#include stdio.h#include stdlib.htypedef FILE stream.docx
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#include <stdio.h> #include <stdlib.h> typedef FILE* stream; typedef char* string; #define MAX_STUDENT 10 typedef struct { int id; char name[30]; }stStud; #define KEY 0xAB const string menu[]={"Add new student", "print Student information", "Quit"}; int displayMenu(void); void fnEncryptDecrypt(unsigned char * inBuf, int numBytes, unsigned char * outBuf); void AddNewStud(stStud * studList, int * count); void PrintStudentList(stStud * studList, int count); void loadStudInfoFromFile(stStud * studList, int * count); void WriteStudInfoToFile(stStud * studList, int count); int main(void) { stStud studList[MAX_STUDENT]; int count = 0; int sel; loadStudInfoFromFile(studList, &count); do { sel = displayMenu(); switch(sel) { case 1: AddNewStud(studList, &count); break; case 2: PrintStudentList(studList, count); break; case 3: // quit break; default: puts("Invalid Selection!!!"); } }while(sel!=3); WriteStudInfoToFile(studList, count); return 0; } int displayMenu(void) { int i; int sel; puts("\t\tStudent Menu"); for(i=0 ; i<sizeof(menu)/sizeof(menu[0]) ; ++i) { printf("%i. %s\n", i+1, menu[i]); } printf("Select: "); scanf("%i", &sel); return sel; } //add students to the array void AddNewStud(stStud * studList, int * count) { printf("Enter Student ID:"); scanf("%i", &studList[*count].id); printf("Enter Student name:"); scanf(" %39[^\n]", studList[*count].name); (*count)++; } void PrintStudentList(stStud * studList, int count) { int i; for(i=0;i<count;++i) { printf("Stud ID: %i, \t Student Name: %s\n", studList[i].id, studList[i].name); } } void loadStudInfoFromFile(stStud * studList, int * count) { int ret; stream in; unsigned char* buf; in = fopen("stud.bin", "rb"); if(in != NULL) { buf = (unsigned char *)malloc (sizeof(int)); ret = fread(buf,sizeof(int),1,in); if(ret == 1) { fnEncryptDecrypt(buf, sizeof(int), (unsigned char *) count); free(buf); buf = (unsigned char *) malloc ((*count) * sizeof(stStud)); fread(buf,(*count) * sizeof(stStud),1,in); fnEncryptDecrypt(buf , (*count) * sizeof(stStud), (unsigned char *) studList); } free(buf); fclose(in); } } void WriteStudInfoToFile(stStud * studList, int count) { stream out; unsigned char * buf; buf = malloc (sizeof(int)); fnEncryptDecrypt((unsigned char *) &count, sizeof(int), buf); out = fopen("stud.bin", "wb"); fwrite(buf,sizeof(int),1,out); free(buf); buf =(unsigned char*) malloc (count * sizeof(stStud)); fnEncryptDecrypt((unsigned char *) studList, count * sizeof(stStud), buf); fwrite(buf,count * sizeof(stStud),1,out); free(buf); fclose(out); } void fnEncryptDecrypt(unsigned char * inBuf, int numBytes, unsigned char * outBuf) { int i; for(i=0; i< numBytes;++i) { outBuf[i] = inBuf[i] ^ KEY; } } /***** ...
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![#include <stdio.h>
#include <stdlib.h>
typedef FILE* stream;
typedef char* string;
#define MAX_STUDENT 10
typedef struct
{
int id;
char name[30];
}stStud;
#define KEY 0xAB
const string menu[]={"Add new student", "print Student
information", "Quit"};](https://image.slidesharecdn.com/includestdio-221031171342-642f641d/85/include-stdio-h-include-stdlib-htypedef-FILE-stream-docx-1-320.jpg)
![int displayMenu(void);
void fnEncryptDecrypt(unsigned char * inBuf, int numBytes,
unsigned char * outBuf);
void AddNewStud(stStud * studList, int * count);
void PrintStudentList(stStud * studList, int count);
void loadStudInfoFromFile(stStud * studList, int * count);
void WriteStudInfoToFile(stStud * studList, int count);
int main(void)
{
stStud studList[MAX_STUDENT];
int count = 0;
int sel;
loadStudInfoFromFile(studList, &count);
do
{](data:image/gif;base64,R0lGODlhAQABAIAAAAAAAP///yH5BAEAAAAALAAAAAABAAEAAAIBRAA7)
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Program of sorting using shell sort #include stdio.h #de.pdf
/* Program of sorting using shell sort */
#include
#define MAX 20
main()
{
int arr[MAX], i,j,k,n,incr;
printf(\"Enter the number of elements : \");
scanf(\"%d\",&n);
for(i=0;i=1)
{
for(j=incr;j= 0 && k < arr[i]; i = i-incr)
arr[i+incr]=arr[i];
arr[i+incr]=k;
}
printf(\"Increment=%d \ \",incr);
for (i = 0; i < n; i++)
printf(\"%d \", arr[i]);
printf(\"\ \");
incr=incr-2; /*Decrease the increment*/
}/*End of while*/
printf(\"Sorted list is :\ \");
for (i = 0; i < n; i++)
printf(\"%d \", arr[i]);
printf(\"\ \");
}/*End of main()*/
redix sort
/*Program of sorting using radix sort*/
# include
# include
struct node
{
int info ;
struct node *link;
}*start=NULL;
main()
{
struct node *tmp,*q;
int i,n,item;
printf(\"Enter the number of elements in the list : \");
scanf(\"%d\", &n);
for(i=0;iinfo=item;
tmp->link=NULL;
if(start==NULL) /* Inserting first element */
start=tmp;
else
{
q=start;
while(q->link!=NULL)
q=q->link;
q->link=tmp;
}
}/*End of for*/
printf(\"Unsorted list is :\ \");
display();
radix_sort();
printf(\"Sorted list is :\ \");
display ();
}/*End of main()*/
display()
{
struct node *p=start;
while( p !=NULL)
{
printf(\"%d \", p->info);
p= p->link;
}
printf(\"\ \");
}/*End of display()*/
radix_sort()
{
int i,k,dig,maxdig,mindig,least_sig,most_sig;
struct node *p, *rear[10], *front[10];
least_sig=1;
most_sig=large_dig(start);
for(k = least_sig; k <= most_sig ; k++)
{
printf(\"PASS %d : Examining %dth digit from right \",k,k);
for(i = 0 ; i <= 9 ; i++)
{
rear[i] = NULL;
front[i] = NULL ;
}
maxdig=0;
mindig=9;
p = start ;
while( p != NULL)
{
/*Find kth digit in the number*/
dig = digit(p->info, k);
if(dig>maxdig)
maxdig=dig;
if(diglink = p ;
rear[dig] = p ;
p=p->link;/*Go to next number in the list*/
}/*End while */
/* maxdig and mindig are the maximum amd minimum
digits of the kth digits of all the numbers*/
printf(\"mindig=%d maxdig=%d\ \",mindig,maxdig);
/*Join all the queues to form the new linked list*/
start=front[mindig];
for(i=mindig;ilink=front[i+1];
else
rear[i+1]=rear[i];
}
rear[maxdig]->link=NULL;
printf(\"New list : \");
display();
}/* End for */
}/*End of radix_sort*/
/* This function finds number of digits in the largest element of the list */
int large_dig()
{
struct node *p=start ;
int large = 0,ndig = 0 ;
while(p != NULL)
{
if(p ->info > large)
large = p->info;
p = p->link ;
}
printf(\"Largest Element is %d , \",large);
while(large != 0)
{
ndig++;
large = large/10 ;
}
printf(\"Number of digits in it are %d\ \",ndig);
return(ndig);
} /*End of large_dig()*/
/*This function returns kth digit of a number*/
int digit(int number, int k)
{
int digit, i ;
for(i = 1 ; i <=k ; i++)
{
digit = number % 10 ;
number = number /10 ;
}
return(digit);
}/*End of digit()*/
Solution
/* Program of sorting using shell sort */
#include
#define MAX 20
main()
{
int arr[MAX], i,j,k,n,incr;
printf(\"Enter the number of elements : \");
scanf(\"%d\",&n);
for(i=0;i=1)
{
for(j=incr;j= 0 && k < arr[i]; i = i-incr)
arr[i+incr]=arr[i];
arr[i+incr]=k;
}
printf(\.
IN C LANGUAGE- I've been trying to finish this program for the last fe.docx
IN C LANGUAGE: I've been trying to finish this program for the last few days, and I'm stuck on one part of it, and I can't figure out what to do.
The goal is to read in a file (separated by commas), put the information into a linked list, and then sort that linked list by either first or last name using function pointers.
Everything I have so far works, aside from the fact that my output is never sorted. I'm doing something wrong with the listSort function, and I don't know what it is.
Can anyone please just take a look at what I have and offer a suggestion or solution? It might not even be an issue with the function itself, but something somewhere else.
I would greatly appreciate it.
Please do not make any changes to the variables in the structs. Month has to be a string, it cannot be an integer.
driver.c:
#include "functions.h"
#include <stdio.h>
#include <stdlib.h>
#include <assert.h>
#include <string.h>
int main(int argc, char* argv[]) {
//creating input and output files
FILE* in = NULL;
FILE* out = NULL;
node_t* head = NULL;
node_t* temp = NULL;
//checking for 4 command line arguments
assert(argc == 4);
//opening input file
in = fopen(argv[1], "r");
assert(in != NULL);
//opening output file
out = fopen(argv[2], "w");
assert(out != NULL);
int sortBy = atoi(argv[4]);
temp = createList(in, &head);
if(sortBy == 1) {
sortList(&head, compare_by_firstname);
}
else if(sortBy == 2) {
sortList(&head, compare_by_lastname);
}
printList(out, temp);
deleteList(&head);
//closing files
fclose(in);
fclose(out);
return 0;
}
functions.c:
#include "functions.h"
#include <stdio.h>
#include <stdlib.h>
#include <assert.h>
#include <string.h>
#include <stdbool.h>
bool isLeapYear(int year) {
if(year % 4 != 0) {
return false;
}
else if(year % 100 != 0) {
return true;
}
else if(year % 400 != 0) {
return false;
}
else {
return true;
}
}
void printBorder(FILE* out)
{
int i = 0;
for(i = 0; i < 80; i++)
fprintf(out,"*");
fprintf(out, "\n");
}
bool checkInvalidDate(node_t *node) {
int year = (int)node->birthday.year;
if(node->birthday.day < 1 || node->birthday.year < 1) {
return true;
}
if(strcmp(node->birthday.month, "February") == 0) {
if(isLeapYear(year)) {
if(node->birthday.day > 29) {
return true;
}
}
else if(node->birthday.day > 28) {
return true;
}
}
else if((strcmp(node->birthday.month, "April") == 0) || (strcmp(node->birthday.month, "June") == 0) ||
(strcmp(node->birthday.month, "September") == 0) || (strcmp(node->birthday.month, "November") == 0)) {
if(node->birthday.day > 30) {
return true;
}
}
else {
if(node->birthday.day > 31) {
return true;
}
}
return false;
}
void add(node_t **node, node_t **head) {
node_t* current = *head;
while(current != NULL) {
//comparing first and last names to check for duplicates; if duplicate the info isn't added to the list
if((strcmp(current->firstName, (*node)->firstName) == 0) && (strcmp(current->lastName, (*node)->lastName) == 0)) {
return;
}
current = current->next;
}
//adding information as a node to the list
(*node.
HERE is a C PROGRAM BELOW EVERYTIME I TRY TO RUN IT ON MATRIX IT GI.pdf
Given the exponential function y = (0.5)^x: a. Complete Table 2. Table 2: End behavior of y =
(0.5)x b. As x rightarrow infinity, (0.5)^x rightarrow c. As x rightarrow -infinity, (0.5)^x
rightarrow
Solution
b. As x tends to infinity, (0.5)^x = (1/2)^infinity = 1/(2^infinity) approaches to zero.
Hence answer is 0
c. As x tends to - infinity, (0.5)^x = (1/2)^(-infinity) = 2^infinity approaches to infinity
Hence answer is infinity..
