Hint: For a complete example of file IO, please see the attached TestData program.
Sample Output:
A string.
Another string.
Yet more text...
DONE
Solution
Answer
import java.io.File;
import java.io.FileOutputStream;
import java.io.IOException;
import java.util.Scanner;
class WriteStringFile {
public static void main(String[] args) {
File file = new File(\"D:/userFile.txt\");
String data;
Scanner input = new Scanner(System.in);
data = input.readLine();
try (FileOutputStream fop = new FileOutputStream(file)) {
// if file doesn\'t exists, then create it
if (!file.exists()) {
file.createNewFile();
}
if (data != \"DONE\") {
// get the content in bytes
byte[] contentInBytes = content.getBytes();
fop.write(contentInBytes);
fop.flush();
fop.close();
} else {
System.out.println(\"User Enter the value Done.. So unable to write into file\");
}
} catch (IOException e) {
e.printStackTrace();
}
}
}
.
Here is my code for a linefile editor import java.io.BufferedRea.pdf
Here is my code for a line/file editor:
import java.io.BufferedReader;
import java.io.File;
import java.io.FileReader;
import java.io.IOException;
import java.util.ArrayList;
import java.util.List;
import java.util.Scanner;
public class FileEdit {
public static List lines = new ArrayList();
public static void main(String[] args) throws IOException
{
Scanner s = new Scanner(System.in);
System.out.print("EDIT: ");
String fileName = s.nextLine();
BufferedReader br = new BufferedReader(new FileReader(new File(fileName)));
String str="";
int count=0;
while((str=br.readLine())!=null)
{
lines.add(count, str);
System.out.println((++count) + "> "+ str);
}
String text="";
String command = "";
System.out.print((++count)+"> ");
command=s.nextLine();
String commandArr[];
int flag;
while(!command.equals("E"))
{
flag=0;
commandArr = command.split("\\s");
if(commandArr[0].equals("I")) //Insertion
{
while(true)
{
if(commandArr.length==1&&flag!=2)
{
System.out.print((count)+"> ");
text = s.nextLine();
insertLine(text,count-1);
}
else if(commandArr.length==2&&flag!=2)
{
System.out.print((commandArr[1])+"> ");
count=Integer.parseInt(commandArr[1]);
text = s.nextLine();
insertLine(text,Integer.parseInt(commandArr[1])-1);
}
else
insertLine(text,count-1);
System.out.print((++count)+"> ");
command=s.nextLine();
commandArr = command.split("\\s");
if(commandArr[0].equals("I")||commandArr[0].equals("L")||commandArr[0].equals("D")||comm
andArr[0].equals("E"))
{
flag=1;
break;
}
else
{
flag=2;
text=command;
}
}
}
else if(commandArr[0].equals("L")) //Listing
{
printList();
}
else if(commandArr[0].equals("D")) //Deletion
{
if(commandArr.length==1)
{
deleteLine(count-1,count-1);
}
else if(commandArr.length==2)
{
deleteLine(Integer.parseInt(commandArr[1])-1,Integer.parseInt(commandArr[1])-1);
}
else if(commandArr.length==3)
{
deleteLine(Integer.parseInt(commandArr[1])-1,Integer.parseInt(commandArr[2])-1);
}
count=lines.size();
}
if(flag!=1)
{
System.out.print((count++)+"> ");
command = s.nextLine();
}
}
}
public static void insertLine(String text,int count)
{
if (count >= lines.size()) {
for(int i = lines.size(); i <= count; i++) {
lines.add(null);
}
}
lines.set(count,text);
}
public static void deleteLine(int start,int end)
{
for(int i=start;i<=end;i++)
lines.remove(i);
}
public static void printList()
{
for(int i=0;i "+lines.get(i));
}
}
Here is my output:
EDIT: textin.txt
1> The First Line
1>
2> And another line
3> I 3
3> The second line
4> One more line
5> L
1> null
2> null
3> The second line
4> One more line
5> D 2
3> L
1> null
2> The second line
3> One more line
4> E
How can I keep it from printing "null" after ">" and it just be blank. I need the output to look like
this:
1> The first line
2>
3> And another line
4> I 3
3> The second line
4> One more line
5> L
1> The first line
2>
3> The second line
4> One more line
5> And another line
5> D 2
4> L
1> The first line
2> The second line
3> One more line
4> And another line
5> E.
Below is my code for a line editor import java.io.BufferedReader;.pdf
Below is my code for a line editor:
import java.io.BufferedReader;
import java.io.File;
import java.io.FileReader;
import java.io.IOException;
import java.util.ArrayList;
import java.util.List;
import java.util.Scanner;
public class Main {
public static List lines = new ArrayList();
public static void main(String[] args) throws IOException
{
Scanner s = new Scanner(System.in);
System.out.print("Input file: ");
String fileName = s.nextLine();
BufferedReader br = new BufferedReader(new FileReader(new File(fileName)));
String str="";
int count=0;
while((str=br.readLine())!=null)
{
lines.add(count, str);
System.out.println((++count) + "> "+ str);
}
String text="";
String command = "";
System.out.print((++count)+"> ");
command=s.nextLine();
String commandArr[];
int flag;
while(!command.equals("E"))
{
flag=0;
commandArr = command.split("\\s");
if(commandArr[0].equals("I")) //Insertion
{
while(true)
{
if(commandArr.length==1&&flag!=2)
{
System.out.print((count)+"> ");
text = s.nextLine();
insertLine(text,count-1);
}
else if(commandArr.length==2&&flag!=2)
{
System.out.print((commandArr[1])+"> ");
count=Integer.parseInt(commandArr[1]);
text = s.nextLine();
insertLine(text,Integer.parseInt(commandArr[1])-1);
}
else
insertLine(text,count-1);
System.out.print((++count)+"> ");
command=s.nextLine();
commandArr = command.split("\\s");
if(commandArr[0].equals("I")||commandArr[0].equals("L")||commandArr[0].equals("D")||comm
andArr[0].equals("E"))
{
flag=1;
break;
}
else
{
flag=2;
text=command;
}
}
}
else if(commandArr[0].equals("L")) //Listing
{
printList();
}
else if(commandArr[0].equals("D")) //Deletion
{
if(commandArr.length==1)
{
deleteLine(count-1,count-1);
}
else if(commandArr.length==2)
{
deleteLine(Integer.parseInt(commandArr[1])-1,Integer.parseInt(commandArr[1])-1);
}
else if(commandArr.length==3)
{
deleteLine(Integer.parseInt(commandArr[1])-1,Integer.parseInt(commandArr[2])-1);
}
count=lines.size();
}
if(flag!=1)
{
System.out.print((count++)+"> ");
command = s.nextLine();
}
}
}
public static void insertLine(String text,int count)
{
lines.add(count,text);
}
public static void deleteLine(int start,int end)
{
for(int i=start;i<=end;i++)
lines.remove(i);
}
public static void printList()
{
for(int i=0;i "+lines.get(i));
}
}
The output should look like this:
EDIT: textin.txt (user inputs textin.txt)
1> The first line
2>
3> And another line
4> I 3
3> The second line
4> One more line
5> L
1> The first line
2>
3> The second line
4> One more line
5> And another line //This is now line 5; not 3;
5> D 2 // line 5, since L was issued from line 5;
4> L //line 4, since one line was deleted;
1> The first line
2> The second line
3> One more line
4> And another line
5> E
Here is a screenshot of the errors I am getting and need help fixing them: FileEdit [Java
Application] /Users/caseysharp/.p2/pool/plugins/org.eclipse.justj.openjdk.hotspot.jre.full.mac
EDIT: textin.txt 1> The First line 1> 2> And another line 3> I 3 3> The second line Exception in
thread "main" java..
DisplayBook.java import java.io.File; import java.io.FileInput.pdf
DisplayBook.java
import java.io.File;
import java.io.FileInputStream;
import java.io.FileOutputStream;
import java.io.IOException;
import java.util.Scanner;
public class DisplayBook {
public static void main(String[] args) throws IOException {
// TODO Auto-generated method stub
Scanner scan = new Scanner(System.in);
File file = new File(\"D:\\\\books.txt\");
if(file.exists()){
System.out.println(\"File exists\");
FileInputStream fis = new FileInputStream(file);
int content;
while ((content = fis.read()) != -1) {
System.out.print((char) content);
}
fis.close();
}
else{
System.out.println(\"File does not exist\");
String titles = \"\";
FileOutputStream fop = new FileOutputStream(file);
while(true){
System.out.println(\"Enter Book Title (none to exit):\");
String title = scan.nextLine();
if(title.equalsIgnoreCase(\"none\")){
break;
}
else{
titles = titles + title +\"\ \";
}
}
byte[] contentInBytes = titles.getBytes();
fop.write(contentInBytes);
fop.flush();
fop.close();
}
}
}
Output:
File does not exist
Enter Book Title (none to exit):
Physics
Enter Book Title (none to exit):
Chemistry
Enter Book Title (none to exit):
Science
Enter Book Title (none to exit):
Maths
Enter Book Title (none to exit):
none
--------------------------------------
File exists
Physics
Chemistry
Science
Maths
Solution
DisplayBook.java
import java.io.File;
import java.io.FileInputStream;
import java.io.FileOutputStream;
import java.io.IOException;
import java.util.Scanner;
public class DisplayBook {
public static void main(String[] args) throws IOException {
// TODO Auto-generated method stub
Scanner scan = new Scanner(System.in);
File file = new File(\"D:\\\\books.txt\");
if(file.exists()){
System.out.println(\"File exists\");
FileInputStream fis = new FileInputStream(file);
int content;
while ((content = fis.read()) != -1) {
System.out.print((char) content);
}
fis.close();
}
else{
System.out.println(\"File does not exist\");
String titles = \"\";
FileOutputStream fop = new FileOutputStream(file);
while(true){
System.out.println(\"Enter Book Title (none to exit):\");
String title = scan.nextLine();
if(title.equalsIgnoreCase(\"none\")){
break;
}
else{
titles = titles + title +\"\ \";
}
}
byte[] contentInBytes = titles.getBytes();
fop.write(contentInBytes);
fop.flush();
fop.close();
}
}
}
Output:
File does not exist
Enter Book Title (none to exit):
Physics
Enter Book Title (none to exit):
Chemistry
Enter Book Title (none to exit):
Science
Enter Book Title (none to exit):
Maths
Enter Book Title (none to exit):
none
--------------------------------------
File exists
Physics
Chemistry
Science
Maths.
Write a Java program that opens the file, reads all the Grade from t.pdf
The Java program opens a file, reads grades within it, and uses a method to calculate and display:
1) The grades with their letter values (A, B, C, etc.)