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Program of sorting using shell sort #include stdio.h #de.pdf
/* Program of sorting using shell sort */
#include
#define MAX 20
main()
{
int arr[MAX], i,j,k,n,incr;
printf(\"Enter the number of elements : \");
scanf(\"%d\",&n);
for(i=0;i=1)
{
for(j=incr;j= 0 && k < arr[i]; i = i-incr)
arr[i+incr]=arr[i];
arr[i+incr]=k;
}
printf(\"Increment=%d \ \",incr);
for (i = 0; i < n; i++)
printf(\"%d \", arr[i]);
printf(\"\ \");
incr=incr-2; /*Decrease the increment*/
}/*End of while*/
printf(\"Sorted list is :\ \");
for (i = 0; i < n; i++)
printf(\"%d \", arr[i]);
printf(\"\ \");
}/*End of main()*/
redix sort
/*Program of sorting using radix sort*/
# include
# include
struct node
{
int info ;
struct node *link;
}*start=NULL;
main()
{
struct node *tmp,*q;
int i,n,item;
printf(\"Enter the number of elements in the list : \");
scanf(\"%d\", &n);
for(i=0;iinfo=item;
tmp->link=NULL;
if(start==NULL) /* Inserting first element */
start=tmp;
else
{
q=start;
while(q->link!=NULL)
q=q->link;
q->link=tmp;
}
}/*End of for*/
printf(\"Unsorted list is :\ \");
display();
radix_sort();
printf(\"Sorted list is :\ \");
display ();
}/*End of main()*/
display()
{
struct node *p=start;
while( p !=NULL)
{
printf(\"%d \", p->info);
p= p->link;
}
printf(\"\ \");
}/*End of display()*/
radix_sort()
{
int i,k,dig,maxdig,mindig,least_sig,most_sig;
struct node *p, *rear[10], *front[10];
least_sig=1;
most_sig=large_dig(start);
for(k = least_sig; k <= most_sig ; k++)
{
printf(\"PASS %d : Examining %dth digit from right \",k,k);
for(i = 0 ; i <= 9 ; i++)
{
rear[i] = NULL;
front[i] = NULL ;
}
maxdig=0;
mindig=9;
p = start ;
while( p != NULL)
{
/*Find kth digit in the number*/
dig = digit(p->info, k);
if(dig>maxdig)
maxdig=dig;
if(diglink = p ;
rear[dig] = p ;
p=p->link;/*Go to next number in the list*/
}/*End while */
/* maxdig and mindig are the maximum amd minimum
digits of the kth digits of all the numbers*/
printf(\"mindig=%d maxdig=%d\ \",mindig,maxdig);
/*Join all the queues to form the new linked list*/
start=front[mindig];
for(i=mindig;ilink=front[i+1];
else
rear[i+1]=rear[i];
}
rear[maxdig]->link=NULL;
printf(\"New list : \");
display();
}/* End for */
}/*End of radix_sort*/
/* This function finds number of digits in the largest element of the list */
int large_dig()
{
struct node *p=start ;
int large = 0,ndig = 0 ;
while(p != NULL)
{
if(p ->info > large)
large = p->info;
p = p->link ;
}
printf(\"Largest Element is %d , \",large);
while(large != 0)
{
ndig++;
large = large/10 ;
}
printf(\"Number of digits in it are %d\ \",ndig);
return(ndig);
} /*End of large_dig()*/
/*This function returns kth digit of a number*/
int digit(int number, int k)
{
int digit, i ;
for(i = 1 ; i <=k ; i++)
{
digit = number % 10 ;
number = number /10 ;
}
return(digit);
}/*End of digit()*/
Solution
/* Program of sorting using shell sort */
#include
#define MAX 20
main()
{
int arr[MAX], i,j,k,n,incr;
printf(\"Enter the number of elements : \");
scanf(\"%d\",&n);
for(i=0;i=1)
{
for(j=incr;j= 0 && k < arr[i]; i = i-incr)
arr[i+incr]=arr[i];
arr[i+incr]=k;
}
printf(\.
IN C LANGUAGE- I've been trying to finish this program for the last fe.docx
IN C LANGUAGE: I've been trying to finish this program for the last few days, and I'm stuck on one part of it, and I can't figure out what to do.
The goal is to read in a file (separated by commas), put the information into a linked list, and then sort that linked list by either first or last name using function pointers.
Everything I have so far works, aside from the fact that my output is never sorted. I'm doing something wrong with the listSort function, and I don't know what it is.
Can anyone please just take a look at what I have and offer a suggestion or solution? It might not even be an issue with the function itself, but something somewhere else.
I would greatly appreciate it.
Please do not make any changes to the variables in the structs. Month has to be a string, it cannot be an integer.
driver.c:
#include "functions.h"
#include <stdio.h>
#include <stdlib.h>
#include <assert.h>
#include <string.h>
int main(int argc, char* argv[]) {
//creating input and output files
FILE* in = NULL;
FILE* out = NULL;
node_t* head = NULL;
node_t* temp = NULL;
//checking for 4 command line arguments
assert(argc == 4);
//opening input file
in = fopen(argv[1], "r");
assert(in != NULL);
//opening output file
out = fopen(argv[2], "w");
assert(out != NULL);
int sortBy = atoi(argv[4]);
temp = createList(in, &head);
if(sortBy == 1) {
sortList(&head, compare_by_firstname);
}
else if(sortBy == 2) {
sortList(&head, compare_by_lastname);
}
printList(out, temp);
deleteList(&head);
//closing files
fclose(in);
fclose(out);
return 0;
}
functions.c:
#include "functions.h"
#include <stdio.h>
#include <stdlib.h>
#include <assert.h>
#include <string.h>
#include <stdbool.h>
bool isLeapYear(int year) {
if(year % 4 != 0) {
return false;
}
else if(year % 100 != 0) {
return true;
}
else if(year % 400 != 0) {
return false;
}
else {
return true;
}
}
void printBorder(FILE* out)
{
int i = 0;
for(i = 0; i < 80; i++)
fprintf(out,"*");
fprintf(out, "\n");
}
bool checkInvalidDate(node_t *node) {
int year = (int)node->birthday.year;
if(node->birthday.day < 1 || node->birthday.year < 1) {
return true;
}
if(strcmp(node->birthday.month, "February") == 0) {
if(isLeapYear(year)) {
if(node->birthday.day > 29) {
return true;
}
}
else if(node->birthday.day > 28) {
return true;
}
}
else if((strcmp(node->birthday.month, "April") == 0) || (strcmp(node->birthday.month, "June") == 0) ||
(strcmp(node->birthday.month, "September") == 0) || (strcmp(node->birthday.month, "November") == 0)) {
if(node->birthday.day > 30) {
return true;
}
}
else {
if(node->birthday.day > 31) {
return true;
}
}
return false;
}
void add(node_t **node, node_t **head) {
node_t* current = *head;
while(current != NULL) {
//comparing first and last names to check for duplicates; if duplicate the info isn't added to the list
if((strcmp(current->firstName, (*node)->firstName) == 0) && (strcmp(current->lastName, (*node)->lastName) == 0)) {
return;
}
current = current->next;
}
//adding information as a node to the list
(*node.
HERE is a C PROGRAM BELOW EVERYTIME I TRY TO RUN IT ON MATRIX IT GI.pdf
Given the exponential function y = (0.5)^x: a. Complete Table 2. Table 2: End behavior of y =
(0.5)x b. As x rightarrow infinity, (0.5)^x rightarrow c. As x rightarrow -infinity, (0.5)^x
rightarrow
Solution
b. As x tends to infinity, (0.5)^x = (1/2)^infinity = 1/(2^infinity) approaches to zero.
Hence answer is 0
c. As x tends to - infinity, (0.5)^x = (1/2)^(-infinity) = 2^infinity approaches to infinity
Hence answer is infinity..
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The document describes a C program to merge two sorted integer files (DATA1 and DATA2) into a single sorted output file (DATA). The program takes the file names as command line arguments. It reads the sorted integers from both input files into arrays, merges the arrays while sorting, and writes the merged sorted list to the output file. Functions are used to sort the input arrays and write the sorted data to the input files. Sample run shows input, intermediate sorted files and final merged output file.
c programming
03 of 03 parts
Get Part 1 from https://www.slideshare.net/ArunUmrao/notes-for-c-programming-for-bca-mca-b-sc-msc-be-amp-btech-1st-year-1
Get Part 2 from https://www.slideshare.net/ArunUmrao/notes-for-c-programming-for-bca-mca-b-sc-msc-be-amp-btech-1st-year-2
C is a general-purpose, procedural computer programming language supporting structured programming, lexical variable scope, and recursion, while a static type system prevents unintended operations. C provides constructs that map efficiently to typical machine instructions and has found lasting use in applications previously coded in assembly language. Such applications include operating systems and various application software for computers, from supercomputers to PLCs and embedded system.
C program
This document contains code snippets for various operations on linked lists and polynomials in C programming language. It includes 9 questions covering topics like:
1. Counting characters, words, digits in a string
2. Squeezing a string by removing spaces
3. Swapping values using pointers
4. Comparing two strings
5. Concatenating two strings
6. Multiplying two matrices
7. Reversing a string
8. Performing insertion, deletion and traversal on singly linked lists
9. Implementing polynomial addition and multiplication by representing polynomials as linked lists
For each question, the C code to implement the operation is provided along with sample input/output.
Presentation for structure in c
Structures in C allow grouping of related data types under a single name. To define a structure, the struct keyword is used followed by the structure name and a list of member variables enclosed in curly braces. Structure variables are declared just like regular variables but with the structure name type. Members can be accessed using the dot (.) operator. An example demonstrates creating structure variables to represent distances with feet and inches members, taking input and adding the distances by manipulating the structure members.
(C++ programming)menu quit, book show, book change, book remo.docx
This document outlines a C++ program for a library management system with the following key features:
1. It allows users to access an admin section or library section with passwords. The admin section allows adding/viewing books and members while the library section allows searching, issuing, returning books.
2. Books, members, issued/returned details are stored in text files. Functions are defined to read/write these files.
3. When a book is issued, its availability status is changed and due date recorded. Returning a book updates its status and records the transaction.
4. Other options include searching for books/members, checking member/book details, viewing books due today.
#include stdio.hint main() { int count; FILE myFi.pdf
#include
int main()
{
int count;
FILE *myFile;
myFile = fopen(\"input.txt\",\"r\");
long numbers[8];
int i;
for(i=0;i<8;i++)
{
fscanf(myFile,\"%ld\",&numbers[i]);
}
for (i=0;i<8;i++)
{
printf(\"Number is: %ld\ \ \", numbers[i]);
}
for(i=0;i<8;i++)
{
count=0; //initialize count=0 for every integer
while(numbers[i]!=0)
{
numbers[i] /= 10; // number[i] = number[i]/10
++count;
}
printf(\"Number of digits in numbers[%d] is %d\ \",i,count);
}
return 0;
}
Output
Number is: 7
Number is: 7
Number is: 5
Number is: 6
Number
is:7
Number is:
43
Number is:
444
Number is:
8
Number of digits in numbers[0]
is 1
Number of digits in numbers[1]
is 1
Number of digits in numbers[2]
is 1
Number of digits in numbers[3]
is 1
Number of digits in numbers[4]
is 1
Number of digits in numbers[5]
is 2
Number of digits in numbers[6]
is 3
Number of digits in numbers[7] is 1
Solution
#include
int main()
{
int count;
FILE *myFile;
myFile = fopen(\"input.txt\",\"r\");
long numbers[8];
int i;
for(i=0;i<8;i++)
{
fscanf(myFile,\"%ld\",&numbers[i]);
}
for (i=0;i<8;i++)
{
printf(\"Number is: %ld\ \ \", numbers[i]);
}
for(i=0;i<8;i++)
{
count=0; //initialize count=0 for every integer
while(numbers[i]!=0)
{
numbers[i] /= 10; // number[i] = number[i]/10
++count;
}
printf(\"Number of digits in numbers[%d] is %d\ \",i,count);
}
return 0;
}
Output
Number is: 7
Number is: 7
Number is: 5
Number is: 6
Number
is:7
Number is:
43
Number is:
444
Number is:
8
Number of digits in numbers[0]
is 1
Number of digits in numbers[1]
is 1
Number of digits in numbers[2]
is 1
Number of digits in numbers[3]
is 1
Number of digits in numbers[4]
is 1
Number of digits in numbers[5]
is 2
Number of digits in numbers[6]
is 3
Number of digits in numbers[7] is 1.
struct procedure { Date dateOfProcedure; int procedureID.pdf
struct procedure
{
Date dateOfProcedure;
int procedureID;
int procedureProviderID;
};
Data.h
#include
#include
class Date
{
friend ostream& operator<<( ostream &, const Date & );
// allows easy output to a ostream
public:
Date( int m = 1, int d = 1, int y = 1900 ); // constructor, note the default values
void setDate( int, int, int ); // set the date const
Date &operator+=( int ); // add days, modify object
bool leapYear( int) const; // is this a leap year?
bool endOfMonth( int ) const; // is this end of month?
int getMonth ( ) const;
int getDay ( ) const;
int getYear ( ) const;
string getMonthString( ) const;
private:
int month;
int day;
int year;
static const int days[]; // array of days per month static const string monthName[]; // array of
month names
void helpIncrement();
// utility function
}; #endif Data.cpp // Member function definitions for Date
class in separate
date.cpp
#include
#include \"date.h\"
#include // Initialize static members at file scope; // one class-wide copy.
const int Date::days[] = { 0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31 }; const string
Date::monthName[] = { \"January\", \"February\", \"March\", \"April\", \"May\", \"June\",
\"July\", \"August\", \"September\", \"October\", \"November\", \"December\" };
// Date constructor
Date::Date(int m, int d, int y)
{
setDate(m, d, y); } // Set the date
void Date::setDate(int mm, int dd, int yy)
{
month = (mm >= 1 && mm <= 12) ? mm : 1; year = (yy >=1900&& yy <= 2100) ? yy
: 1900; // test for a leap year
if (month == 2 && leapYear(year)) day = (dd >= 1 && dd <= 29) ? dd : 1;
else
day =(dd >= 1 && dd <= days[month]) ? dd : 1;
} // Add a specific number of days to a date
const Date &Date::operator+=(int additionalDays)
{
for (int i = 0; i < additionalDays; i++) helpIncrement();
return *this; // enables cascading
} // If the year is leap year, return true; //
otherwise, return false
bool Date::leapYear(int testYear)
const
{
if (testYear % 400 == 0 || (testYear %100 != 0 && testYear % 4 == 0)) return true; // a
leap year
else
return false; // not a leap year
} // Determine if the day is the end of the month
bool Date::endOfMonth(int testDay)
const
{
if (month == 2 && leapYear(year))
return (testDay == 29); // last day of Feb. in leap year
else
return (testDay == days[month]);
} // Function to help increment the date
void Date::helpIncrement()
{
if (!endOfMonth(day))
{
// date is not at the end of the month
day++;
}
else if (month < 12)
{
// date is at the end of the month, but month < 12 day = 1; ++month;
}
else
{
// end of month and year: last day of the year
day = 1;
month = 1;
++year;
} }
// Overloaded output operator
ostream &operator<<(ostream& output, const Date &d)
{
output << d.monthName[d.month] << \' \' << d.day << \", \"<< d.year;
return output; // enables cascading
}
int Date::getMonth() const //
{
Public:
int month[20];
If(month[]!=”/0”)
Cout<<”Enter Month :-<>month;
Elseif (month[]<12)
{
Cout<>month[];
else
return month;
}
int Date::getDay() const //
{
Pu.
DSA.pdf
Data Structures are the programmatic way of storing data so that data can be used efficiently. Almost every enterprise application uses various types of data structures in one or the other way. This tutorial will give you a great understanding on Data Structures needed to understand the complexity of enterprise level applications and need of algorithms, and data structures.
9.C Programming
The document contains code snippets demonstrating various programming concepts in C including:
- Hello world program
- Infinite loops using while and for loops
- For and while loops to iterate from 0 to 5
- Increment and decrement operators
- Functions to add two numbers and using pointers
- Conditional operators to check odd/even
- Switch statements
- Checking leap years
- Palindrome, perfect, and factorial numbers using recursion
- Greatest common divisor and least common multiple
- Converting between binary and decimal
- Permutations and combinations using functions
- Pattern matching using loops and variables
(Meta 5) ejemplo vectores dev c++
This document contains code snippets in C programming language for different programs:
1) A nested for loop program that prints increasing numbers on separate lines
2) A program that takes integer input for an array and prints the values
3) A for loop program that takes integer input for an array and prints the index values
4) A for loop program that prints increasing numbers from 0 to the user input number
(Meta 5) ejemplo vectores 2 dev c++
This document contains code snippets in C programming language for different programs. The first program uses nested for loops to print the number 3 in a triangular pattern. The second program prompts the user to input values into an array and then prints the values. The third program uses a for loop to input values into an array and print the final value of the counter. The last program uses a for loop to print numbers from 0 to the user-input value.
#include stdafx.h#include iostream#include Stringusing.pdf
#include \"stdafx.h\"
#include
#include
using namespace std;
const int MAX_ARRAY_SIZE = 4;
class memberType {
public:
string name; // Name of the person
int id; // Member ID
int n_books; // Number of books
double amount; // Total amount spent
public:
memberType();
memberType(string ID, string first, string last, int books, double amount);
//
void setMemberInfo(string ID, string fName, string lName, int bPurchased, double amount);
// member Name access methods
void setMemberID(string);
void setName(string, string); // set new name
string getName(); // get Name
bool isMemberID(string) const;
int getBooksPurchased() const;
double getTotalAmountSpent() const;
void purchaseBook(double amount);
void resetbooksBoughtAndAmount();
void printMemberID() const;
void printName() const;
void printInfo() const;
private:
string memberID;
string firstName;
string lastName;
int booksPurchased;
double amountSpent;
};
class bookType
{
public:
void setTitle(string); //functions dealing with the title
string getTitle();
bool compareTitle(string);
//functions dealing with the copies of book
void setCopies(int);
void showCopies();
void updateCopies(int);
int getCopies();
void setPublisher(string); //functions dealing with publisher of book
void showPublisher();
void updatePublisher(string);
string getPublisher();
void setISBN(string); //functions dealing with ISBN of book
void showISBN();
void updateISBN(string);
string getISBN();
bool compareISBN(string);
void setPrice(double bookPrice); //functions dealing with price of book
void showPrice();
void updatePrice(double compPrice);
double getPrice();
void setAuthors(string); //functions dealing with author of book
void showAuthors();
void updateAuthor(string compAuthor);
string getAuthors(int i); //ith number of author will return
bookType();
private:
string title;
string authors[4]; //objects of type bookType can contain up to 4 authors
string publisher;
string ISBN;
double price;
int copies;
int authorsNo;
};
bookType::bookType()
{
title = \"\";
for (int i = 0; i < 4; i++)
authors[i] = \"\";
publisher = \"\";
ISBN = \"\";
price = 0;
copies = 0;
authorsNo = 0;
}
void bookType::setTitle(string myTitle)
{
title = myTitle;
}
string bookType::getTitle()
{
return title;
}
bool bookType::compareTitle(string otherTitle)
{
return (title.compare(otherTitle) == 0);
}
void bookType::setAuthors(string myAuthor)
{
authorsNo = authorsNo % 4;
if (myAuthor.compare(\"\") == 0)
return;
else
{
authors[authorsNo] = myAuthor; //store the author name
authorsNo++; //keep track of authors count
}
}
void bookType::showAuthors()
{
for (int i = 0; i < authorsNo; i++)
cout << authors[i] << \", \";
cout << \"\ \ \";
}
void bookType::updateAuthor(string myAuthor)
{
setAuthors(myAuthor);
}
string bookType::getAuthors(int i)
{
return authors[i];
}
void bookType::setCopies(int myCopies)
{
copies = myCopies;
}
void bookType::showCopies()
{
cout << \"\ \\tThe number of copies \" << copies;
}
void bookType::updateCopies(int myCopies)
{
copies = myCopie.
sodapdf-converted into ppt presentation(1).pdf
The document contains code snippets and questions related to C programming. It includes questions on functions to check birthdays, find syntax errors, count odd divisor numbers, check if a number is prime, swap bits, calculate array sums, find maximum of 3 numbers, rotate linked lists, calculate unlocking keys from locking keys, and insert strings into ternary search trees. Solutions to the questions are also provided as code snippets.
Complete DB code following the instructions Implement the D.pdf
Complete DB code following the instructions:
Implement the DB and use the unittest_db() function to test it. DO NOT CHANGE THIS CODE
FOR YOUR SUBMISSION, as it will be used to test your code against the answers (with different
input files).
a) The student ID should be the base class's key member.
b) The student ID should be automatically assigned such that the ID/key should be n if the student
is the nth student to join the school. If the student later leaves (i.e., deleted from the BST), the ID
does NOT get reassigned to another student. Thus, the student ID of the last student to join the
school should reflect the TOTAL number of students that have joined this school since its
reception (regardless of whether some have left).
5. Test your code against the provided input and output files.
a) The provided answer for the BST unit test is in "unittest_ans_t100_s100.txt". The s100 refers to
the seed of 100 (-s 100), and t100 refers to the number of elements to add to the BST (-t 100).
b) The provided answer for the DB is in "students_1_ans.txt" for the "students_1.txt" input file.
6. Make sure your code has no memory leaks (using valgrind).
7. Your code should work beyond the provided unit tests. That is, even if it does work for all the
given tests, if the code has an identifiable bug (i.e., by reading the source code), points WILL be
deducted.
For example, if I were to change
unittest_bst(num_test, seed, cout, 5); ->
unittest_bst(num_test, seed, cout, 100);
it should still work.
db.h
#ifndef DB_H_
#define DB_H_
#include <iostream>
#include "bst.h"
using namespace std;
class SNode : public Node {
private:
string first;
string last;
unsigned int age;
static unsigned int num_students;
public:
// Constructors and destructor
SNode();
SNode(string f_, string l_, unsigned int a_);
~SNode();
// public interface
void change_first(string f_);
void change_last(string l_);
string get_first();
string get_last();
unsigned int get_age();
void print_info(ostream& to);
};
#endif
main.cc
#include <iostream>
#include <fstream>
#include <string>
#include <sstream>
#include <vector>
#include <algorithm>
#include <getopt.h>
#include "bst.h"
#include "db.h"
using namespace std;
const char* const short_opt = ":ht:s:i:o:";
const struct option long_opt[] = {
{"help", 0, NULL, 'h'},
{"test", 1, NULL, 't'},
{"seed", 1, NULL, 's'},
{"input", 1, NULL, 'i'},
{"output", 1, NULL, 'o'},
{NULL, 0, NULL, 0}
};
void usage(char* argv);
void unittest_bst(unsigned int n, unsigned int seed, ostream& to,
unsigned int remain);
bool read_db(const string file_name, vector<string>& firsts,
vector<string>& lasts, vector<unsigned int>& ages);
void create_db(BST& db, const vector<string> f_, const vector<string> l_,
const vector<unsigned int> a_);
bool unittest_db(string ifile_name, string ofile_name);
/* Main function
*/
int main(int argc, char **argv)
{
int c = 0;
string test = "0";
string seed_str = "0";
string ifile_name;
string ofile_name;
if(argc < 2) {
usage(argv[0]);
return 0;
}
whil.