2) The sum of all grades in the file
3) The average of all grades in the file
Factors.javaimport java.io.; import java.util.Scanner; class .pdf
Factors.java
import java.io.*;
import java.util.Scanner;
class Factors
{
public static void main(String args[])
{
int a,i;
Scanner in = new Scanner(System.in);
System.out.println(\"Enter an integer\");
a = in.nextInt();
System.out.println(\"You entered integer \"+a);
System.out.print(\"\ \");
System.out.print(\"The factors are : \");
for(i=1;i<=a/2;i++)
{
if(a%i==0)
System.out.print(i+\",\");
}
System.out.print(a);
}
}
Output:
Enter an integer
44
You entered integer 44
The factors are : 1,2,4,11,22,44
Solution
Factors.java
import java.io.*;
import java.util.Scanner;
class Factors
{
public static void main(String args[])
{
int a,i;
Scanner in = new Scanner(System.in);
System.out.println(\"Enter an integer\");
a = in.nextInt();
System.out.println(\"You entered integer \"+a);
System.out.print(\"\ \");
System.out.print(\"The factors are : \");
for(i=1;i<=a/2;i++)
{
if(a%i==0)
System.out.print(i+\",\");
}
System.out.print(a);
}
}
Output:
Enter an integer
44
You entered integer 44
The factors are : 1,2,4,11,22,44.
I can not get my code to comply it gets mad that I have run declare mo.docx
I can not get my code to comply it gets mad that I have run declare more than once. Please help me figure out how to get this to work
package projectOne;
import java.io.File;
import java.io.FileNotFoundException;
import java.util.ArrayList;
import java.util.Scanner;
import java.util.*;
import java.io.*;
public class Main
{ public static void main(String[] pArgs){
Main mainObject = new Main();
mainObject.run();
} private void run() {
public FileReader readFile() throws FileNotFoundException {
FileReader file = new FileReader(\"p01-in.txt\");
return file;
}
public ArrayList<Integer> getData(FileReader file) throws FileNotFoundException {
Scanner data_read = new Scanner(new BufferedReader(file));
ArrayList<Integer> data = new ArrayList<Integer>();
while (data_read.hasNext()) {
int item = data_read.nextInt();
data.add(item);
}
return data;
}
public ArrayList<Integer> calcRuns(ArrayList<Integer> data) {
ArrayList<Integer> runs = new ArrayList<Integer>();
Iterator<Integer> i_data = data.iterator();
int run = 0;
int temp = i_data.next();
while (i_data.hasNext()) {
int t1 = i_data.next();
if (temp < t1)
run++;
else {
runs.add(run);
run = 0;
}
temp = t1;
}
i_data = data.iterator();
temp = i_data.next();
run = 0;
while (i_data.hasNext()) {
int t1 = i_data.next();
if (temp > t1)
run++;
else {
runs.add(run);
run = 0;
}
temp = t1;
}
return runs;
}
public void writeOutput(ArrayList<Integer> runs) throws IOException {
PrintWriter out = new PrintWriter(new File(\"p01-runs.txt\"));
ArrayList<Integer> run_count = new ArrayList<Integer>();
int i = 0;
int r_count = Collections.frequency(run_count, i);
while (i < 100) {
run_count.add(r_count);
i++;
r_count = Collections.frequency(runs, i);
}
Iterator<Integer> i_runs = run_count.iterator();
i = 0;
while (i_runs.hasNext()) {
int run = i_runs.next();
String line = \"run_\" + i + \" ,\" + run;
i++;
out.println(line);
}
}
public void run() throws Exception {
FileReader fin = this.readFile();
ArrayList<Integer> data = this.getData(fin);
ArrayList<Integer> runs = this.calcRuns(data);
this.writeOutput(runs);
}
}
Solution
In your program you have to change the syntax errors at method declaration.so your not getting output.
.
Write a program that will count the number of characters- words- and l.docx
Write a program that will count the number of characters, words, and lines in a file. Words are separated by whitespace characters. The file name should be passed as a command-line argument, as shown;
c:\\exercise>java Exercise12_13 Loan.java
File Loan.java has
1919 characters
210 words
71 lines
c:\\exercise>
Solution
import java.io.File;
import java.io.FileNotFoundException;
import java.util.Scanner;
import java.util.StringTokenizer;
public class ass1
{
public static void main(String[] args) throws FileNotFoundException
{
//Â Â Scanner console = new Scanner(System.in);
System.out.println(args[0]);
//System.out.println(args[1]);
//System.out.println(args[2]);
String inputFile=args[0];
File file = new File(inputFile);
Scanner in = new Scanner(file);
int words = 0;
int lines = 0;
int chars = 0;
while(in.hasNextLine()) {
lines++;
String line = in.nextLine();
chars += line.length();
words += new StringTokenizer(line, \" \").countTokens();
}
System.out.println(\"characters are \"+chars);
System.out.println(\"words are \"+words);
System.out.println(\"lines are \"+lines);
}
}
.
Java Program I keep receiving the following error in my code- Can you.pdf
Java Program
I keep receiving the following error in my code. Can you please fix this error and test to make
sure it does not show a error? I've placed bracets but keep receiving the error "reached end of file
while parsing private static void displayStudent()" See below
Error:
sh -c javac -classpath .:target/dependency/* -d . $(find . -type f -name '*.java')
./Main.java:121: error: reached end of file while parsing
private static void displayStudent()
^
1 error
exit status 1
------------------------------------------
Code:
import java.util.Scanner;
import java.io.File;
import java.io.FileWriter;
import java.io.IOException;
public class StudentCourseProgram {
public static void main(String[] args) {
Scanner input = new Scanner(System.in);
int choice;
do {
System.out.println("Please choose an option:");
System.out.println("1. Add student");
System.out.println("2. Add course");
System.out.println("3. Display student");
System.out.println("4. Display course");
System.out.println("5. Exit");
choice = input.nextInt();
switch (choice) {
case 1:
addStudent();
break;
case 2:
addCourse();
break;
case 3:
displayStudent();
break;
case 4:
displayCourse();
break;
case 5:
System.out.println("Exiting program...");
break;
default:
System.out.println("Invalid option. Please try again.");
}
} while (choice != 5);
}
private static void addStudent() {
Scanner input = new Scanner(System.in);
System.out.println("Adding a student...");
System.out.print("Enter student ID: ");
int id = input.nextInt();
System.out.print("Enter student first name: ");
String firstName = input.next();
System.out.print("Enter student last name: ");
String lastName = input.next();
System.out.print("Enter student address: ");
String address = input.next();
System.out.print("Enter student city: ");
String city = input.next();
System.out.print("Enter student state: ");
String state = input.next();
System.out.print("Enter student zip code: ");
int zip = input.nextInt();
try {
FileWriter writer = new FileWriter("student.txt", true);
Scanner reader = new Scanner(new File("student.txt"));
while (reader.hasNextLine()) {
String line = reader.nextLine();
String[] parts = line.split(",");
if (Integer.parseInt(parts[0]) == id) {
System.out.println("Error: Student with that ID already exists.");
return;
}
}
writer.write(id + "," + firstName + "," + lastName + "," + address + "," + city + "," + state + "," +
zip + "\n");
System.out.println("Student added successfully.");
reader.close();
writer.close();
} catch (IOException e) {
System.out.println("Error: " + e.getMessage());
}
}
private static void addCourse() {
Scanner input = new Scanner(System.in);
System.out.println("Adding a course...");
System.out.print("Enter course ID: ");
int id = input.nextInt();
System.out.print("Enter course name: ");
String name = input.next();
System.out.print("Enter course description: ");
String description = input.next();
try {
FileWriter writer = new FileWriter("course.txt", true);
Scanner reader = new Scanner(new Fil.
File Handling.pptx
This document provides an introduction to file handling in Java. It discusses how file handling allows programs to permanently store output data by writing it to files on secondary storage devices like hard disks. It covers key concepts like input and output streams that represent the flow of data into and out of a program. It also discusses how to create, write to, read from, and delete files in Java using classes like File, FileWriter, FileReader and Scanner. Common file methods like getName(), getAbsolutePath(), exists() are also outlined.
54240326 (1)
The document contains code for several Java programs that implement common data structures and client-server applications. Specifically, it includes programs to:
1. Implement a stack using an array and perform push and pop operations.
2. Implement a queue using an array and perform insertion, deletion, and display operations.
3. Implement a singly linked list and perform operations like creation, addition, deletion, and display of nodes.
4. Implement a producer-consumer problem using threads and synchronization to add and remove elements from a shared queue.
54240326 copy
The document contains code for implementing various data structures and algorithms in Java, including a stack, queue, linked list, producer-consumer problem, scrolling text applet, and client-server file transfer program. The stack, queue, and linked list code demonstrates how to create the data structures and perform common operations like push, pop, insert, delete, and display. The producer-consumer code uses threads and synchronization to model the scenario. The scrolling text applet animates text moving across the screen. The client-server code allows a client to request a file from a server, which then sends the file contents to display on the client.
The sample program for above series in JAVA will be like belowimpo.pdf
The sample program for above series in JAVA will be like below
import java.util.Scan;
public class PIVALUE {
public static void main(String[] args) {
// Scan object creation
Scan input = new Scan (System.in);
// ask user for input
System.out.println(\"Enter num.\");
double j = input.nextDouble(); // user entered value of j
double Tot = 0;
for(j=0; j<10000; j++){
if(j%2 == 0) // check remainder of `j/2` is 0
Tot += -1 / ( 2 * j - 1);
else
Tot += 1 / (2 * j - 1);
}
System.out.println(Tot);
}
}
Solution
The sample program for above series in JAVA will be like below
import java.util.Scan;
public class PIVALUE {
public static void main(String[] args) {
// Scan object creation
Scan input = new Scan (System.in);
// ask user for input
System.out.println(\"Enter num.\");
double j = input.nextDouble(); // user entered value of j
double Tot = 0;
for(j=0; j<10000; j++){
if(j%2 == 0) // check remainder of `j/2` is 0
Tot += -1 / ( 2 * j - 1);
else
Tot += 1 / (2 * j - 1);
}
System.out.println(Tot);
}
}.
The sample program for above series in JAVA will be like belowimpo.pdf
The sample program for above series in JAVA will be like below
import java.util.Scan;
public class PIVALUE {
public static void main(String[] args) {
// Scan object creation
Scan input = new Scan (System.in);
// ask user for input
System.out.println(\"Enter num.\");
double j = input.nextDouble(); // user entered value of j
double Tot = 0;
for(j=0; j<10000; j++){
if(j%2 == 0) // check remainder of `j/2` is 0
Tot += -1 / ( 2 * j - 1);
else
Tot += 1 / (2 * j - 1);
}
System.out.println(Tot);
}
}
Solution
The sample program for above series in JAVA will be like below
import java.util.Scan;
public class PIVALUE {
public static void main(String[] args) {
// Scan object creation
Scan input = new Scan (System.in);
// ask user for input
System.out.println(\"Enter num.\");
double j = input.nextDouble(); // user entered value of j
double Tot = 0;
for(j=0; j<10000; j++){
if(j%2 == 0) // check remainder of `j/2` is 0
Tot += -1 / ( 2 * j - 1);
else
Tot += 1 / (2 * j - 1);
}
System.out.println(Tot);
}
}.
2. Write a program which will open an input file and write to an out.pdf
2. Write a program which will open an input file and write to an output file. Use exception
handling (try & Catch to display appropriate error handling).