__MACOSX._assign3assign3.DS_Store__MACOSXassign3._.D.docx
__MACOSX/._assign3
assign3/.DS_Store
__MACOSX/assign3/._.DS_Store
assign3/assign3_part2/list.c
#include <stdio.h>
#include "list.h"
/* list helper functions */
int list_size(thread_info_list *list)
{
int cnt = 0;
if (!list) return -1;
pthread_mutex_lock(&list->lock);
list_elem *le = list->head;
while (le) {
cnt++;
le = le->next;
}
pthread_mutex_unlock(&list->lock);
return cnt;
}
int list_insert_head(thread_info_list *list, list_elem *new)
{
if (!list || !new) return -1;
pthread_mutex_lock(&list->lock);
new->next = list->head;
new->prev = 0;
if (new->next) {
new->next->prev = new;
}
list->head = new;
if (list->tail == 0) {
list->tail = new;
}
pthread_mutex_unlock(&list->lock);
return 0;
}
int list_insert_tail(thread_info_list *list, list_elem *new)
{
if (!list || !new) return -1;
pthread_mutex_lock(&list->lock);
new->prev = list->tail;
new->next = 0;
if (new->prev) {
new->prev->next = new;
}
list->tail = new;
if (list->head == 0) {
list->head = new;
}
pthread_mutex_unlock(&list->lock);
return 0;
}
int list_remove(thread_info_list *list, list_elem *old)
{
if (!old || !list) return -1;
pthread_mutex_lock(&list->lock);
if (old->next) {
old->next->prev = old->prev;
}
if (old->prev) {
old->prev->next = old->next;
}
if (list->tail == old) {
list->tail = old->prev;
}
if (list->head == old) {
list->head = old->next;
}
old->next = old->prev = 0;
pthread_mutex_unlock(&list->lock);
return 0;
}
void print_list(thread_info_list *list)
{
pthread_mutex_lock(&list->lock);
list_elem *le = list->head;
while (le) {
printf("0x%X,", (unsigned int)le->info);
le = le->next;
}
pthread_mutex_unlock(&list->lock);
printf("\n");
}
__MACOSX/assign3/assign3_part2/._list.c
assign3/assign3_part2/worker.c
#include <stdio.h>
#include <errno.h>
#include <stdlib.h>
#include <string.h>
#include <signal.h>
#include "scheduler.h"
/*******************************************************************************
*
* Implement these functions.
*
******************************************************************************/
/* Handler for SIGTERM signal */
void cancel_thread()
{
printf("Thread %u: terminating.\n", (unsigned int)pthread_self());
/* signal that done in queue */
sem_post(&queue_sem);
pthread_exit(NULL);
}
/* TODO: Handle the SIGUSR1 signal */
void suspend_thread()
{
printf("Thread %u: suspending.\n", (unsigned int)pthread_self());
/*add your code here to wait for a resume signal from the scheduler*/
printf("Thread %u: resuming.\n",(unsigned int) pthread_self());
}
/*******************************************************************************
*
*
*
******************************************************************************/
/*
* waits to gain access to the scheduler queue.
*/
static int enter_scheduler_queue(thread_info_t *info)
{
/*
* wait for available room in queue.
* create a new list entry for this thread
* store this thread info ...
#include iostream #include fstream #include map #include.pdf
#include
#include
#include
#include
#include
using namespace std;
//Used from
int index_of_largest(const int array[], int startIndex, const int size);
void selectSort(int a[], const int size);
void swap(int& i1, int& it2);
void fileOption();
void inputOption();
void unique(const int inputArray[], int uniqueArray[], int countArray[], const int size);
void printFileScreen(const int uniqueArray[], const int countArray[], const int size);
int main()
{
// int size;
char userInput[5];
//
cout << \"Read file from Input file? (Y/N) \";
cin >> userInput;
if(toupper(userInput[0]) == \'Y\')
{
fileOption();
}
else
{
inputOption();
}
return 0;
}
void selectSort(int array[], const int size)
{
int smallestIndex;
for(int i = 0;i < size - 1;i++)
{
smallestIndex = index_of_largest(array , i , size);
swap(array[i] , array[smallestIndex]);
}
}
void swap(int& i1, int& i2)
{
int temp;
temp = i1;
i1 = i2;
i2 = temp;
}
int index_of_largest(const int array[], int startIndex, const int size)
{
int min = array[startIndex] , index_of_min = startIndex;
for(int i = startIndex + 1;i < size;i++)
if(array[i] > min)
{
min = array[i];
index_of_min = i;
}
return index_of_min;
}
void unique(const int inputArray[], int uniqueArray[], int countArray[], const int size)
{
int number;
int j = 0;
int index;
for(int i = 0;i < size;i++)
{
int flag = 0;
number = inputArray[i];
//for each element in inputArray, check if it is there in uniqueArray
for (int k = 0; k < j; ++k)
{
if(number == uniqueArray[k])
{
flag = 1;
index = k;
}
}
if(flag == 0)
{
//if so, increment the countArray by 1
countArray[j]++;
// copy the element to uniqueArray
uniqueArray[j] = number;
j++;
}
else
countArray[index]++;
}
}
void fileOption()
{
int inputArray[50];
int countArray[50] = {0};
int uniqueArray[50]= {0};
ifstream in_stream;
char fileName[16];
cout << \"What file would you like to obtain input from? \";
cin >> fileName;
in_stream.open(fileName);
if(in_stream.fail())
{
cout << \"Error: Could not open file!\" << endl;
exit(1);
}
int size = 0;
while(in_stream >> inputArray[size])
{
size++;
}
selectSort(inputArray , size);
unique(inputArray , uniqueArray , countArray , size);
printFileScreen(uniqueArray , countArray , size);
}
void inputOption()
{
int inputArray[50];
int countArray[50] = {0};
int uniqueArray[50] = {0};
int size;
cout << \"How many numbers would you like to enter? \";
cin >> size;
cout << \"Enter numbers:\" << endl;
for(int i = 0;i < size;i++)
{
cin >> inputArray[i];
}
selectSort(inputArray , size);
unique(inputArray , uniqueArray , countArray , size);
printFileScreen(uniqueArray , countArray , size);
}
void printFileScreen(const int uniqueArray[], const int countArray[], const int size)
{
//cout << \"jaja\";
ofstream output_file;
output_file.open(\"lab3Exercise2.txt\");
if(output_file.fail())
{
cout << \"Error: Could not open file!\" << endl;
exit(1);
}
output_file << \"N\\t\" << \"count\ \";
for(int i = 0;i < size;i++)
{
if(uniqueArray[i] != 0)
{
output_file << uniqueArray[i] << \"\\t\";
outpu.
program on string in java Lab file 2 (3-year)
This document contains C code snippets demonstrating various string and file handling functions in C including:
1. Functions to get the length of a string, concatenate two strings, and search for a substring within a string.
2. Code implementing a B+ tree data structure with functions for initialization, traversal, insertion, and deletion.
3. A threaded binary tree implementation with functions for insertion, deletion, search, and printing the tree.
4. Additional examples of copying one string to another, reading from and writing to binary files, and counting the number of words in a file.
I have the following code and I need to know why I am receiving the .pdf
Edwards Manufacturing Company purchases two component parts from three different suppliers.
The suppliers have limited capacity, and no one supplier can meet all the company’s needs. In
addition, the suppliers charge different prices for the components. Component price data (in
price per unit) are as follows:
Each supplier has a limited capacity in terms of the total number of components it can supply.
However, as long as Edwards provides sufficient advance orders, each supplier can devote its
capacity to component 1, component 2, or any combination of the two components, if the total
number of units ordered is within its capacity. Supplier capacities are as follows:
If the Edwards production plan for the next period includes 1000 units of component 1 and 800
units of component 2, what purchases do you recommend? That is, how many units of each
component should be ordered from each supplier? Round your answers to the nearest whole
number.
What is the total purchase cost for the components? Round your answer to the nearest dollar.
$ ????SupplierComponent1231$11$10$132$10$11$11
Solution
For component 1:
supplier 2 have minimum cost and capacity 1050.
So all the component 1 (1000) can be purchased from supplier 2.
So, supplier 2 have 50 more capacity.
Now, supplier 1 have minimum cost for component 2 and have capacity of 650.
so, number of component 2 from supplier1 can be 650.
Now, we need 150 more component 2.
supplier 2 and 3 have same cost for component 2.
supllier have only 50 capacity now and supplier have 850.
So, 150 component 2 can be purchased from supplier 3.
Let y be the number of components fromSupplier123TotalcostComponent
1010000100010000Component 265001508008150Total cost18150.
cmdfile.txtsleep 5ls -latrsleep 3pwdsleep 1wc .docx
cmdfile.txt
sleep 5
ls -latr
sleep 3
pwd
sleep 1
wc /etc/passwd
CommandNode.c
#include <unistd.h>
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include "CommandNode.h"
/**
*CommandNode.c
*The function bodies for a linked list of commands
**/
/**
Example of linked list initialization in code:
//First command node (head of list)
headCommand = (CommandNode*)malloc(sizeof(CommandNode));
CreateCommandNode(headCommand, command, index, NULL);
index = 1;
.....
//Later command nodes
while ((linesize = getline(&line, &len, fp)) >= 0) {
...
index++;
nextCommand1 = (CommandNode*)malloc(sizeof(CommandNode));
CreateCommandNode(nextCommand1, command, index, NULL);
InsertCommandAfter(headCommand, nextCommand1);
}
**/
//create a new command node. usually nextCmd can be NULL and function InsertCommandAfter can be called to insert after head node.
void CreateCommandNode(CommandNode* thisNode, char cmd[20][20], int ind, CommandNode* nextCmd) {
//this is useful if you store a string (char *): strcpy(thisNode->command, cmd);
for (int i = 0; i < 20; i++)
for (int j = 0; j < 20; j++)
thisNode->command[i][j] = cmd[i][j];
thisNode->index = ind;
thisNode->nextCommandPtr = nextCmd;
return;
}
//insert node newNode after thisNode
void InsertCommandAfter(CommandNode* thisNode, CommandNode* newNode) {
CommandNode* tmpNext = NULL;
tmpNext = thisNode->nextCommandPtr;
thisNode->nextCommandPtr = newNode;
newNode->nextCommandPtr = tmpNext;
return;
}
//get next command node in linked list
CommandNode* GetNextCommand(CommandNode* thisNode) {
return thisNode->nextCommandPtr;
}
//find a command based on the pid
CommandNode* FindCommand(CommandNode* cmd, int pid) {
CommandNode* tmpNext = cmd;
while (tmpNext != NULL) {
if (tmpNext->PID == pid) { return tmpNext; }
tmpNext = tmpNext->nextCommandPtr;
}
return NULL;
}
CommandNode.h
/**
*CommandNode.h
*The struct for a Command Node and function prototypes for a linked list of commands
**/
typedef struct command_struct {
char command[20][20];
int index;
int PID;
int starttime;
struct command_struct* nextCommandPtr;
} CommandNode;
void CreateCommandNode(CommandNode* thisNode, char cmd[20][20], int ind, CommandNode* nextCmd);
void InsertCommandAfter(CommandNode* thisNode, CommandNode* newNode);
CommandNode* GetNextCommand(CommandNode* thisNode);
CommandNode* FindCommand(CommandNode* cmd, int pid);
HWdetails&helperFile/HW_details1.jpg
HWdetails&helperFile/HW_details1_1.jpg
HWdetails&helperFile/read_parse_file.c
#define _GNU_SOURCE
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
int main(void)
{
FILE * fp;
char * line = NULL;
size_t len = 0;
ssize_t read;
char newString[20][20];
int i,j,ctr;
fp = fopen("commands.txt", "r");
if (fp == NULL)
exit.
selection_sort..c#include stdio.hheader file input output func.pdf
selection_sort..c
#include //header file input output function
#define N 10//default value N=10
void selection_sort(int a[], int n);//function declaration
int *find_largest(int a[], int n);//function declaration
void swap(int *p, int *q);//function declaration
int main(void)
{//main function
int i;
int a[N];//variable declaration
int x,y;
printf(\"Enter %d numbers to be sorted: \", N);
for (i = 0; i < N; i++)
scanf(\"%d\", &a[i]);//keyboard inputting
selection_sort(a, N);//calling method
printf(\"In sorted order:\");
for (i = 0; i < N; i++)
printf(\" %d\", a[i]);
printf(\"\ \");
printf(\"largest element in the array :\");
*find_largest(a,N);//calling method
printf(\"\ Enter two variables for swapping :\");
scanf(\"%d\",&x);
scanf(\"%d\",&y);
swap(&x,&y);
return 0;
}
void selection_sort(int a[], int n)
{//function definition
int i, largest = 0, temp;
if (n == 1)
return;
for (i = 1; i < n; i++)
if (a[i] > a[largest])
largest = i;//selection sort logic
if (largest < n - 1) {
temp = a[n-1];
a[n-1] = a[largest];
a[largest] = temp;
}
selection_sort(a, n - 1);
}
int *find_largest(int *a, int n)
{//function definition
for(int i=1;i if you want to find smallest element*/
*a=*(a+i);
}
printf(\"%d\",*a);
}
void swap(int *p, int *q)
{//function definition
*p = *p + *q; // x now becomes 15
*q = *p - *q; // y becomes 10
*p = *p - *q; // x becomes 5
printf(\"After swapping of values :\");
printf(\"%d %d\ \",*p,*q);
}
output
Enter 10 numbers to be sorted: 562 1 344 6 4 3 2 4 67 8
In sorted order: 1 2 3 4 4 6 8 67 344 562
largest element in the array :562
Enter two variables for swapping :10 20
After swapping of
values :20 10
Solution
selection_sort..c
#include //header file input output function
#define N 10//default value N=10
void selection_sort(int a[], int n);//function declaration
int *find_largest(int a[], int n);//function declaration
void swap(int *p, int *q);//function declaration
int main(void)
{//main function
int i;
int a[N];//variable declaration
int x,y;
printf(\"Enter %d numbers to be sorted: \", N);
for (i = 0; i < N; i++)
scanf(\"%d\", &a[i]);//keyboard inputting
selection_sort(a, N);//calling method
printf(\"In sorted order:\");
for (i = 0; i < N; i++)
printf(\" %d\", a[i]);
printf(\"\ \");
printf(\"largest element in the array :\");
*find_largest(a,N);//calling method
printf(\"\ Enter two variables for swapping :\");
scanf(\"%d\",&x);
scanf(\"%d\",&y);
swap(&x,&y);
return 0;
}
void selection_sort(int a[], int n)
{//function definition
int i, largest = 0, temp;
if (n == 1)
return;
for (i = 1; i < n; i++)
if (a[i] > a[largest])
largest = i;//selection sort logic
if (largest < n - 1) {
temp = a[n-1];
a[n-1] = a[largest];
a[largest] = temp;
}
selection_sort(a, n - 1);
}
int *find_largest(int *a, int n)
{//function definition
for(int i=1;i if you want to find smallest element*/
*a=*(a+i);
}
printf(\"%d\",*a);
}
void swap(int *p, int *q)
{//function definition
*p = *p + *q; // x now becomes 15
*q = *p - *q; // y becomes 10
*p = *p - *q; // x bec.
Data structures
The document contains code snippets related to data structures and pointers in C programming. It includes examples of using pointers to access array elements, dynamically allocating memory using functions like malloc(), calloc(), and realloc(). It also demonstrates writing and reading data from binary files using pointers, structs, and file I/O functions. Additional examples include concatenating files, counting words and characters in a file, and implementing employee records using file operations.
After reading chapter 4, evaluate the history of the Data Encryp.docx
DES was once the standard for encryption but has since been replaced due to advances in computing power allowing for brute force attacks to break the encryption. The history of DES and its replacement demonstrates how cryptography must continually evolve to stay ahead of increasing computational abilities by enhancing encryption algorithms and key lengths.
After reading Chapter 2 and the Required Resources please discuss th.docx
After reading Chapter 2 and the Required Resources please discuss the following:
Place yourself in the US Government in the 1960s when the Internet concept was being developed.
Provide at least two ideas that would serve as controls on the general public when using the internet.
Base these rules on what you have seen people do with the Internet today.
Make sure that you properly support your post and cite the e-text or valid sources.
.
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c programming
03 of 03 parts
Get Part 1 from https://www.slideshare.net/ArunUmrao/notes-for-c-programming-for-bca-mca-b-sc-msc-be-amp-btech-1st-year-1
Get Part 2 from https://www.slideshare.net/ArunUmrao/notes-for-c-programming-for-bca-mca-b-sc-msc-be-amp-btech-1st-year-2
C is a general-purpose, procedural computer programming language supporting structured programming, lexical variable scope, and recursion, while a static type system prevents unintended operations. C provides constructs that map efficiently to typical machine instructions and has found lasting use in applications previously coded in assembly language. Such applications include operating systems and various application software for computers, from supercomputers to PLCs and embedded system.
C program
This document contains code snippets for various operations on linked lists and polynomials in C programming language. It includes 9 questions covering topics like:
1. Counting characters, words, digits in a string
2. Squeezing a string by removing spaces
3. Swapping values using pointers
4. Comparing two strings
5. Concatenating two strings
6. Multiplying two matrices
7. Reversing a string
8. Performing insertion, deletion and traversal on singly linked lists
9. Implementing polynomial addition and multiplication by representing polynomials as linked lists
For each question, the C code to implement the operation is provided along with sample input/output.
Presentation for structure in c
Structures in C allow grouping of related data types under a single name. To define a structure, the struct keyword is used followed by the structure name and a list of member variables enclosed in curly braces. Structure variables are declared just like regular variables but with the structure name type. Members can be accessed using the dot (.) operator. An example demonstrates creating structure variables to represent distances with feet and inches members, taking input and adding the distances by manipulating the structure members.
(C++ programming)menu quit, book show, book change, book remo.docx
This document outlines a C++ program for a library management system with the following key features:
1. It allows users to access an admin section or library section with passwords. The admin section allows adding/viewing books and members while the library section allows searching, issuing, returning books.
2. Books, members, issued/returned details are stored in text files. Functions are defined to read/write these files.
3. When a book is issued, its availability status is changed and due date recorded. Returning a book updates its status and records the transaction.
4. Other options include searching for books/members, checking member/book details, viewing books due today.
#include stdio.hint main() { int count; FILE myFi.pdf
#include
int main()
{
int count;
FILE *myFile;
myFile = fopen(\"input.txt\",\"r\");
long numbers[8];
int i;
for(i=0;i<8;i++)
{
fscanf(myFile,\"%ld\",&numbers[i]);
}
for (i=0;i<8;i++)
{
printf(\"Number is: %ld\ \ \", numbers[i]);
}
for(i=0;i<8;i++)
{
count=0; //initialize count=0 for every integer
while(numbers[i]!=0)
{
numbers[i] /= 10; // number[i] = number[i]/10
++count;
}
printf(\"Number of digits in numbers[%d] is %d\ \",i,count);
}
return 0;
}
Output
Number is: 7
Number is: 7
Number is: 5
Number is: 6
Number
is:7
Number is:
43
Number is:
444
Number is:
8
Number of digits in numbers[0]
is 1
Number of digits in numbers[1]
is 1
Number of digits in numbers[2]
is 1
Number of digits in numbers[3]
is 1
Number of digits in numbers[4]
is 1
Number of digits in numbers[5]
is 2
Number of digits in numbers[6]
is 3
Number of digits in numbers[7] is 1
Solution
#include
int main()
{
int count;
FILE *myFile;
myFile = fopen(\"input.txt\",\"r\");
long numbers[8];
int i;
for(i=0;i<8;i++)
{
fscanf(myFile,\"%ld\",&numbers[i]);
}
for (i=0;i<8;i++)
{
printf(\"Number is: %ld\ \ \", numbers[i]);
}
for(i=0;i<8;i++)
{
count=0; //initialize count=0 for every integer
while(numbers[i]!=0)
{
numbers[i] /= 10; // number[i] = number[i]/10
++count;
}
printf(\"Number of digits in numbers[%d] is %d\ \",i,count);
}
return 0;
}
Output
Number is: 7
Number is: 7
Number is: 5
Number is: 6
Number
is:7
Number is:
43
Number is:
444
Number is:
8
Number of digits in numbers[0]
is 1
Number of digits in numbers[1]
is 1
Number of digits in numbers[2]
is 1
Number of digits in numbers[3]
is 1
Number of digits in numbers[4]
is 1
Number of digits in numbers[5]
is 2
Number of digits in numbers[6]
is 3
Number of digits in numbers[7] is 1.
struct procedure { Date dateOfProcedure; int procedureID.pdf
struct procedure
{
Date dateOfProcedure;
int procedureID;
int procedureProviderID;
};
Data.h
#include
#include
class Date
{
friend ostream& operator<<( ostream &, const Date & );
// allows easy output to a ostream
public:
Date( int m = 1, int d = 1, int y = 1900 ); // constructor, note the default values
void setDate( int, int, int ); // set the date const
Date &operator+=( int ); // add days, modify object
bool leapYear( int) const; // is this a leap year?
bool endOfMonth( int ) const; // is this end of month?
int getMonth ( ) const;
int getDay ( ) const;
int getYear ( ) const;
string getMonthString( ) const;
private:
int month;
int day;
int year;
static const int days[]; // array of days per month static const string monthName[]; // array of
month names
void helpIncrement();
// utility function
}; #endif Data.cpp // Member function definitions for Date
class in separate
date.cpp
#include
#include \"date.h\"
#include // Initialize static members at file scope; // one class-wide copy.
const int Date::days[] = { 0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31 }; const string
Date::monthName[] = { \"January\", \"February\", \"March\", \"April\", \"May\", \"June\",
\"July\", \"August\", \"September\", \"October\", \"November\", \"December\" };
// Date constructor
Date::Date(int m, int d, int y)
{
setDate(m, d, y); } // Set the date
void Date::setDate(int mm, int dd, int yy)
{
month = (mm >= 1 && mm <= 12) ? mm : 1; year = (yy >=1900&& yy <= 2100) ? yy
: 1900; // test for a leap year
if (month == 2 && leapYear(year)) day = (dd >= 1 && dd <= 29) ? dd : 1;
else
day =(dd >= 1 && dd <= days[month]) ? dd : 1;
} // Add a specific number of days to a date
const Date &Date::operator+=(int additionalDays)
{
for (int i = 0; i < additionalDays; i++) helpIncrement();
return *this; // enables cascading
} // If the year is leap year, return true; //
otherwise, return false
bool Date::leapYear(int testYear)
const
{
if (testYear % 400 == 0 || (testYear %100 != 0 && testYear % 4 == 0)) return true; // a
leap year
else
return false; // not a leap year
} // Determine if the day is the end of the month
bool Date::endOfMonth(int testDay)
const
{
if (month == 2 && leapYear(year))
return (testDay == 29); // last day of Feb. in leap year
else
return (testDay == days[month]);
} // Function to help increment the date
void Date::helpIncrement()
{
if (!endOfMonth(day))
{
// date is not at the end of the month
day++;
}
else if (month < 12)
{
// date is at the end of the month, but month < 12 day = 1; ++month;
}
else
{
// end of month and year: last day of the year
day = 1;
month = 1;
++year;
} }
// Overloaded output operator
ostream &operator<<(ostream& output, const Date &d)
{
output << d.monthName[d.month] << \' \' << d.day << \", \"<< d.year;
return output; // enables cascading
}
int Date::getMonth() const //
{
Public:
int month[20];
If(month[]!=”/0”)
Cout<<”Enter Month :-<>month;
Elseif (month[]<12)
{
Cout<>month[];
else
return month;
}
int Date::getDay() const //
{
Pu.
DSA.pdf
Data Structures are the programmatic way of storing data so that data can be used efficiently. Almost every enterprise application uses various types of data structures in one or the other way. This tutorial will give you a great understanding on Data Structures needed to understand the complexity of enterprise level applications and need of algorithms, and data structures.
9.C Programming
The document contains code snippets demonstrating various programming concepts in C including:
- Hello world program
- Infinite loops using while and for loops
- For and while loops to iterate from 0 to 5
- Increment and decrement operators
- Functions to add two numbers and using pointers
- Conditional operators to check odd/even
- Switch statements
- Checking leap years
- Palindrome, perfect, and factorial numbers using recursion
- Greatest common divisor and least common multiple
- Converting between binary and decimal
- Permutations and combinations using functions
- Pattern matching using loops and variables
(Meta 5) ejemplo vectores dev c++
This document contains code snippets in C programming language for different programs:
1) A nested for loop program that prints increasing numbers on separate lines
2) A program that takes integer input for an array and prints the values
3) A for loop program that takes integer input for an array and prints the index values
4) A for loop program that prints increasing numbers from 0 to the user input number
(Meta 5) ejemplo vectores 2 dev c++
This document contains code snippets in C programming language for different programs. The first program uses nested for loops to print the number 3 in a triangular pattern. The second program prompts the user to input values into an array and then prints the values. The third program uses a for loop to input values into an array and print the final value of the counter. The last program uses a for loop to print numbers from 0 to the user-input value.