Solution
herer is code:
import java.io.*;
public class Copytxt{
public static void main (String[] args)throws IOException {
int i;
FileInputStream fin;
FileOutputStream fout;
try {
// open input file
try {
fin = new FileInputStream(\"input.txt\"); // pass the input location file here
}
catch (FileNotFoundException e) { // if file not found
System.out.println(\"Input File Not Found\");
return;
}
// open output file
try {
fout = new FileOutputStream(\"output.txt\"); // pass the output location file here
}
catch (FileNotFoundException e) { // if file not found
System.out.println(\"Error Opening Output File\");
return;
}
}
catch (ArrayIndexOutOfBoundsException e) {
System.out.println(\"Usage: CopyFile From To\");
return;
}
// Copy File
try {
byte[] buffer = new byte[1024];
int length;
/*copying the contents from input stream to
* output stream using read and write methods
*/
while ((length = fin.read(buffer)) > 0){
fout.write(buffer, 0, length);
}
}
catch (IOException e) {
System.out.println(\"File Error\");
}
fin.close();
fout.close();
}
}.
java input & output statements
The document provides examples of Java code for printing "Hello World", getting different types of user input including integer, float, double, string, and character, and printing outputs. It includes code snippets using classes, importing packages like Scanner, and using methods like println, nextInt, nextFloat etc. to get input and display output in Java programs.
ExtraFileIO.pptx
The document discusses file input and output in Java. It covers the File class, opening files for reading using a Scanner, handling exceptions, reading data line-by-line and tokenizing the lines. It also discusses opening files for writing using a PrintWriter and outputting data to files. The key topics are file I/O, exception handling, and using Scanners and PrintWriters to read from and write to files in Java.
Please help me with a UML class diagram for the following code im.pdf
Please help me with a UML class diagram for the following code:
import java.io.BufferedReader;
import java.io.File;
import java.io.FileReader;
import java.io.IOException;
import java.util.ArrayList;
import java.util.List;
import java.util.Scanner;
public class FileEdit {
public static List lines = new ArrayList();
public static void main(String[] args) throws IOException
{
Scanner s = new Scanner(System.in);
System.out.print("EDIT: ");
String fileName = s.nextLine();
BufferedReader br = new BufferedReader(new FileReader(new File(fileName)));
String str="";
int count=0;
while((str=br.readLine())!=null)
{
lines.add(count, str);
System.out.println((++count) + "> "+ str);
}
String text="";
String command = "";
System.out.print((++count)+"> ");
command=s.nextLine();
String commandArr[];
int flag;
while(!command.equals("E"))
{
flag=0;
commandArr = command.split("\\s");
if(commandArr[0].equals("I")) //Insertion
{
while(true)
{
if(commandArr.length==1&&flag!=2)
{
System.out.print((count)+"> ");
text = s.nextLine();
insertLine(text,count-1);
}
else if(commandArr.length==2&&flag!=2)
{
System.out.print((commandArr[1])+"> ");
count=Integer.parseInt(commandArr[1]);
text = s.nextLine();
insertLine(text,Integer.parseInt(commandArr[1])-1);
}
else
insertLine(text,count-1);
System.out.print((++count)+"> ");
command=s.nextLine();
commandArr = command.split("\\s");
if(commandArr[0].equals("I")||commandArr[0].equals("L")||commandArr[0].equals("D")||comm
andArr[0].equals("E"))
{
flag=1;
break;
}
else
{
flag=2;
text=command;
}
}
}
else if(commandArr[0].equals("L")) //Listing
{
printList();
}
else if(commandArr[0].equals("D")) //Deletion
{
if(commandArr.length==1)
{
deleteLine(count-1,count-1);
}
else if(commandArr.length==2)
{
deleteLine(Integer.parseInt(commandArr[1])-1,Integer.parseInt(commandArr[1])-1);
}
else if(commandArr.length==3)
{
deleteLine(Integer.parseInt(commandArr[1])-1,Integer.parseInt(commandArr[2])-1);
}
count=lines.size();
}
if(flag!=1)
{
System.out.print((count++)+"> ");
command = s.nextLine();
}
}
}
public static void insertLine(String text,int count)
{
if (count >= lines.size()) {
for(int i = lines.size(); i <= count; i++) {
lines.add(null);
}
}
lines.set(count,text);
}
public static void deleteLine(int start,int end)
{
for(int i=start;i<=end;i++)
lines.remove(i);
}
public static void printList()
{
for(int i=0;i "+lines.get(i));
}
}.
Step1 First compound has 2 double bonds ; Second .pdf
Step1 First compound has 2 double bonds ; Second compound has 5 double bonds;
Third compound has 1 double bond. Step2 Second molecule requires maximum energy due to 5
double bonds.This is likely to to have UV/ Visible absorption.
Solution
Step1 First compound has 2 double bonds ; Second compound has 5 double bonds;
Third compound has 1 double bond. Step2 Second molecule requires maximum energy due to 5
double bonds.This is likely to to have UV/ Visible absorption..
Molarity = molesvolume9liter) Molarity of NH4Cl.pdf
Molarity = moles/volume9liter) Molarity of NH4Cl = 4.50 mol /2.00L=2.25 M
Solution
Molarity = moles/volume9liter) Molarity of NH4Cl = 4.50 mol /2.00L=2.25 M.
More Related Content
Similar to The Java Program for the above given isimport java.io.File;impo.pdf
Modify the Total program so that it writes the values in two columns.pdf
Modify the Total program so that it writes the values in two columns, like this:
Previous answer Chegg provided displayed the correct output for in.txt but now, how do I make
the empty blank as shown?
Previous solution provided by Chegg:
import java.io.File;
import java.io.FileNotFoundException;
import java.io.PrintWriter;
import java.util.Scanner;
public class Total {
public static void main(String[] args) throws FileNotFoundException {
Scanner console = new Scanner(System.in);
System.out.print("Input file: ");
String inputFileName = console.next();
String outputFileName = "out.txt";
File inputFile = new File(inputFileName);
Scanner in = new Scanner(inputFile);
PrintWriter out = new PrintWriter(outputFileName);
double total = 0;
while (in.hasNext()) {
double value1 = in.hasNextDouble() ? in.nextDouble() : 0;
double value2 = in.hasNextDouble() ? in.nextDouble() : 0;
out.printf("%8.2f %7.2f%n", value1, value2);
total += (value1 + value2);
}
out.printf("Total: %10.2f%n", total);
in.close();
out.close();
}
} Your file content Your program produced no output Expected file content
1: Compare output Output differs. See highlights below. Special character legend 2: Compare
output A Input in2.txt.
import java.io.BufferedReader;import java.io.File;import java.io.pdf
import java.io.BufferedReader;
import java.io.File;
import java.io.FileReader;
import java.io.IOException;
import java.util.ArrayList;
import java.util.List;
import java.util.Scanner;
public class FileEdit {
public static List lines = new ArrayList();
public static void main(String[] args) throws IOException
{
Scanner s = new Scanner(System.in);
System.out.print("EDIT: ");
String fileName = s.nextLine();
BufferedReader br = new BufferedReader(new FileReader(new File(fileName)));
String str="";
int count=0;
while((str=br.readLine())!=null)
{
lines.add(count, str);
System.out.println((++count) + "> "+ str);
}
String text="";
String command = "";
System.out.print((++count)+"> ");
command=s.nextLine();
String commandArr[];
int flag;
while(!command.equals("E"))
{
flag=0;
commandArr = command.split("\\s");
if(commandArr[0].equals("I")) //Insertion
{
while(true)
{
if(commandArr.length==1&&flag!=2)
{
System.out.print((count)+"> ");
text = s.nextLine();
insertLine(text,count-1);
}
else if(commandArr.length==2&&flag!=2)
{
System.out.print((commandArr[1])+"> ");
count=Integer.parseInt(commandArr[1]);
text = s.nextLine();
insertLine(text,Integer.parseInt(commandArr[1])-1);
}
else
insertLine(text,count-1);
System.out.print((++count)+"> ");
command=s.nextLine();
commandArr = command.split("\\s");
if(commandArr[0].equals("I")||commandArr[0].equals("L")||commandArr[0].equals("D")||comm
andArr[0].equals("E"))
{
flag=1;
break;
}
else
{
flag=2;
text=command;
}
}
}
else if(commandArr[0].equals("L")) //Listing
{
printList();
}
else if(commandArr[0].equals("D")) //Deletion
{
if(commandArr.length==1)
{
deleteLine(count-1,count-1);
}
else if(commandArr.length==2)
{
deleteLine(Integer.parseInt(commandArr[1])-1,Integer.parseInt(commandArr[1])-1);
}
else if(commandArr.length==3)
{
deleteLine(Integer.parseInt(commandArr[1])-1,Integer.parseInt(commandArr[2])-1);
}
count=lines.size();
}
if(flag!=1)
{
System.out.print((count++)+"> ");
command = s.nextLine();
}
}
}
public static void insertLine(String text,int count)
{
if (count >= lines.size()) {
for(int i = lines.size(); i <= count; i++) {
lines.add(null);
}
}
lines.set(count,text);
}
public static void deleteLine(int start,int end)
{
for(int i=start;i<=end;i++)
lines.remove(i);
}
public static void printList()
{
for(int i=0;i "+lines.get(i));
}
}
Here is my output:
EDIT: textin.txt
1> The First Line
1>
2> And another line
3> I 3
3> The second line
4> One more line
5> L
1> null
2> null
3> The second line
4> One more line
5> D 2
3> L
1> null
2> The second line
3> One more line
4> E
How can I keep it from printing "null" after ">" and it just be blank. I need the output to look like
this:
1> The first line
2>
3> And another line
4> I 3
3> The second line
4> One more line
5> L
1> The first line
2>
3> The second line
4> One more line
5> And another line
5> D 2
4> L
1> The first line
2> The second line
3> One more line
4> And another line
5> E.
JAVA Q2- Write a program that reads strings from the user and writes t.docx
JAVA
Q2: Write a program that reads strings from the user and writes them to an output file called userStrings.txt. Stop processing when the user enters the string \"DONE\". Do not write the sentinel string (\"DONE\") to the output file. [15 points] Hint: For a complete example of file IO, please see the attached TestData program.
Sample Output:
A string.
Another string.
Yet more text...
DONE
Solution
Answer
import java.io.File;
import java.io.FileOutputStream;
import java.io.IOException;
import java.util.Scanner;
class WriteStringFile {
public static void main(String[] args) {
File file = new File(\"D:/userFile.txt\");
String data;
Scanner input = new Scanner(System.in);
data = input.readLine();
try (FileOutputStream fop = new FileOutputStream(file)) {
// if file doesn\'t exists, then create it
if (!file.exists()) {
file.createNewFile();
}
if (data != \"DONE\") {
// get the content in bytes
byte[] contentInBytes = content.getBytes();
fop.write(contentInBytes);
fop.flush();
fop.close();
} else {
System.out.println(\"User Enter the value Done.. So unable to write into file\");
}
} catch (IOException e) {
e.printStackTrace();
}
}
}
.
Here is my code for a linefile editor import java.io.BufferedRea.pdf
Here is my code for a line/file editor:
import java.io.BufferedReader;
import java.io.File;
import java.io.FileReader;
import java.io.IOException;
import java.util.ArrayList;
import java.util.List;
import java.util.Scanner;
public class FileEdit {
public static List lines = new ArrayList();
public static void main(String[] args) throws IOException
{
Scanner s = new Scanner(System.in);
System.out.print("EDIT: ");
String fileName = s.nextLine();
BufferedReader br = new BufferedReader(new FileReader(new File(fileName)));
String str="";
int count=0;
while((str=br.readLine())!=null)
{
lines.add(count, str);
System.out.println((++count) + "> "+ str);
}
String text="";
String command = "";
System.out.print((++count)+"> ");
command=s.nextLine();
String commandArr[];
int flag;
while(!command.equals("E"))
{
flag=0;
commandArr = command.split("\\s");
if(commandArr[0].equals("I")) //Insertion
{
while(true)
{
if(commandArr.length==1&&flag!=2)
{
System.out.print((count)+"> ");
text = s.nextLine();
insertLine(text,count-1);
}
else if(commandArr.length==2&&flag!=2)
{
System.out.print((commandArr[1])+"> ");
count=Integer.parseInt(commandArr[1]);
text = s.nextLine();
insertLine(text,Integer.parseInt(commandArr[1])-1);
}
else
insertLine(text,count-1);
System.out.print((++count)+"> ");
command=s.nextLine();
commandArr = command.split("\\s");
if(commandArr[0].equals("I")||commandArr[0].equals("L")||commandArr[0].equals("D")||comm
andArr[0].equals("E"))
{
flag=1;
break;
}
else
{
flag=2;
text=command;
}
}
}
else if(commandArr[0].equals("L")) //Listing
{
printList();
}
else if(commandArr[0].equals("D")) //Deletion
{
if(commandArr.length==1)
{
deleteLine(count-1,count-1);
}
else if(commandArr.length==2)
{
deleteLine(Integer.parseInt(commandArr[1])-1,Integer.parseInt(commandArr[1])-1);
}
else if(commandArr.length==3)
{
deleteLine(Integer.parseInt(commandArr[1])-1,Integer.parseInt(commandArr[2])-1);
}
count=lines.size();
}
if(flag!=1)
{
System.out.print((count++)+"> ");
command = s.nextLine();
}
}
}
public static void insertLine(String text,int count)
{
if (count >= lines.size()) {
for(int i = lines.size(); i <= count; i++) {
lines.add(null);
}
}
lines.set(count,text);
}
public static void deleteLine(int start,int end)
{
for(int i=start;i<=end;i++)
lines.remove(i);
}
public static void printList()
{
for(int i=0;i "+lines.get(i));
}
}
Here is my output:
EDIT: textin.txt
1> The First Line
1>
2> And another line
3> I 3
3> The second line
4> One more line
5> L
1> null
2> null
3> The second line
4> One more line
5> D 2
3> L
1> null
2> The second line
3> One more line
4> E
How can I keep it from printing "null" after ">" and it just be blank. I need the output to look like
this:
1> The first line
2>
3> And another line
4> I 3
3> The second line
4> One more line
5> L
1> The first line
2>
3> The second line
4> One more line
5> And another line
5> D 2
4> L
1> The first line
2> The second line
3> One more line
4> And another line
5> E.
Below is my code for a line editor import java.io.BufferedReader;.pdf
Below is my code for a line editor:
import java.io.BufferedReader;
import java.io.File;
import java.io.FileReader;
import java.io.IOException;
import java.util.ArrayList;
import java.util.List;
import java.util.Scanner;
public class Main {
public static List lines = new ArrayList();
public static void main(String[] args) throws IOException
{
Scanner s = new Scanner(System.in);
System.out.print("Input file: ");
String fileName = s.nextLine();
BufferedReader br = new BufferedReader(new FileReader(new File(fileName)));
String str="";
int count=0;
while((str=br.readLine())!=null)
{
lines.add(count, str);
System.out.println((++count) + "> "+ str);
}
String text="";
String command = "";
System.out.print((++count)+"> ");
command=s.nextLine();
String commandArr[];
int flag;
while(!command.equals("E"))
{
flag=0;
commandArr = command.split("\\s");
if(commandArr[0].equals("I")) //Insertion
{
while(true)
{
if(commandArr.length==1&&flag!=2)
{
System.out.print((count)+"> ");
text = s.nextLine();
insertLine(text,count-1);
}
else if(commandArr.length==2&&flag!=2)
{
System.out.print((commandArr[1])+"> ");
count=Integer.parseInt(commandArr[1]);
text = s.nextLine();
insertLine(text,Integer.parseInt(commandArr[1])-1);
}
else
insertLine(text,count-1);
System.out.print((++count)+"> ");
command=s.nextLine();
commandArr = command.split("\\s");
if(commandArr[0].equals("I")||commandArr[0].equals("L")||commandArr[0].equals("D")||comm
andArr[0].equals("E"))
{
flag=1;
break;
}
else
{
flag=2;
text=command;
}
}
}
else if(commandArr[0].equals("L")) //Listing
{
printList();
}
else if(commandArr[0].equals("D")) //Deletion
{
if(commandArr.length==1)
{
deleteLine(count-1,count-1);
}
else if(commandArr.length==2)
{
deleteLine(Integer.parseInt(commandArr[1])-1,Integer.parseInt(commandArr[1])-1);
}
else if(commandArr.length==3)
{
deleteLine(Integer.parseInt(commandArr[1])-1,Integer.parseInt(commandArr[2])-1);
}
count=lines.size();
}
if(flag!=1)
{
System.out.print((count++)+"> ");
command = s.nextLine();
}
}
}
public static void insertLine(String text,int count)
{
lines.add(count,text);
}
public static void deleteLine(int start,int end)
{
for(int i=start;i<=end;i++)
lines.remove(i);
}
public static void printList()
{
for(int i=0;i "+lines.get(i));
}
}
The output should look like this:
EDIT: textin.txt (user inputs textin.txt)
1> The first line
2>
3> And another line
4> I 3
3> The second line
4> One more line
5> L
1> The first line
2>
3> The second line
4> One more line
5> And another line //This is now line 5; not 3;
5> D 2 // line 5, since L was issued from line 5;
4> L //line 4, since one line was deleted;
1> The first line
2> The second line
3> One more line
4> And another line
5> E
Here is a screenshot of the errors I am getting and need help fixing them: FileEdit [Java
Application] /Users/caseysharp/.p2/pool/plugins/org.eclipse.justj.openjdk.hotspot.jre.full.mac
EDIT: textin.txt 1> The First line 1> 2> And another line 3> I 3 3> The second line Exception in
thread "main" java..
DisplayBook.java import java.io.File; import java.io.FileInput.pdf
DisplayBook.java
import java.io.File;
import java.io.FileInputStream;
import java.io.FileOutputStream;
import java.io.IOException;
import java.util.Scanner;
public class DisplayBook {
public static void main(String[] args) throws IOException {
// TODO Auto-generated method stub
Scanner scan = new Scanner(System.in);
File file = new File(\"D:\\\\books.txt\");
if(file.exists()){
System.out.println(\"File exists\");
FileInputStream fis = new FileInputStream(file);
int content;
while ((content = fis.read()) != -1) {
System.out.print((char) content);
}
fis.close();
}
else{
System.out.println(\"File does not exist\");
String titles = \"\";
FileOutputStream fop = new FileOutputStream(file);
while(true){
System.out.println(\"Enter Book Title (none to exit):\");
String title = scan.nextLine();
if(title.equalsIgnoreCase(\"none\")){
break;
}
else{
titles = titles + title +\"\ \";
}
}
byte[] contentInBytes = titles.getBytes();
fop.write(contentInBytes);
fop.flush();
fop.close();
}
}
}
Output:
File does not exist
Enter Book Title (none to exit):
Physics
Enter Book Title (none to exit):
Chemistry
Enter Book Title (none to exit):
Science
Enter Book Title (none to exit):
Maths
Enter Book Title (none to exit):
none
--------------------------------------
File exists
Physics
Chemistry
Science
Maths
Solution
DisplayBook.java
import java.io.File;
import java.io.FileInputStream;
import java.io.FileOutputStream;
import java.io.IOException;
import java.util.Scanner;
public class DisplayBook {
public static void main(String[] args) throws IOException {
// TODO Auto-generated method stub
Scanner scan = new Scanner(System.in);
File file = new File(\"D:\\\\books.txt\");
if(file.exists()){
System.out.println(\"File exists\");
FileInputStream fis = new FileInputStream(file);
int content;
while ((content = fis.read()) != -1) {
System.out.print((char) content);
}
fis.close();
}
else{
System.out.println(\"File does not exist\");
String titles = \"\";
FileOutputStream fop = new FileOutputStream(file);
while(true){
System.out.println(\"Enter Book Title (none to exit):\");
String title = scan.nextLine();
if(title.equalsIgnoreCase(\"none\")){
break;
}
else{
titles = titles + title +\"\ \";
}
}
byte[] contentInBytes = titles.getBytes();
fop.write(contentInBytes);
fop.flush();
fop.close();
}
}
}
Output:
File does not exist
Enter Book Title (none to exit):
Physics
Enter Book Title (none to exit):
Chemistry
Enter Book Title (none to exit):
Science
Enter Book Title (none to exit):
Maths
Enter Book Title (none to exit):
none
--------------------------------------
File exists
Physics
Chemistry
Science
Maths.
Write a Java program that opens the file, reads all the Grade from t.pdf
The Java program opens a file, reads grades within it, and uses a method to calculate and display:
1) The grades with their letter values (A, B, C, etc.)
2) The sum of all grades in the file
3) The average of all grades in the file
Factors.javaimport java.io.; import java.util.Scanner; class .pdf
Factors.java
import java.io.*;
import java.util.Scanner;
class Factors
{
public static void main(String args[])
{
int a,i;
Scanner in = new Scanner(System.in);
System.out.println(\"Enter an integer\");
a = in.nextInt();
System.out.println(\"You entered integer \"+a);
System.out.print(\"\ \");
System.out.print(\"The factors are : \");
for(i=1;i<=a/2;i++)
{
if(a%i==0)
System.out.print(i+\",\");
}
System.out.print(a);
}
}
Output:
Enter an integer
44
You entered integer 44
The factors are : 1,2,4,11,22,44
Solution
Factors.java
import java.io.*;
import java.util.Scanner;
class Factors
{
public static void main(String args[])
{
int a,i;
Scanner in = new Scanner(System.in);
System.out.println(\"Enter an integer\");
a = in.nextInt();
System.out.println(\"You entered integer \"+a);
System.out.print(\"\ \");
System.out.print(\"The factors are : \");
for(i=1;i<=a/2;i++)
{
if(a%i==0)
System.out.print(i+\",\");
}
System.out.print(a);
}
}
Output:
Enter an integer
44
You entered integer 44
The factors are : 1,2,4,11,22,44.