#include stdafx.h#include iostream#include Stringusing.pdf
#include \"stdafx.h\"
#include
#include
using namespace std;
const int MAX_ARRAY_SIZE = 4;
class memberType {
public:
string name; // Name of the person
int id; // Member ID
int n_books; // Number of books
double amount; // Total amount spent
public:
memberType();
memberType(string ID, string first, string last, int books, double amount);
//
void setMemberInfo(string ID, string fName, string lName, int bPurchased, double amount);
// member Name access methods
void setMemberID(string);
void setName(string, string); // set new name
string getName(); // get Name
bool isMemberID(string) const;
int getBooksPurchased() const;
double getTotalAmountSpent() const;
void purchaseBook(double amount);
void resetbooksBoughtAndAmount();
void printMemberID() const;
void printName() const;
void printInfo() const;
private:
string memberID;
string firstName;
string lastName;
int booksPurchased;
double amountSpent;
};
class bookType
{
public:
void setTitle(string); //functions dealing with the title
string getTitle();
bool compareTitle(string);
//functions dealing with the copies of book
void setCopies(int);
void showCopies();
void updateCopies(int);
int getCopies();
void setPublisher(string); //functions dealing with publisher of book
void showPublisher();
void updatePublisher(string);
string getPublisher();
void setISBN(string); //functions dealing with ISBN of book
void showISBN();
void updateISBN(string);
string getISBN();
bool compareISBN(string);
void setPrice(double bookPrice); //functions dealing with price of book
void showPrice();
void updatePrice(double compPrice);
double getPrice();
void setAuthors(string); //functions dealing with author of book
void showAuthors();
void updateAuthor(string compAuthor);
string getAuthors(int i); //ith number of author will return
bookType();
private:
string title;
string authors[4]; //objects of type bookType can contain up to 4 authors
string publisher;
string ISBN;
double price;
int copies;
int authorsNo;
};
bookType::bookType()
{
title = \"\";
for (int i = 0; i < 4; i++)
authors[i] = \"\";
publisher = \"\";
ISBN = \"\";
price = 0;
copies = 0;
authorsNo = 0;
}
void bookType::setTitle(string myTitle)
{
title = myTitle;
}
string bookType::getTitle()
{
return title;
}
bool bookType::compareTitle(string otherTitle)
{
return (title.compare(otherTitle) == 0);
}
void bookType::setAuthors(string myAuthor)
{
authorsNo = authorsNo % 4;
if (myAuthor.compare(\"\") == 0)
return;
else
{
authors[authorsNo] = myAuthor; //store the author name
authorsNo++; //keep track of authors count
}
}
void bookType::showAuthors()
{
for (int i = 0; i < authorsNo; i++)
cout << authors[i] << \", \";
cout << \"\ \ \";
}
void bookType::updateAuthor(string myAuthor)
{
setAuthors(myAuthor);
}
string bookType::getAuthors(int i)
{
return authors[i];
}
void bookType::setCopies(int myCopies)
{
copies = myCopies;
}
void bookType::showCopies()
{
cout << \"\ \\tThe number of copies \" << copies;
}
void bookType::updateCopies(int myCopies)
{
copies = myCopie.
sodapdf-converted into ppt presentation(1).pdf
The document contains code snippets and questions related to C programming. It includes questions on functions to check birthdays, find syntax errors, count odd divisor numbers, check if a number is prime, swap bits, calculate array sums, find maximum of 3 numbers, rotate linked lists, calculate unlocking keys from locking keys, and insert strings into ternary search trees. Solutions to the questions are also provided as code snippets.
Complete DB code following the instructions Implement the D.pdf
Complete DB code following the instructions:
Implement the DB and use the unittest_db() function to test it. DO NOT CHANGE THIS CODE
FOR YOUR SUBMISSION, as it will be used to test your code against the answers (with different
input files).
a) The student ID should be the base class's key member.
b) The student ID should be automatically assigned such that the ID/key should be n if the student
is the nth student to join the school. If the student later leaves (i.e., deleted from the BST), the ID
does NOT get reassigned to another student. Thus, the student ID of the last student to join the
school should reflect the TOTAL number of students that have joined this school since its
reception (regardless of whether some have left).
5. Test your code against the provided input and output files.
a) The provided answer for the BST unit test is in "unittest_ans_t100_s100.txt". The s100 refers to
the seed of 100 (-s 100), and t100 refers to the number of elements to add to the BST (-t 100).
b) The provided answer for the DB is in "students_1_ans.txt" for the "students_1.txt" input file.
6. Make sure your code has no memory leaks (using valgrind).
7. Your code should work beyond the provided unit tests. That is, even if it does work for all the
given tests, if the code has an identifiable bug (i.e., by reading the source code), points WILL be
deducted.
For example, if I were to change
unittest_bst(num_test, seed, cout, 5); ->
unittest_bst(num_test, seed, cout, 100);
it should still work.
db.h
#ifndef DB_H_
#define DB_H_
#include <iostream>
#include "bst.h"
using namespace std;
class SNode : public Node {
private:
string first;
string last;
unsigned int age;
static unsigned int num_students;
public:
// Constructors and destructor
SNode();
SNode(string f_, string l_, unsigned int a_);
~SNode();
// public interface
void change_first(string f_);
void change_last(string l_);
string get_first();
string get_last();
unsigned int get_age();
void print_info(ostream& to);
};
#endif
main.cc
#include <iostream>
#include <fstream>
#include <string>
#include <sstream>
#include <vector>
#include <algorithm>
#include <getopt.h>
#include "bst.h"
#include "db.h"
using namespace std;
const char* const short_opt = ":ht:s:i:o:";
const struct option long_opt[] = {
{"help", 0, NULL, 'h'},
{"test", 1, NULL, 't'},
{"seed", 1, NULL, 's'},
{"input", 1, NULL, 'i'},
{"output", 1, NULL, 'o'},
{NULL, 0, NULL, 0}
};
void usage(char* argv);
void unittest_bst(unsigned int n, unsigned int seed, ostream& to,
unsigned int remain);
bool read_db(const string file_name, vector<string>& firsts,
vector<string>& lasts, vector<unsigned int>& ages);
void create_db(BST& db, const vector<string> f_, const vector<string> l_,
const vector<unsigned int> a_);
bool unittest_db(string ifile_name, string ofile_name);
/* Main function
*/
int main(int argc, char **argv)
{
int c = 0;
string test = "0";
string seed_str = "0";
string ifile_name;
string ofile_name;
if(argc < 2) {
usage(argv[0]);
return 0;
}
whil.
__MACOSX._assign3assign3.DS_Store__MACOSXassign3._.D.docx
__MACOSX/._assign3
assign3/.DS_Store
__MACOSX/assign3/._.DS_Store
assign3/assign3_part2/list.c
#include <stdio.h>
#include "list.h"
/* list helper functions */
int list_size(thread_info_list *list)
{
int cnt = 0;
if (!list) return -1;
pthread_mutex_lock(&list->lock);
list_elem *le = list->head;
while (le) {
cnt++;
le = le->next;
}
pthread_mutex_unlock(&list->lock);
return cnt;
}
int list_insert_head(thread_info_list *list, list_elem *new)
{
if (!list || !new) return -1;
pthread_mutex_lock(&list->lock);
new->next = list->head;
new->prev = 0;
if (new->next) {
new->next->prev = new;
}
list->head = new;
if (list->tail == 0) {
list->tail = new;
}
pthread_mutex_unlock(&list->lock);
return 0;
}
int list_insert_tail(thread_info_list *list, list_elem *new)
{
if (!list || !new) return -1;
pthread_mutex_lock(&list->lock);
new->prev = list->tail;
new->next = 0;
if (new->prev) {
new->prev->next = new;
}
list->tail = new;
if (list->head == 0) {
list->head = new;
}
pthread_mutex_unlock(&list->lock);
return 0;
}
int list_remove(thread_info_list *list, list_elem *old)
{
if (!old || !list) return -1;
pthread_mutex_lock(&list->lock);
if (old->next) {
old->next->prev = old->prev;
}
if (old->prev) {
old->prev->next = old->next;
}
if (list->tail == old) {
list->tail = old->prev;
}
if (list->head == old) {
list->head = old->next;
}
old->next = old->prev = 0;
pthread_mutex_unlock(&list->lock);
return 0;
}
void print_list(thread_info_list *list)
{
pthread_mutex_lock(&list->lock);
list_elem *le = list->head;
while (le) {
printf("0x%X,", (unsigned int)le->info);
le = le->next;
}
pthread_mutex_unlock(&list->lock);
printf("\n");
}
__MACOSX/assign3/assign3_part2/._list.c
assign3/assign3_part2/worker.c
#include <stdio.h>
#include <errno.h>
#include <stdlib.h>
#include <string.h>
#include <signal.h>
#include "scheduler.h"
/*******************************************************************************
*
* Implement these functions.
*
******************************************************************************/
/* Handler for SIGTERM signal */
void cancel_thread()
{
printf("Thread %u: terminating.\n", (unsigned int)pthread_self());
/* signal that done in queue */
sem_post(&queue_sem);
pthread_exit(NULL);
}
/* TODO: Handle the SIGUSR1 signal */
void suspend_thread()
{
printf("Thread %u: suspending.\n", (unsigned int)pthread_self());
/*add your code here to wait for a resume signal from the scheduler*/
printf("Thread %u: resuming.\n",(unsigned int) pthread_self());
}
/*******************************************************************************
*
*
*
******************************************************************************/
/*
* waits to gain access to the scheduler queue.
*/
static int enter_scheduler_queue(thread_info_t *info)
{
/*
* wait for available room in queue.
* create a new list entry for this thread
* store this thread info ...
#include iostream #include fstream #include map #include.pdf
#include
#include
#include
#include
#include
using namespace std;
//Used from
int index_of_largest(const int array[], int startIndex, const int size);
void selectSort(int a[], const int size);
void swap(int& i1, int& it2);
void fileOption();
void inputOption();
void unique(const int inputArray[], int uniqueArray[], int countArray[], const int size);
void printFileScreen(const int uniqueArray[], const int countArray[], const int size);
int main()
{
// int size;
char userInput[5];
//
cout << \"Read file from Input file? (Y/N) \";
cin >> userInput;
if(toupper(userInput[0]) == \'Y\')
{
fileOption();
}
else
{
inputOption();
}
return 0;
}
void selectSort(int array[], const int size)
{
int smallestIndex;
for(int i = 0;i < size - 1;i++)
{
smallestIndex = index_of_largest(array , i , size);
swap(array[i] , array[smallestIndex]);
}
}
void swap(int& i1, int& i2)
{
int temp;
temp = i1;
i1 = i2;
i2 = temp;
}
int index_of_largest(const int array[], int startIndex, const int size)
{
int min = array[startIndex] , index_of_min = startIndex;
for(int i = startIndex + 1;i < size;i++)
if(array[i] > min)
{
min = array[i];
index_of_min = i;
}
return index_of_min;
}
void unique(const int inputArray[], int uniqueArray[], int countArray[], const int size)
{
int number;
int j = 0;
int index;
for(int i = 0;i < size;i++)
{
int flag = 0;
number = inputArray[i];
//for each element in inputArray, check if it is there in uniqueArray
for (int k = 0; k < j; ++k)
{
if(number == uniqueArray[k])
{
flag = 1;
index = k;
}
}
if(flag == 0)
{
//if so, increment the countArray by 1
countArray[j]++;
// copy the element to uniqueArray
uniqueArray[j] = number;
j++;
}
else
countArray[index]++;
}
}
void fileOption()
{
int inputArray[50];
int countArray[50] = {0};
int uniqueArray[50]= {0};
ifstream in_stream;
char fileName[16];
cout << \"What file would you like to obtain input from? \";
cin >> fileName;
in_stream.open(fileName);
if(in_stream.fail())
{
cout << \"Error: Could not open file!\" << endl;
exit(1);
}
int size = 0;
while(in_stream >> inputArray[size])
{
size++;
}
selectSort(inputArray , size);
unique(inputArray , uniqueArray , countArray , size);
printFileScreen(uniqueArray , countArray , size);
}
void inputOption()
{
int inputArray[50];
int countArray[50] = {0};
int uniqueArray[50] = {0};
int size;
cout << \"How many numbers would you like to enter? \";
cin >> size;
cout << \"Enter numbers:\" << endl;
for(int i = 0;i < size;i++)
{
cin >> inputArray[i];
}
selectSort(inputArray , size);
unique(inputArray , uniqueArray , countArray , size);
printFileScreen(uniqueArray , countArray , size);
}
void printFileScreen(const int uniqueArray[], const int countArray[], const int size)
{
//cout << \"jaja\";
ofstream output_file;
output_file.open(\"lab3Exercise2.txt\");
if(output_file.fail())
{
cout << \"Error: Could not open file!\" << endl;
exit(1);
}
output_file << \"N\\t\" << \"count\ \";
for(int i = 0;i < size;i++)
{
if(uniqueArray[i] != 0)
{
output_file << uniqueArray[i] << \"\\t\";
outpu.
program on string in java Lab file 2 (3-year)
This document contains C code snippets demonstrating various string and file handling functions in C including:
1. Functions to get the length of a string, concatenate two strings, and search for a substring within a string.
2. Code implementing a B+ tree data structure with functions for initialization, traversal, insertion, and deletion.
3. A threaded binary tree implementation with functions for insertion, deletion, search, and printing the tree.
4. Additional examples of copying one string to another, reading from and writing to binary files, and counting the number of words in a file.