I can not get my code to comply it gets mad that I have run declare mo.docx
I can not get my code to comply it gets mad that I have run declare more than once. Please help me figure out how to get this to work
package projectOne;
import java.io.File;
import java.io.FileNotFoundException;
import java.util.ArrayList;
import java.util.Scanner;
import java.util.*;
import java.io.*;
public class Main
{ public static void main(String[] pArgs){
Main mainObject = new Main();
mainObject.run();
} private void run() {
public FileReader readFile() throws FileNotFoundException {
FileReader file = new FileReader(\"p01-in.txt\");
return file;
}
public ArrayList<Integer> getData(FileReader file) throws FileNotFoundException {
Scanner data_read = new Scanner(new BufferedReader(file));
ArrayList<Integer> data = new ArrayList<Integer>();
while (data_read.hasNext()) {
int item = data_read.nextInt();
data.add(item);
}
return data;
}
public ArrayList<Integer> calcRuns(ArrayList<Integer> data) {
ArrayList<Integer> runs = new ArrayList<Integer>();
Iterator<Integer> i_data = data.iterator();
int run = 0;
int temp = i_data.next();
while (i_data.hasNext()) {
int t1 = i_data.next();
if (temp < t1)
run++;
else {
runs.add(run);
run = 0;
}
temp = t1;
}
i_data = data.iterator();
temp = i_data.next();
run = 0;
while (i_data.hasNext()) {
int t1 = i_data.next();
if (temp > t1)
run++;
else {
runs.add(run);
run = 0;
}
temp = t1;
}
return runs;
}
public void writeOutput(ArrayList<Integer> runs) throws IOException {
PrintWriter out = new PrintWriter(new File(\"p01-runs.txt\"));
ArrayList<Integer> run_count = new ArrayList<Integer>();
int i = 0;
int r_count = Collections.frequency(run_count, i);
while (i < 100) {
run_count.add(r_count);
i++;
r_count = Collections.frequency(runs, i);
}
Iterator<Integer> i_runs = run_count.iterator();
i = 0;
while (i_runs.hasNext()) {
int run = i_runs.next();
String line = \"run_\" + i + \" ,\" + run;
i++;
out.println(line);
}
}
public void run() throws Exception {
FileReader fin = this.readFile();
ArrayList<Integer> data = this.getData(fin);
ArrayList<Integer> runs = this.calcRuns(data);
this.writeOutput(runs);
}
}
Solution
In your program you have to change the syntax errors at method declaration.so your not getting output.
.
Write a program that will count the number of characters- words- and l.docx
Write a program that will count the number of characters, words, and lines in a file. Words are separated by whitespace characters. The file name should be passed as a command-line argument, as shown;
c:\\exercise>java Exercise12_13 Loan.java
File Loan.java has
1919 characters
210 words
71 lines
c:\\exercise>
Solution
import java.io.File;
import java.io.FileNotFoundException;
import java.util.Scanner;
import java.util.StringTokenizer;
public class ass1
{
public static void main(String[] args) throws FileNotFoundException
{
//Â Â Scanner console = new Scanner(System.in);
System.out.println(args[0]);
//System.out.println(args[1]);
//System.out.println(args[2]);
String inputFile=args[0];
File file = new File(inputFile);
Scanner in = new Scanner(file);
int words = 0;
int lines = 0;
int chars = 0;
while(in.hasNextLine()) {
lines++;
String line = in.nextLine();
chars += line.length();
words += new StringTokenizer(line, \" \").countTokens();
}
System.out.println(\"characters are \"+chars);
System.out.println(\"words are \"+words);
System.out.println(\"lines are \"+lines);
}
}
.
Java Program I keep receiving the following error in my code- Can you.pdf
Java Program
I keep receiving the following error in my code. Can you please fix this error and test to make
sure it does not show a error? I've placed bracets but keep receiving the error "reached end of file
while parsing private static void displayStudent()" See below
Error:
sh -c javac -classpath .:target/dependency/* -d . $(find . -type f -name '*.java')
./Main.java:121: error: reached end of file while parsing
private static void displayStudent()
^
1 error
exit status 1
------------------------------------------
Code:
import java.util.Scanner;
import java.io.File;
import java.io.FileWriter;
import java.io.IOException;
public class StudentCourseProgram {
public static void main(String[] args) {
Scanner input = new Scanner(System.in);
int choice;
do {
System.out.println("Please choose an option:");
System.out.println("1. Add student");
System.out.println("2. Add course");
System.out.println("3. Display student");
System.out.println("4. Display course");
System.out.println("5. Exit");
choice = input.nextInt();
switch (choice) {
case 1:
addStudent();
break;
case 2:
addCourse();
break;
case 3:
displayStudent();
break;
case 4:
displayCourse();
break;
case 5:
System.out.println("Exiting program...");
break;
default:
System.out.println("Invalid option. Please try again.");
}
} while (choice != 5);
}
private static void addStudent() {
Scanner input = new Scanner(System.in);
System.out.println("Adding a student...");
System.out.print("Enter student ID: ");
int id = input.nextInt();
System.out.print("Enter student first name: ");
String firstName = input.next();
System.out.print("Enter student last name: ");
String lastName = input.next();
System.out.print("Enter student address: ");
String address = input.next();
System.out.print("Enter student city: ");
String city = input.next();
System.out.print("Enter student state: ");
String state = input.next();
System.out.print("Enter student zip code: ");
int zip = input.nextInt();
try {
FileWriter writer = new FileWriter("student.txt", true);
Scanner reader = new Scanner(new File("student.txt"));
while (reader.hasNextLine()) {
String line = reader.nextLine();
String[] parts = line.split(",");
if (Integer.parseInt(parts[0]) == id) {
System.out.println("Error: Student with that ID already exists.");
return;
}
}
writer.write(id + "," + firstName + "," + lastName + "," + address + "," + city + "," + state + "," +
zip + "\n");
System.out.println("Student added successfully.");
reader.close();
writer.close();
} catch (IOException e) {
System.out.println("Error: " + e.getMessage());
}
}
private static void addCourse() {
Scanner input = new Scanner(System.in);
System.out.println("Adding a course...");
System.out.print("Enter course ID: ");
int id = input.nextInt();
System.out.print("Enter course name: ");
String name = input.next();
System.out.print("Enter course description: ");
String description = input.next();
try {
FileWriter writer = new FileWriter("course.txt", true);
Scanner reader = new Scanner(new Fil.
File Handling.pptx
This document provides an introduction to file handling in Java. It discusses how file handling allows programs to permanently store output data by writing it to files on secondary storage devices like hard disks. It covers key concepts like input and output streams that represent the flow of data into and out of a program. It also discusses how to create, write to, read from, and delete files in Java using classes like File, FileWriter, FileReader and Scanner. Common file methods like getName(), getAbsolutePath(), exists() are also outlined.
54240326 (1)
The document contains code for several Java programs that implement common data structures and client-server applications. Specifically, it includes programs to:
1. Implement a stack using an array and perform push and pop operations.
2. Implement a queue using an array and perform insertion, deletion, and display operations.
3. Implement a singly linked list and perform operations like creation, addition, deletion, and display of nodes.
4. Implement a producer-consumer problem using threads and synchronization to add and remove elements from a shared queue.
54240326 copy
The document contains code for implementing various data structures and algorithms in Java, including a stack, queue, linked list, producer-consumer problem, scrolling text applet, and client-server file transfer program. The stack, queue, and linked list code demonstrates how to create the data structures and perform common operations like push, pop, insert, delete, and display. The producer-consumer code uses threads and synchronization to model the scenario. The scrolling text applet animates text moving across the screen. The client-server code allows a client to request a file from a server, which then sends the file contents to display on the client.
The sample program for above series in JAVA will be like belowimpo.pdf
The sample program for above series in JAVA will be like below
import java.util.Scan;
public class PIVALUE {
public static void main(String[] args) {
// Scan object creation
Scan input = new Scan (System.in);
// ask user for input
System.out.println(\"Enter num.\");
double j = input.nextDouble(); // user entered value of j
double Tot = 0;
for(j=0; j<10000; j++){
if(j%2 == 0) // check remainder of `j/2` is 0
Tot += -1 / ( 2 * j - 1);
else
Tot += 1 / (2 * j - 1);
}
System.out.println(Tot);
}
}
Solution
The sample program for above series in JAVA will be like below
import java.util.Scan;
public class PIVALUE {
public static void main(String[] args) {
// Scan object creation
Scan input = new Scan (System.in);
// ask user for input
System.out.println(\"Enter num.\");
double j = input.nextDouble(); // user entered value of j
double Tot = 0;
for(j=0; j<10000; j++){
if(j%2 == 0) // check remainder of `j/2` is 0
Tot += -1 / ( 2 * j - 1);
else
Tot += 1 / (2 * j - 1);
}
System.out.println(Tot);
}
}.
The sample program for above series in JAVA will be like belowimpo.pdf
The sample program for above series in JAVA will be like below
import java.util.Scan;
public class PIVALUE {
public static void main(String[] args) {
// Scan object creation
Scan input = new Scan (System.in);
// ask user for input
System.out.println(\"Enter num.\");
double j = input.nextDouble(); // user entered value of j
double Tot = 0;
for(j=0; j<10000; j++){
if(j%2 == 0) // check remainder of `j/2` is 0
Tot += -1 / ( 2 * j - 1);
else
Tot += 1 / (2 * j - 1);
}
System.out.println(Tot);
}
}
Solution
The sample program for above series in JAVA will be like below
import java.util.Scan;
public class PIVALUE {
public static void main(String[] args) {
// Scan object creation
Scan input = new Scan (System.in);
// ask user for input
System.out.println(\"Enter num.\");
double j = input.nextDouble(); // user entered value of j
double Tot = 0;
for(j=0; j<10000; j++){
if(j%2 == 0) // check remainder of `j/2` is 0
Tot += -1 / ( 2 * j - 1);
else
Tot += 1 / (2 * j - 1);
}
System.out.println(Tot);
}
}.
2. Write a program which will open an input file and write to an out.pdf
2. Write a program which will open an input file and write to an output file. Use exception
handling (try & Catch to display appropriate error handling).
Solution
herer is code:
import java.io.*;
public class Copytxt{
public static void main (String[] args)throws IOException {
int i;
FileInputStream fin;
FileOutputStream fout;
try {
// open input file
try {
fin = new FileInputStream(\"input.txt\"); // pass the input location file here
}
catch (FileNotFoundException e) { // if file not found
System.out.println(\"Input File Not Found\");
return;
}
// open output file
try {
fout = new FileOutputStream(\"output.txt\"); // pass the output location file here
}
catch (FileNotFoundException e) { // if file not found
System.out.println(\"Error Opening Output File\");
return;
}
}
catch (ArrayIndexOutOfBoundsException e) {
System.out.println(\"Usage: CopyFile From To\");
return;
}
// Copy File
try {
byte[] buffer = new byte[1024];
int length;
/*copying the contents from input stream to
* output stream using read and write methods
*/
while ((length = fin.read(buffer)) > 0){
fout.write(buffer, 0, length);
}
}
catch (IOException e) {
System.out.println(\"File Error\");
}
fin.close();
fout.close();
}
}.
java input & output statements
The document provides examples of Java code for printing "Hello World", getting different types of user input including integer, float, double, string, and character, and printing outputs. It includes code snippets using classes, importing packages like Scanner, and using methods like println, nextInt, nextFloat etc. to get input and display output in Java programs.
ExtraFileIO.pptx
The document discusses file input and output in Java. It covers the File class, opening files for reading using a Scanner, handling exceptions, reading data line-by-line and tokenizing the lines. It also discusses opening files for writing using a PrintWriter and outputting data to files. The key topics are file I/O, exception handling, and using Scanners and PrintWriters to read from and write to files in Java.
Please help me with a UML class diagram for the following code im.pdf
Please help me with a UML class diagram for the following code:
import java.io.BufferedReader;
import java.io.File;
import java.io.FileReader;
import java.io.IOException;
import java.util.ArrayList;
import java.util.List;
import java.util.Scanner;
public class FileEdit {
public static List lines = new ArrayList();
public static void main(String[] args) throws IOException
{
Scanner s = new Scanner(System.in);
System.out.print("EDIT: ");
String fileName = s.nextLine();
BufferedReader br = new BufferedReader(new FileReader(new File(fileName)));
String str="";
int count=0;
while((str=br.readLine())!=null)
{
lines.add(count, str);
System.out.println((++count) + "> "+ str);
}
String text="";
String command = "";
System.out.print((++count)+"> ");
command=s.nextLine();
String commandArr[];
int flag;
while(!command.equals("E"))
{
flag=0;
commandArr = command.split("\\s");
if(commandArr[0].equals("I")) //Insertion
{
while(true)
{
if(commandArr.length==1&&flag!=2)
{
System.out.print((count)+"> ");
text = s.nextLine();
insertLine(text,count-1);
}
else if(commandArr.length==2&&flag!=2)
{
System.out.print((commandArr[1])+"> ");
count=Integer.parseInt(commandArr[1]);
text = s.nextLine();
insertLine(text,Integer.parseInt(commandArr[1])-1);
}
else
insertLine(text,count-1);
System.out.print((++count)+"> ");
command=s.nextLine();
commandArr = command.split("\\s");
if(commandArr[0].equals("I")||commandArr[0].equals("L")||commandArr[0].equals("D")||comm
andArr[0].equals("E"))
{
flag=1;
break;
}
else
{
flag=2;
text=command;
}
}
}
else if(commandArr[0].equals("L")) //Listing
{
printList();
}
else if(commandArr[0].equals("D")) //Deletion
{
if(commandArr.length==1)
{
deleteLine(count-1,count-1);
}
else if(commandArr.length==2)
{
deleteLine(Integer.parseInt(commandArr[1])-1,Integer.parseInt(commandArr[1])-1);
}
else if(commandArr.length==3)
{
deleteLine(Integer.parseInt(commandArr[1])-1,Integer.parseInt(commandArr[2])-1);
}
count=lines.size();
}
if(flag!=1)
{
System.out.print((count++)+"> ");
command = s.nextLine();
}
}
}
public static void insertLine(String text,int count)
{
if (count >= lines.size()) {
for(int i = lines.size(); i <= count; i++) {
lines.add(null);
}
}
lines.set(count,text);
}
public static void deleteLine(int start,int end)
{
for(int i=start;i<=end;i++)
lines.remove(i);
}
public static void printList()
{
for(int i=0;i "+lines.get(i));
}
}.
Similar to The Java Program for the above given isimport java.io.File;impo.pdf (20)
Modify the Total program so that it writes the values in two columns.pdf
Modify the Total program so that it writes the values in two columns.pdf
import java.io.BufferedReader;import java.io.File;import java.io.pdf
import java.io.BufferedReader;import java.io.File;import java.io.pdf
JAVA Q2- Write a program that reads strings from the user and writes t.docx
JAVA Q2- Write a program that reads strings from the user and writes t.docx
Here is my code for a linefile editor import java.io.BufferedRea.pdf
Here is my code for a linefile editor import java.io.BufferedRea.pdf
Below is my code for a line editor import java.io.BufferedReader;.pdf
Below is my code for a line editor import java.io.BufferedReader;.pdf
DisplayBook.java import java.io.File; import java.io.FileInput.pdf
DisplayBook.java import java.io.File; import java.io.FileInput.pdf
Write a Java program that opens the file, reads all the Grade from t.pdf
Write a Java program that opens the file, reads all the Grade from t.pdf
Factors.javaimport java.io.; import java.util.Scanner; class .pdf
Factors.javaimport java.io.; import java.util.Scanner; class .pdf
I can not get my code to comply it gets mad that I have run declare mo.docx
I can not get my code to comply it gets mad that I have run declare mo.docx
Write a program that will count the number of characters- words- and l.docx
Write a program that will count the number of characters- words- and l.docx
Java Program I keep receiving the following error in my code- Can you.pdf
Java Program I keep receiving the following error in my code- Can you.pdf
The sample program for above series in JAVA will be like belowimpo.pdf
The sample program for above series in JAVA will be like belowimpo.pdf
The sample program for above series in JAVA will be like belowimpo.pdf
The sample program for above series in JAVA will be like belowimpo.pdf
2. Write a program which will open an input file and write to an out.pdf
2. Write a program which will open an input file and write to an out.pdf
Please help me with a UML class diagram for the following code im.pdf
Please help me with a UML class diagram for the following code im.pdf
More from anjanacottonmills
Step1 First compound has 2 double bonds ; Second .pdf
Step1 First compound has 2 double bonds ; Second compound has 5 double bonds;
Third compound has 1 double bond. Step2 Second molecule requires maximum energy due to 5
double bonds.This is likely to to have UV/ Visible absorption.
Solution
Step1 First compound has 2 double bonds ; Second compound has 5 double bonds;
Third compound has 1 double bond. Step2 Second molecule requires maximum energy due to 5
double bonds.This is likely to to have UV/ Visible absorption..
Molarity = molesvolume9liter) Molarity of NH4Cl.pdf
Molarity = moles/volume9liter) Molarity of NH4Cl = 4.50 mol /2.00L=2.25 M
Solution
Molarity = moles/volume9liter) Molarity of NH4Cl = 4.50 mol /2.00L=2.25 M.
HOCH2CH2OH is more soluble in water. more OH grou.pdf
HOCH2CH2OH is more soluble in water. more OH groups the alcohol has, more
easily to form hydrogen bond with water molecules. As a result, it is more soluble.
Solution
HOCH2CH2OH is more soluble in water. more OH groups the alcohol has, more
easily to form hydrogen bond with water molecules. As a result, it is more soluble..
x = -2Solutionx = -2.pdf
The variable x is defined as -2. The solution is simply that x equals -2, as there are no other operations or calculations shown in the short document. In one line, it defines x and states the solution that x = -2.
z=(2n+1)4where n=all integersSolutionz=(2n+1)4where n=al.pdf
The formula z=(2n+1)/4 is given, where n can be any integer value. This formula is used to calculate the value of z based on the integer value given for n, where z is equal to (2 times n plus 1) divided by 4.
This above plant name is Solidago sempervirens, belong to genus Soli.pdf
This above plant name is Solidago sempervirens, belong to genus Solidago. Species
S.sempervirens commonly know as seaside goldenrod. Authority L.
Solidago sempervirens is a succulent, herbaceous perennial that reaches heights of 4–6 feet (120-
180 cm). It is unusual in the genus in having toothless, hair-less leaves, thicker than those of
most other Solidago species. Flower heads are found in a large paniculiforminflorescence at the
top of the plant, often with branches that bend backwards towards the base. This species blooms
in late summer and well into the fall, later in the season than most of its relatives. Its fruits are
wind-dispersed achenes. They are yellow often, and have sprouts of buds at the end of the short
branchesKingdomPlantaePlants, but not fungi, lichens, or algae (from Stearn’s Botanical
Latin)SubkingdomTracheobiontaVascular plants—plants with a “circulatory system” for
delivering water and nutrientsDivisionMagnoliophytaFlowering plants, also known as
angiospermsClassMagnoliopsidaDicotyledons—plants with two initial seed
leavesSubclassAsteridaeA large class that encompasses astersOrderAsteralesFlowering plants
with a central disk flower and surrounding petals, like daisiesFamilyAsteraceaeThe aster family,
which also includes daisies and sunflowers; from the Greek , “star,” for the star-shaped
flowersGenusSolidagoFrom Latin solido, “to make whole or heal,” because it was believed
these species had healing propertiesSpeciessempervirens\"Evergreen\"
Solution
This above plant name is Solidago sempervirens, belong to genus Solidago. Species
S.sempervirens commonly know as seaside goldenrod. Authority L.
Solidago sempervirens is a succulent, herbaceous perennial that reaches heights of 4–6 feet (120-
180 cm). It is unusual in the genus in having toothless, hair-less leaves, thicker than those of
most other Solidago species. Flower heads are found in a large paniculiforminflorescence at the
top of the plant, often with branches that bend backwards towards the base. This species blooms
in late summer and well into the fall, later in the season than most of its relatives. Its fruits are
wind-dispersed achenes. They are yellow often, and have sprouts of buds at the end of the short
branchesKingdomPlantaePlants, but not fungi, lichens, or algae (from Stearn’s Botanical
Latin)SubkingdomTracheobiontaVascular plants—plants with a “circulatory system” for
delivering water and nutrientsDivisionMagnoliophytaFlowering plants, also known as
angiospermsClassMagnoliopsidaDicotyledons—plants with two initial seed
leavesSubclassAsteridaeA large class that encompasses astersOrderAsteralesFlowering plants
with a central disk flower and surrounding petals, like daisiesFamilyAsteraceaeThe aster family,
which also includes daisies and sunflowers; from the Greek , “star,” for the star-shaped
flowersGenusSolidagoFrom Latin solido, “to make whole or heal,” because it was believed
these species had healing propertiesSpeciessempervirens\"Evergreen\".
The fidelity of DNA replication determines the genome stability and .pdf
The fidelity of DNA replication determines the genome stability and it plays a main role in
evolution of species. The DNA replication fidelity is a key factor of mutations which in turn
leads to many human diseases.
The high fidelity of DNA replication in E.coli is achieved by DNA polymerase III
holoenzyme, which is a complex of 17 proteins. This is responsible for simultaneous replication
of both leading and lagging DNA strand with high speed and high fidelity. There are also four
other accessory DNA polymerases present in E.coli such as Pol I, II, IV and V.
Pol II plays an important role in proofreading of DNA intra-strand crosslinks, DNA
strand exposed to UV irradiation, oxidation, stress-induced mutagenesis. This enzyme is
responsible for translesion synthesis. Pol IV and V are specific to lagging stand. Pol IV is
responsible for proceeding the DNA replication even when there is a mismatch. The Pol IV has a
high constitutive intracellular concentration. Pol V is induced constitutively in the absence of
exogenous DNA damage. The primary role of Pol I in DNA replication is to remove RNA
primers and fill the resulting gap during maturation of Okazaki fragments in the lagging strand.