I have the following code and I need to know why I am receiving the .pdf
Edwards Manufacturing Company purchases two component parts from three different suppliers.
The suppliers have limited capacity, and no one supplier can meet all the company’s needs. In
addition, the suppliers charge different prices for the components. Component price data (in
price per unit) are as follows:
Each supplier has a limited capacity in terms of the total number of components it can supply.
However, as long as Edwards provides sufficient advance orders, each supplier can devote its
capacity to component 1, component 2, or any combination of the two components, if the total
number of units ordered is within its capacity. Supplier capacities are as follows:
If the Edwards production plan for the next period includes 1000 units of component 1 and 800
units of component 2, what purchases do you recommend? That is, how many units of each
component should be ordered from each supplier? Round your answers to the nearest whole
number.
What is the total purchase cost for the components? Round your answer to the nearest dollar.
$ ????SupplierComponent1231$11$10$132$10$11$11
Solution
For component 1:
supplier 2 have minimum cost and capacity 1050.
So all the component 1 (1000) can be purchased from supplier 2.
So, supplier 2 have 50 more capacity.
Now, supplier 1 have minimum cost for component 2 and have capacity of 650.
so, number of component 2 from supplier1 can be 650.
Now, we need 150 more component 2.
supplier 2 and 3 have same cost for component 2.
supllier have only 50 capacity now and supplier have 850.
So, 150 component 2 can be purchased from supplier 3.
Let y be the number of components fromSupplier123TotalcostComponent
1010000100010000Component 265001508008150Total cost18150.
cmdfile.txtsleep 5ls -latrsleep 3pwdsleep 1wc .docx
cmdfile.txt
sleep 5
ls -latr
sleep 3
pwd
sleep 1
wc /etc/passwd
CommandNode.c
#include <unistd.h>
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include "CommandNode.h"
/**
*CommandNode.c
*The function bodies for a linked list of commands
**/
/**
Example of linked list initialization in code:
//First command node (head of list)
headCommand = (CommandNode*)malloc(sizeof(CommandNode));
CreateCommandNode(headCommand, command, index, NULL);
index = 1;
.....
//Later command nodes
while ((linesize = getline(&line, &len, fp)) >= 0) {
...
index++;
nextCommand1 = (CommandNode*)malloc(sizeof(CommandNode));
CreateCommandNode(nextCommand1, command, index, NULL);
InsertCommandAfter(headCommand, nextCommand1);
}
**/
//create a new command node. usually nextCmd can be NULL and function InsertCommandAfter can be called to insert after head node.
void CreateCommandNode(CommandNode* thisNode, char cmd[20][20], int ind, CommandNode* nextCmd) {
//this is useful if you store a string (char *): strcpy(thisNode->command, cmd);
for (int i = 0; i < 20; i++)
for (int j = 0; j < 20; j++)
thisNode->command[i][j] = cmd[i][j];
thisNode->index = ind;
thisNode->nextCommandPtr = nextCmd;
return;
}
//insert node newNode after thisNode
void InsertCommandAfter(CommandNode* thisNode, CommandNode* newNode) {
CommandNode* tmpNext = NULL;
tmpNext = thisNode->nextCommandPtr;
thisNode->nextCommandPtr = newNode;
newNode->nextCommandPtr = tmpNext;
return;
}
//get next command node in linked list
CommandNode* GetNextCommand(CommandNode* thisNode) {
return thisNode->nextCommandPtr;
}
//find a command based on the pid
CommandNode* FindCommand(CommandNode* cmd, int pid) {
CommandNode* tmpNext = cmd;
while (tmpNext != NULL) {
if (tmpNext->PID == pid) { return tmpNext; }
tmpNext = tmpNext->nextCommandPtr;
}
return NULL;
}
CommandNode.h
/**
*CommandNode.h
*The struct for a Command Node and function prototypes for a linked list of commands
**/
typedef struct command_struct {
char command[20][20];
int index;
int PID;
int starttime;
struct command_struct* nextCommandPtr;
} CommandNode;
void CreateCommandNode(CommandNode* thisNode, char cmd[20][20], int ind, CommandNode* nextCmd);
void InsertCommandAfter(CommandNode* thisNode, CommandNode* newNode);
CommandNode* GetNextCommand(CommandNode* thisNode);
CommandNode* FindCommand(CommandNode* cmd, int pid);
HWdetails&helperFile/HW_details1.jpg
HWdetails&helperFile/HW_details1_1.jpg
HWdetails&helperFile/read_parse_file.c
#define _GNU_SOURCE
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
int main(void)
{
FILE * fp;
char * line = NULL;
size_t len = 0;
ssize_t read;
char newString[20][20];
int i,j,ctr;
fp = fopen("commands.txt", "r");
if (fp == NULL)
exit.
selection_sort..c#include stdio.hheader file input output func.pdf
selection_sort..c
#include //header file input output function
#define N 10//default value N=10
void selection_sort(int a[], int n);//function declaration
int *find_largest(int a[], int n);//function declaration
void swap(int *p, int *q);//function declaration
int main(void)
{//main function
int i;
int a[N];//variable declaration
int x,y;
printf(\"Enter %d numbers to be sorted: \", N);
for (i = 0; i < N; i++)
scanf(\"%d\", &a[i]);//keyboard inputting
selection_sort(a, N);//calling method
printf(\"In sorted order:\");
for (i = 0; i < N; i++)
printf(\" %d\", a[i]);
printf(\"\ \");
printf(\"largest element in the array :\");
*find_largest(a,N);//calling method
printf(\"\ Enter two variables for swapping :\");
scanf(\"%d\",&x);
scanf(\"%d\",&y);
swap(&x,&y);
return 0;
}
void selection_sort(int a[], int n)
{//function definition
int i, largest = 0, temp;
if (n == 1)
return;
for (i = 1; i < n; i++)
if (a[i] > a[largest])
largest = i;//selection sort logic
if (largest < n - 1) {
temp = a[n-1];
a[n-1] = a[largest];
a[largest] = temp;
}
selection_sort(a, n - 1);
}
int *find_largest(int *a, int n)
{//function definition
for(int i=1;i if you want to find smallest element*/
*a=*(a+i);
}
printf(\"%d\",*a);
}
void swap(int *p, int *q)
{//function definition
*p = *p + *q; // x now becomes 15
*q = *p - *q; // y becomes 10
*p = *p - *q; // x becomes 5
printf(\"After swapping of values :\");
printf(\"%d %d\ \",*p,*q);
}
output
Enter 10 numbers to be sorted: 562 1 344 6 4 3 2 4 67 8
In sorted order: 1 2 3 4 4 6 8 67 344 562
largest element in the array :562
Enter two variables for swapping :10 20
After swapping of
values :20 10
Solution
selection_sort..c
#include //header file input output function
#define N 10//default value N=10
void selection_sort(int a[], int n);//function declaration
int *find_largest(int a[], int n);//function declaration
void swap(int *p, int *q);//function declaration
int main(void)
{//main function
int i;
int a[N];//variable declaration
int x,y;
printf(\"Enter %d numbers to be sorted: \", N);
for (i = 0; i < N; i++)
scanf(\"%d\", &a[i]);//keyboard inputting
selection_sort(a, N);//calling method
printf(\"In sorted order:\");
for (i = 0; i < N; i++)
printf(\" %d\", a[i]);
printf(\"\ \");
printf(\"largest element in the array :\");
*find_largest(a,N);//calling method
printf(\"\ Enter two variables for swapping :\");
scanf(\"%d\",&x);
scanf(\"%d\",&y);
swap(&x,&y);
return 0;
}
void selection_sort(int a[], int n)
{//function definition
int i, largest = 0, temp;
if (n == 1)
return;
for (i = 1; i < n; i++)
if (a[i] > a[largest])
largest = i;//selection sort logic
if (largest < n - 1) {
temp = a[n-1];
a[n-1] = a[largest];
a[largest] = temp;
}
selection_sort(a, n - 1);
}
int *find_largest(int *a, int n)
{//function definition
for(int i=1;i if you want to find smallest element*/
*a=*(a+i);
}
printf(\"%d\",*a);
}
void swap(int *p, int *q)
{//function definition
*p = *p + *q; // x now becomes 15
*q = *p - *q; // y becomes 10
*p = *p - *q; // x bec.
Data structures
The document contains code snippets related to data structures and pointers in C programming. It includes examples of using pointers to access array elements, dynamically allocating memory using functions like malloc(), calloc(), and realloc(). It also demonstrates writing and reading data from binary files using pointers, structs, and file I/O functions. Additional examples include concatenating files, counting words and characters in a file, and implementing employee records using file operations.
Similar to #include stdio.h#include stdlib.htypedef FILE stream.docx (20)
(C++ programming)menu quit, book show, book change, book remo.docx
(C++ programming)menu quit, book show, book change, book remo.docx
#include stdio.hint main() { int count; FILE myFi.pdf
#include stdio.hint main() { int count; FILE myFi.pdf
struct procedure { Date dateOfProcedure; int procedureID.pdf
struct procedure { Date dateOfProcedure; int procedureID.pdf
#include stdafx.h#include iostream#include Stringusing.pdf
#include stdafx.h#include iostream#include Stringusing.pdf
Complete DB code following the instructions Implement the D.pdf
Complete DB code following the instructions Implement the D.pdf
__MACOSX._assign3assign3.DS_Store__MACOSXassign3._.D.docx
__MACOSX._assign3assign3.DS_Store__MACOSXassign3._.D.docx
#include iostream #include fstream #include map #include.pdf
#include iostream #include fstream #include map #include.pdf
I have the following code and I need to know why I am receiving the .pdf
I have the following code and I need to know why I am receiving the .pdf
cmdfile.txtsleep 5ls -latrsleep 3pwdsleep 1wc .docx
cmdfile.txtsleep 5ls -latrsleep 3pwdsleep 1wc .docx
selection_sort..c#include stdio.hheader file input output func.pdf
selection_sort..c#include stdio.hheader file input output func.pdf
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Advisory from Professionals
Preparing Information Systems (IS) Graduates to Meet the
Challenges of Global IT Security: Some Suggestions
Jeff Sauls
IT Operations Professional
Austin, TX, USA
Naveen Gudigantala
Operations and Technology Management
University of Portland
Portland, OR 97203, USA
[email protected]
ABSTRACT
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Problem
In no more than 250 words, describe the problem, who is affected, and the current ramifications. Explain the consequences if the issue continues.
Idea for Addressing
Solution
In no more than 250 words, outline your idea for addressing the issue and explain why legislation is the best course for advocacy.
Research the Issue
Perform research and compile information for your idea. Present substantive evidence-based findings that support your idea for addressing the problem (studies, research, and reports). Include any similar legislation introduced or passed in other states.
Evidence 1
Evidence 2
Stakeholder Support
Discuss the stakeholders who would support the proposed idea and explain why they would be in support.