In eukaryotes, the replication of leading and lagging strands takes place by Pol and Pol
respectively, which leads to fidelity differences. Whereas in E.coli the replication of both leading
and lagging strands takes place by same enzyme polymerase III. This is also one of the reason
why E.coli has high fidelity of DNA replication.
The tau subunit of Pol III plays a central role in DNA replication fidelity, as it interacts with both
leading and lagging strand. This subunit also contributes for polymerase switching during
replication.
Solution
The fidelity of DNA replication determines the genome stability and it plays a main role in
evolution of species. The DNA replication fidelity is a key factor of mutations which in turn
leads to many human diseases.
The high fidelity of DNA replication in E.coli is achieved by DNA polymerase III
holoenzyme, which is a complex of 17 proteins. This is responsible for simultaneous replication
of both leading and lagging DNA strand with high speed and high fidelity. There are also four
other accessory DNA polymerases present in E.coli such as Pol I, II, IV and V.
Pol II plays an important role in proofreading of DNA intra-strand crosslinks, DNA
strand exposed to UV irradiation, oxidation, stress-induced mutagenesis. This enzyme is
responsible for translesion synthesis. Pol IV and V are specific to lagging stand. Pol IV is
responsible for proceeding the DNA replication even when there is a mismatch. The Pol IV has a
high constitutive intracellular concentration. Pol V is induced constitutively in the absence of
exogenous DNA damage. The primary role of Pol I in DNA replication is to remove RNA
primers and fill the resulting gap during maturation of Okazaki fragments in the lagging strand.
In eukaryotes, the replication of leading .
E= 0.28V reactants are favored .pdf
This short document discusses a chemical reaction with an electrode potential (E) of 0.28 volts. At this electrode potential, the reactants of the reaction are favored to form, rather than the products. The document simply states that with an E value of 0.28V, the reactants are favored in the chemical reaction.
The differences between ipv4 and ipv6 are as below1. Header1.Ch.pdf
The differences between ipv4 and ipv6 are as below
1. Header:
1.Checksum and options feild are present and the length of header with 20 bytes in ipv4.
2.extension header is present with length 40 bytes in ipv6.
2.Addressing scheme:
1. A 32-bit permiting a amont of 2^32 in ipv4.
2. The adress of ipv6 is 128 bit.
3.Translation: Ipv4 is using 4 bytes and ipv6 is is utilizing 16 bytes to transit from ipv4 to ipv6.
4.Data flow: The ipv4 is unable to recognise packet flow for QOS handling.In ipv6 flow label is
present to describe packet flow for QOS handling.
5.Duplicate adress detection: in ipv4 ARP is used to detect the duplicate adress and in ipv6
SLAAC +DHPCPV6 is used.
6.Multicast and routing: ipv6 is healthier in multicast routing when compared to ipv6.in ipv4
with TCP protocol is about 576 bytes.where as in ipv6 with 120 bytes.
7.current status of ipv6 adoption: Gooogle dignified ipv6 adoption at arriving 10% positioning
spot.
Solution
The differences between ipv4 and ipv6 are as below
1. Header:
1.Checksum and options feild are present and the length of header with 20 bytes in ipv4.
2.extension header is present with length 40 bytes in ipv6.
2.Addressing scheme:
1. A 32-bit permiting a amont of 2^32 in ipv4.
2. The adress of ipv6 is 128 bit.
3.Translation: Ipv4 is using 4 bytes and ipv6 is is utilizing 16 bytes to transit from ipv4 to ipv6.
4.Data flow: The ipv4 is unable to recognise packet flow for QOS handling.In ipv6 flow label is
present to describe packet flow for QOS handling.
5.Duplicate adress detection: in ipv4 ARP is used to detect the duplicate adress and in ipv6
SLAAC +DHPCPV6 is used.
6.Multicast and routing: ipv6 is healthier in multicast routing when compared to ipv6.in ipv4
with TCP protocol is about 576 bytes.where as in ipv6 with 120 bytes.
7.current status of ipv6 adoption: Gooogle dignified ipv6 adoption at arriving 10% positioning
spot..
TCP - TCP breaks data into manageable packets and tracks information.pdf
TCP - TCP breaks data into manageable packets and tracks information such as source and
destination of packets. It is able to reroute packets and is responsible for guaranteed delivery of
the data.
IP - This is a connectionless protocol, which means that a session is not created before sending
data. IP is responsible for addressing and routing of packets between computers. It does not
guarantee delivery and does not give acknowledgement of packets that are lost or sent out of
order as this is the responsibility of higher layer protocols such as TCP.
UDP - A connectionless, datagram service that provides an unreliable, best-effort delivery.
ICMP - Internet Control Message Protocol enables systems on a TCP/IP network to share status
and error information such as with the use of PING and TRACERT utilities.
SMTP - Used to reliably send and receive mail over the Internet.
FTP - File transfer protocol is used for transferring files between remote systems. Must resolve
host name to IP address to establish communication. It is connection oriented (i.e. verifies that
packets reach destination).
TFTP - Same as FTP but not connection oriented.
ARP - provides IP-address to MAC address resolution for IP packets. A MAC address is your
computer\'s unique hardware number and appears in the form 00-A0-F1-27-64-E1 (for example).
Each computer stores an ARP cache of other computers ARP-IP combinations.
POP3 - Post Office Protocol. A POP3 mail server holds mail until the workstation is ready to
receive it.
IMAP - Like POP3, Internet Message Access Protocol is a standard protocol for accessing e-
mail from your local server. IMAP (the latest version is IMAP4) is a client/server protocol in
which e-mail is received and held for you by your Internet server.
TELNET - Provides a virtual terminal or remote login across the network that is connection-
based. The remote server must be running a Telnet service for clients to connect.
HTTP - The Hypertext Transfer Protocol is the set of rules for exchanging files (text, graphic
images, sound, video, and other multimedia files) on the World Wide Web. It is the protocol
controlling the transfer and addressing of HTTP requests and responses.
HTTPS - Signifies that a web page is using the Secure Sockets Layer (SSL) protocol and is
providing a secure connection. This is used for secure internet business transactions.
NTP - Network Time Protocol is a protocol that is used to synchronize computer clock times in
a network of computers.
SNMP - Stands for Simple Network Management Protocol and is used for monitoring and status
information on a network. SNMP can be used to monitor any device that is SNMP capable and
this can include computers, printers, routers, servers, gateways and many more using agents on
the target systems. The agents report information back to the management systems by the use of
“traps” which capture snapshot data of the system. This trap information could be system errors,
resource information, or other info.
Static arrays are structures whose size is fixed at compile time and.pdf
Static arrays are structures whose size is fixed at compile time and therefore cannot be extended
or reduced to fit the data set. A dynamic array can be extended by doubling the size but there is
overhead associated with the operation of copying old data and freeing the memory associated
with the old data structure. One potential problem of using arrays for storing data is that arrays
require a contiguous block of memory which may not be available, if the requested contiguous
block is too large. However the advantages of using arrays are that each element in the array can
be accessed very efficiently using an index. However, for applications that can be better
managed without using contiguous memory we define a concept called “linked lists”.
A linked list is a collection of objects linked together by references from one object to another
object. By convention these objects are named as nodes. So the basic linked list is collection of
nodes where each node contains one or more data fields AND a reference to the next node. The
last node points to a NULL reference to indicate the end of the list.
Types of Linked Lists
Linked lists are widely used in many applications because of the flexibility it provides. Unlike
arrays that are dynamically assigned, linked lists do not require memory from a contiguous
block. This makes it very appealing to store data in a linked list, when the data set is large or
device (eg: PDA) has limited memory. One of the disadvantages of linked lists is that they are
not random accessed like arrays. To find information in a linked list one must start from the head
of the list and traverse the list sequentially until it finds (or not find) the node. Another advantage
of linked lists over arrays is that when a node is inserted or deleted, there is no need to “adjust”
the array.
There are few different types of linked lists. A singly linked list as described above provides
access to the list from the head node. Traversal is allowed only one way and there is no going
back. A doubly linked list is a list that has two references, one to the next node and another to
previous node. Doubly linked list also starts from head node, but provide access both ways. That
is one can traverse forward or backward from any node. A multilinked list (see figures 1 & 2) is
a more general linked list with multiple links from nodes. For examples, we can define a Node
that has two references, age pointer and a name pointer. With this structure it is possible to
maintain a single list, where if we follow the name pointer we can traverse the list in alphabetical
order of names and if we traverse the age pointer, we can traverse the list sorted by ages. This
type of node organization may be useful for maintaining a customer list in a bank where same
list can be traversed in any order (name, age, or any other criteria) based on the need.
Designing the Node of a Linked List
Linked list is a collection of linked nodes. A node is a struct with at least a.
D) II, III, IV, and V only .pdf
This document provides the answer choice "D) II, III, IV, and V only" as the solution to a problem or question. No other context or details are given about the problem or question being answered. The answer simply states that choices II, III, IV, and V are the correct options.
NumberList.java (implements the linked list)public class NumberLis.pdf
NumberList.java (implements the linked list)
public class NumberList{
Node first;
Node last;
public NumberList(){
first = null;
last = null;
}
public NumberList(Node node){
this.first = node;
this.last = node;
}
public boolean isEmpty(){
return first == null;
}
public void setLast(Node node){
this.last = node;
}
public void insert(Node node){
if (first==null){
this.first = node;
this.last = node;
}
else{
node.previous = this.last;
node.previous.next = node;
node.next = null;
this.last = node;
}
}
public boolean inList(String num){
Node l = first;
while (l.next != null){
if (l.number.equals(num)){
return true;
}
else{
l = l.next;
}
}
return false;
}
public void printList(){
Node l = first;
while(l.next!=null){
System.out.print(l.number+\" \");
l = l.next;
}
System.out.println();
}
}
Node.java (implements a single node in the linked list; stores number and pointers)
public class Node{
String number;
Node previous;
Node next;
public Node(String num){
number = num;
previous = null;
next = null;
}
public Node(String num, Node p, Node n){
number = num;
previous = p;
next = n;
}
}
Main.java (contains the main method and the helper methods to solve the questions given)
import java.util.Scanner;
import java.util.List;
import java.util.ArrayList;
public class Main{
static boolean checkHappiness(String num){
Node current = new Node(num, null, null);
NumberList numberList = new NumberList(current);
int len = num.length();
int resultant = 0;
for (int i=0; i happyNumbersfrom1to10000(){
ArrayList numbers = new ArrayList();
for (int j=0; j<10000; j++){
if (checkHappiness(String.valueOf(j+1))){
numbers.add(String.valueOf(j+1));
}
}
return numbers;
}
static void happyNumbersfrom9001to10000(){
for (int j=9001; j<=10000; j++){
//System.out.println(j);
if (checkHappiness(String.valueOf(j))){
System.out.println(String.valueOf(j));
}
}
}
static String getLargeHappyNumber(){
for (long i=10000000000000000000L; i<1000000000000000000000L; i++){
if (checkHappiness(String.valueOf(i))){
return String.valueOf(i);
}
return \"-1\";
}
}
static String getLargeUnhappyNumber(){
for (long i=10000000000000000000L; i<1000000000000000000000L; i++){
if (!checkHappiness(String.valueOf(i))){
return String.valueOf(i);
}
return \"-1\";
}
}
public static void main(String[] args){
happyNumbersfrom9001to10000();
System.out.println(happyNumbersfrom1to10000());
}
}
Solution
NumberList.java (implements the linked list)
public class NumberList{
Node first;
Node last;
public NumberList(){
first = null;
last = null;
}
public NumberList(Node node){
this.first = node;
this.last = node;
}
public boolean isEmpty(){
return first == null;
}
public void setLast(Node node){
this.last = node;
}
public void insert(Node node){
if (first==null){
this.first = node;
this.last = node;
}
else{
node.previous = this.last;
node.previous.next = node;
node.next = null;
this.last = node;
}
}
public boolean inList(String num){
Node l = first;
while (l.next != null){
if (l.number.equals(num)){
return true;
}
else{
l = l.
c) H2(g) + 12 O2(g) -H2O (l) as all are in the.pdf
c) H2(g) + 1/2 O2(g) ->H2O (l) as all are in there natural state. therefore it will
represent the enthalpy of formation
Solution
c) H2(g) + 1/2 O2(g) ->H2O (l) as all are in there natural state. therefore it will
represent the enthalpy of formation.
MagicSquareTest.java import java.util.Scanner;public class Mag.pdf
MagicSquareTest.java
import java.util.Scanner;
public class MagicSquareTest {
/**
* @param args
*/
public static void main(String[] args) {
// TODO Auto-generated method stub
Scanner scan = new Scanner(System.in);
System.out.println(\"Enter Number of rows:\");
int rows = scan.nextInt();
System.out.println(\"Enter Nummber of columns:\");
int cols = scan.nextInt();
int a[][] = new int[rows][cols];
System.out.println(\"Enter matrix elements;\");
for(int i=0; i
Solution
MagicSquareTest.java
import java.util.Scanner;
public class MagicSquareTest {
/**
* @param args
*/
public static void main(String[] args) {
// TODO Auto-generated method stub
Scanner scan = new Scanner(System.in);
System.out.println(\"Enter Number of rows:\");
int rows = scan.nextInt();
System.out.println(\"Enter Nummber of columns:\");
int cols = scan.nextInt();
int a[][] = new int[rows][cols];
System.out.println(\"Enter matrix elements;\");
for(int i=0; i.
Interphase.This is the phase where the cell prepares for the next .pdf
Interphase.
This is the phase where the cell prepares for the next round of division. It has three phases
further, Gap1, synthesis, and gap 2. During Gap1, the cell acquired nutrients and increases in
size. The checkpoints ensure that the cell is normal and all the things necessary for DNA
synthesis are prepared. During synthesis phase, DNA replicates. During the Gap 2 phase, the cell
grow further and the checkpoints ensure that the cell is correctly prepared for mitosis phase. The
cell spends most of its time in G1 phase, followed by S phase. Remaining phases comprise only
very short period of the cell cycle.
Solution
Interphase.
This is the phase where the cell prepares for the next round of division. It has three phases
further, Gap1, synthesis, and gap 2. During Gap1, the cell acquired nutrients and increases in
size. The checkpoints ensure that the cell is normal and all the things necessary for DNA
synthesis are prepared. During synthesis phase, DNA replicates. During the Gap 2 phase, the cell
grow further and the checkpoints ensure that the cell is correctly prepared for mitosis phase. The
cell spends most of its time in G1 phase, followed by S phase. Remaining phases comprise only
very short period of the cell cycle..
B.the rotation will be zero as L and D will cance.pdf
B.the rotation will be zero as L and D will cancel out each other..
Solution
B.the rotation will be zero as L and D will cancel out each other...
In the above conversation it is belonging to stereotypes.Stereotyp.pdf
In the above conversation it is belonging to stereotypes.
Stereotypes are a form of generalization. When we generalize, we group or classify people,
places, or things according to the traits they have in common. For example, we may say most
Masai tribesmen are unusually tall or Scandinavians are usually fair-skinned. If our observations
are careless or too limited, the generalization may be faulty, as when someone says, \"Hollywood
hasn\'t produced any quality movies in the past fifteen years.\"
But stereotypes are more serious than mere faulty generalizations. They are fixed, unbending
generalizations about people, places, or things. When a stereotype is challenged, the person who
holds it is unlikely to modify or discard it, because it is based on a distortion of perception. As
Walter Lippmann explains, when we stereotype,\" we do not so much see this man and that
sunset; rather we notice that the thing is man or sunset, and then see chiefly what our mind is full
of on those subjects.\"75
The most common kinds of stereotyping are ethnic and religious. Jews are shrewd and cunning,
clannish, have a financial genius matched only by their greed. Italians are hot-tempered, coarse,
and sensual. The Irish, like the Poles, are big and stupid; in addition, they brawl, lust after heavy
liquor and light conversation. Blacks are primitive and slowwitted. (Often each of these
stereotypes includes a virtue or two – Jews are good family members, Italians artistic, Poles
brave, the Irish devout, blacks athletic.)
Beyond these stereotypes are numerous other, less common ones; Swedish women, foreign film
directors, Southern senators, physical education instructors, fundamentalist clergymen, agnostics,
atheists, democrats, republicans, Mexicans, scientist, prostitutes, politicians, English teachers,
psychiatrists, construction workers, black militants, college dropouts, homosexuals, and society
matrons. There are stereotypes of institutions as well: marriage, the church, government, the
military, the Founding Fathers, Western culture, the Judeo-Christian tradition. (A full list would
include even God and mother.)
FACTS DON\'T MATTER
It is pleasant to assume that when the facts are known, stereotypes disappear. However, that is
seldom the case. The late Harvard psychologist Gordon Allport, in The nature of Prejudice,
pointed out that \"it is possible for a stereotype to grow in defiance of all evidence…\"76 People
who stereotype don\'t dust accept the facts that are offered to them. They measure those facts
against what they already \"know.\" That is, they measure them against the stereotype itself.
Instead of seeing that the stereotype is false and therefore dismissing it, they reject the unfamiliar
facts.
People who think in terms of stereotypes tend to be selective in their perceptions. They reject
conditions that challenge their preformed judgment and retain those that reinforce it. Thus a
person can notice the Jewish employer promoting another Jew but ign.
In general, the objective of an audit is to assess the risk of mater.pdf
In general, the objective of an audit is to assess the risk of material misstatements in the financial
statements. Material misstatements can arise from inadequacies in internal controls and from
inaccurate management assertions. Thus, testing the validity of the various implicit managerial
assertions is a key objective of an auditor.
Existence and Completeness
Auditing standards require that auditors test basic underlying management assertions implicit in
the financial statements. Key among these various assertions are existence or occurrence, which
describe a singular concept: Journal entries are not fiction. As the name implies, an auditor will
conduct various procedures to verify that assets do in fact exist and that recorded transactions did
in fact occur. Additionally, an auditor will seek evidence of completeness, so that the financial
statements include all material transactions that occurred, and so the records do not omit material
transactions for any reason.
Rights and Obligations
The various rights and obligations of the company are important management assertions inherent
in the financial statements. Thus, an auditor will obtain evidence regarding a company\'s rights,
such as proper title to assets and status of intellectual property. An auditor will be concerned
with assertions relating to the company\'s obligations, such as accounts payable balances, long-
term debts and tax liabilities. Thus, the audit objectives will be fulfilled upon validating these
specific assertions.
Valuation or Allocation
Valuation or allocation are managerial assertions which are often material to the financial
statements; thus, an auditor will diligently conduct audit procedures relating to these objectives.
Generally accepted accounting principles, or GAAP, require that certain balance sheet items be
presented using different valuation methodologies. Meeting these standards is a key audit
objective, as the risk of material misstatement is low in probability, but high in magnitude. Thus,
among other things, the historical cost of assets is verified, depreciation methods are scrutinized
and the fair value of investments are calculated to satisfy this objective.
Presentation and Disclosure
Another specific audit objective is validating the presentation of the financial statements and the
adequacy of the disclosures therein. Financial statements should conform to certain requirements
and expectations, and should include the balance sheet, income statement, statement of cash
flows and the statement of owner\'s equity. Relating to disclosure, the auditor will consider the
sufficiency and clarity of footnotes and the transparency in management discussion and analysis,
so he can assess the risk of material misstatement and fulfill the audit objective.
Specific audit objectives are the application of the general audit objectives to a given class of
transactions, account balance, or presentation and disclosure. There must be at least one specific
audit object.
if one can understand a few things it is easier to solve these kind .pdf
if one can understand a few things it is easier to solve these kind of questions.
first comes to I which codes for repressor. there are 3 types - I+ Wild type and is normal and
make Z and Y products in the presence of lactose only. I- is a recessive mutation will make
products of Z and Y in the presence and the absence of lactose. Is is a dominant mutation will
stop both the products Z and Y. if you have atlease one Is in the question neither the products
will from no matter whether the lactose is presence or not.
then comes to Operator (O) ,O+ is a wild type and can produce both the products of Z and Y. in
the presence of lactose.and Oc is the mutation and can produce both Z and Y products in the
Presence and even in the absence of lactose.
next comes to Z+ represents production of -galactosidase and Z- represents no production of -
galactosidase.
1.Ans. I+P-O+Z+Y+
In this haploid genotype the transcription of operon genes totally absent. Transcription of
structural genes is not occured. because here P- indicates the mutation in the promoter site, the
RNA polymerase unable to bind to the Promoter and initiate transcription of structural genes. so
these cells cannot able to grow on media which contain lactose as sole carbon source. the Strain
is lac-
2.Ans.I+P+OcZ+Y-
here Oc is the mutation and can produce both Z and Y products in the Presence and even in the
absence of lactose means constitutively. but here only the product beta galctosidase produced
constitutively and the permease is not produced because Permease gene is is mutated (Y-). these
cells also cannot able to grow on media which contain lactose as sole carbon source.because the
absence of functional permease enzyme so the Strain is lac-
3.Ans. I-P+OcZ+Y+
here I- is a recessive mutation will make products of Z and Y in the presence and the absence of
lactose and Oc is the mutation and can produce both Z and Y products in the Presence and even
in the absence of lactose. so this haploid genotype transcribes operon genes constitutively. and
both the products of Z and Y expressed continuously, so these cells able to grow on the media
which contain lactose as sole carbon source. the strain is lac+
b. partially diploid cap-I+P+O+Z-Y+/Cap+I-P+O+Z+Y-
these cells can able to grow on lactose medium because these cells have presence of wild
promoter which can able to active the transcription of structural genes. so these cells can able to
grow on the media which contain lactose as sole carbon source.
ii. Transcription of beta-galactosidase and permease is Inducible
though diploid cell have mutated inducer region which can able to activate constitutive
expression but Wild type (I+) is dominant over I- . so it is inducible operon.
Solution
if one can understand a few things it is easier to solve these kind of questions.
first comes to I which codes for repressor. there are 3 types - I+ Wild type and is normal and
make Z and Y products in the presence of lactose only. I- is a recessive mutation .
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if one can understand a few things it is easier to solve these kind .pdf
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