Stakeholder(s) Supporting 1
Stakeholder(s) Supporting 2
Stakeholder Opposition
Discuss the stakeholders who would oppose the proposed idea. Explain why they would be in opposition and how you would prepare to debate or converse about these considerations.
Stakeholder(s) Opposed 1
Stakeholder(s) Opposed 2
Financial Incentives/Costs
In no more than 250 words, summarize the financial impact for the issue and the idea (added costs, cost savings, increased revenue, etc.). Provided support.
Legislature: Information Needed and Process for Proposal
Discuss the how to advocate for your proposal using legislation. Include the following:
Provide the name and complete contact information for the legislator.
Describe the steps for how you would present this to your legislator.
Outline the process if your legislator chooses to introduce your idea as a bill to congress.
Christian Principles and Nursing Advocacy
In no more than 250 words, discuss how principles of a Christian worldview lend support to legislative advocacy in health care without bias. Be specific as to how these principles help advocate for inclusiveness and positive health outcomes for all populations, including those more vulnerable, without regard to gender, sexual orientation, culture, race, religion/belief, etc.
© 2019. Grand Canyon University. All Rights Reserved.
2
.
Advanced pathoRespond to Stacy and Sonia 1 day agoStacy A.docx
A 16-year-old boy presented to the clinic with symptoms of strep throat including a sore throat for 3 days. On examination, he exhibited signs of strep throat including tonsil swelling and exudate. A rapid strep test was positive, confirming strep throat. He was prescribed penicillin but had an anaphylactic reaction. This type of severe allergic reaction is caused by IgE antibodies produced in response to environmental allergens like penicillin, which is a fungus. While he had no known drug allergies, he likely had a previous sensitization to penicillin that caused the anaphylactic reaction upon administration.
After completing the reading this week, we reflect on a few ke.docx
After completing the reading this week, we reflect on a few key concepts this week:
Discuss Howell and Mendez’s three perspectives on followership. Note how these behaviors influence work productivity.
Please be sure to answer all the questions above in the initial post.Please ensure the initial post and two response posts are substantive. Substantive posts will do at least TWO of the following:
Ask an interesting, thoughtful question pertaining to the topic
Expand on the topic, by adding additional thoughtful information
Answer a question posted by another student in detail
Share an applicable personal experience
Provide an outside source
Make an argument
At least one scholarly (peer-reviewed) resource should be used in the initial discussion thread. Please ensure to use information from your readings and other sources from the UC Library. Use APA references and in-text citations.Please be sure to engage by Wednesday at 11:59pm ET and then engage on two more days throughout the week (for a total of three days of engagement, before Sunday at 11:59pm, ET.
.
Adopting Enterprise Risk Management inToday’s Wo.docx
Adopting Enterprise Risk Management in
Today’s World:
An Evidence-Based Guide for Implementation
by
Dr. Steven Deck
COPYRIGHT © Steven Deck, 2017
Biography
Dr. Steven Deck has over 25 years of experience developing and
implementing risk management, environmental health and safety,
international safety and security, emergency response, and continuity of
operation programs and processes in higher education and in biomedical and
pharmaceutical industries. Dr. Deck has also lead efforts to identify and treat
risks associated with implementing a strategic plan at a large research
university. Hence, he has experience managing risks at both the operational
and strategic level. He holds a doctorate in management, an MBA, and a
bachelor’s degree in safety and industrial hygiene management. Dr. Deck
also holds an associate in risk management and is a certified industrial
hygienist, safety professional, and hazardous materials manager.
Dedication
This book is dedicated to the people who tirelessly work to reduce
risks organizations face in today’s fast-paced world. Their efforts sometimes
go unnoticed since, if successful, risk managers prevent adverse events from
occurring or significantly reduce their impact on the organization. A good
day for a risk manager is one that is uneventful with operations continuing
without interruption. However, their work is critical to an organization’s
ability to achieve its mission. By reducing risks that threaten an
organization’s survival, risk managers preserve the organization’s ability to
offer people opportunities to earn a living and provide for their families.
Indeed, a risk manager role is critical to the success of society even if their
work sometimes goes unnoticed.
Acknowledgments
First, I would like to thank my advisory committee for my dissertation,
Dr. Thomas Mierzwa and Dr. Denise Breckon. The research for my
dissertation served as the foundation for the writing of this book. Their hard
work and commitment to my growth as a scholar enabled me to grow
intellectually and develop the skills needed to write this book. I would also
like to acknowledge Dr. Roger Ward, Senior Vice President for Operations
and Institutional Effectiveness and Vice Dean for the Graduate School at the
University of Maryland Baltimore for encouraging me to pursue my doctoral
degree and continuing to support me throughout my career. Thanks also goes
to Dr. Lauren Sweetman for her guidance and editing of this book. Last, and
most importantly, I would like to thank my wife, Bonnie, for her patience and
support as I fulfilled the demanding requirements of a doctoral program and
writing this book.
Table of Contents
Introduction
Part 1: Understanding Organizational Risk and Risk Management
Chapter 1: Organizational Risk
Chapter 2: Traditional Risk Management
Chapter 3: Frameworks for ERM
Part 2: Management Science and ERM: From Theory to Practice
Chapter 4: Organiz.
Addisons diseaseYou may use the textbook as one reference a.docx
- Addison's disease is a disorder that was chosen to research. The key aspects that must be discussed include the pathophysiology, most common presenting symptoms, how it is diagnosed, and a standard treatment plan. National screening and treatment guidelines should also be referenced if available.
- Additional scholarly sources beyond the course materials should be used. When researching, look for unique information that may surprise the reader rather than just common facts.
- The discussion post will be evaluated based on criteria such as the quality of the initial post and peer responses, use of scholarly sources, organization, grammar and APA style.
AdultGeriatric DepressionIntroduction According to Mace.docx
Adult/Geriatric Depression
Introduction
According to Mace et al. (2017), geriatric depression disorder is a health condition that affects adult people with the main symptoms see for a patient with the condition being the frequent change in moods and the sadness. The condition is as well common among younger people although this is not a common health issue. As a result of the impact of the condition, there is a need for getting an understanding of the condition and the best medical process that is to be followed. With the treatment of the patients there is a need for understanding the fact that if one medication does not succeed, there is a need for implementation of a new medication or change of the prescription for the patient. This paper focuses on the options chosen for managing the conditions for a patient.
Symptoms of adult/geriatric depression disorder
A patient with geriatric depression disorder will be seen to have issues of mood swings as the main symptoms of the condition. The other symptoms that will be evident will include issues of the feeling of despair and sad and the patient will have issues of pain and aches in all parts of the body. Laird et al. (2019) allude that a patient with the health condition will have risks of loss of appetite and lack of hope or lacking any willingness to get help. There are risks of the patient as well as lacking any morale in improving their condition.
Causes
Vlasova et al. (2019) allude that one of the main risk factors that increased the chance of a patient getting geriatric depression is old age. As such, a patient who is old will be at high risk of being affected by the health condition. The other risk factor is that women are at higher risk of suffering from the health issue as compared to their male counterparts. The people who have a low-level education as well as those that have physical illnesses and influencing the standard of thinking of the person.
Thus, there is a need for healthcare providers to help in the education of the patients who are at risk of suffering from the health condition. The patients who are found to be making use of various drugs and that have psychological stressors are at high risk of being affected by the health concern. The patients as well may lead to affecting the patients who have white matter changes.
Patient case
This is a case of 32-year-old patients of Hispanic origin and who came to the US for his education. The patient has his mother passing on while he was in school and ended up being admitted to the healthcare facility as a result of depression issues. On having an assessment, it is found that the patient does not suffer from any other health issue and the patient has presented that he has had pain and stiff parts of the body.
There are various signs and symptoms expressed by the patient with the main issue faced by the patient being undermined as a result of the color of his skin. There is evidence of the patient having little socializa.
Adopt-a-Plant Project guidelinesOverviewThe purpose of this.docx
Adopt-a-Plant Project guidelines
Overview:
The purpose of this project is for you to choose a plant, conduct online research into the biology of the plant, and communicate what you have learned. You will be preparing an annotated bibliography on the plant you choose. The entire project is worth 50 points
Annotated Bibliography (50 points)
You will prepare an annotated bibliography with a list of the top 10 most interesting facts about your plant.
· Each fact should be paraphrased (i.e. written in your own words, no quotations allowed).
· Then tell me why this is interesting to you – make connections to your life or to currents issues in our world.
· Finally, give a full citation and tell me why you think this is a reliable, trustworthy source. Use this libguide to help you come up with reasons why your source is trustworthy.
· At least one of your sources should be from a peer-reviewed, science journal article.
Here is an example:
Fact 1: Taxol is a chemotherapy agent derived from the bark of the Pacific Yew Tree. The chemical itself is derived from a fungal endophtye within the bark. I thought this was very interesting, because the Pacific Yew tree is native to the state of Washington, and my aunt Jane received Taxol while undergoing chemotherapy for ovarian cancer. I also thought it was interesting because of the mutualistic relationship between the plant and the fungus.Citation: Plant natural products from cultured multipotent cells
Roberts, Susan; Kolewe, Martin. Nature Biotechnology28.11 (Nov 2010): 1175-6.
This is a reliable source because it is published in a peer-reviewed science journal article, written by two PhDs that are providing a review of the current literature on the topic
To complete the assignment, you should first choose a plant, gather articles discussing your plant, read the articles sufficiently enough to discuss the plant, and finally write the annotated bibliography. You are expected to produce original work, and any plagiarism will receive a zero. The paper should be double-spaced, and typed in 12 point font size, with normal margins. The instructions for how to properly cite your sources are at the end of this handout.
*** Reminder: The scientific name of a plant should always be typed in italics, with the first letter of the Genus capitalized. For ex.: Digitalis lanata. When you search for information on your plant online, make sure to use the scientific name, which will bring back a wider variety of results
The bibliography is worth 50 points and will be graded on:
1. Effort
• Quality of references
•Depth/breadth/quality of material covered
2. Following directions/ requirements
I will use the following rubric to grade your bibliography:
Research, Critical Reading and Documentation
Balanced, authoritative sources; correctly cited sources; effectively integrated outside sources. Most sources from science journals
10 pts
Effective sources, correctly cited, Could have a few more.
Adolescent development is broad and wide-ranging, including phys.docx
Adolescent development is broad and wide-ranging, including physical, socioemotional, and cognitive growth. It is important to have an understanding of cognitive and social development in order to meet the diverse needs of adolescent students.
For this benchmark assignment, consider information learned regarding physical development and integrate it with your understanding of the social/emotional and cognitive changes occurring during adolescence.
Write a 750-1,000 word essay addressing the following concerns:
In what ways do cognitive and social development affect learning? How do individuality, identity development, and personal behavior affect learning differences and development?
How can diverse strengths, interests, and needs of adolescents be accounted for when designing developmentally appropriate instruction that advances learning?
What is the significance of providing an environment that brings multiple perspectives into the discussion of content? What considerations should be made for the cultural norms and the personal, family, and community experiences of students?
Why are school, community, and family supports critical to student learning and the overall well-being of adolescents? How can ongoing support from these stakeholders be assured?
How can technology reinforce learning experiences? How can a teacher ensure appropriate use of digital tools in the classroom? In what ways might digital tools support student success and creativity in a collaborative, student-focused classroom environment?
Support your essay with a minimum of three scholarly resources.
Prepare this assignment according to the APA guidelines found in the APA Style Guide, located in the Student Success Center. An abstract is not required.
Benchmark Information:
MA in Secondary Education
This benchmark assignment assesses the following programmatic competencies [and professional standards]:
1.1: Create developmentally appropriate instruction that takes into account individual students’ strengths, interests, and needs and that enables each student to advance and accelerate his or her learning. [InTASC 1(a), 1(b), 1(d), 1(e), 1(h), 1(i), 1(j); ISTE-T 2b; MC3]
1.3: Bring multiple perspectives to the discussion of content, including attention to students’ personal, family, and community experiences and cultural norms. [InTASC 2(d), 2(j), 2(n), 4(m), 5(p), 10(m); MC1, MC3, MC5]
4.3: Working collaboratively with school colleagues, teacher candidates build ongoing connections with community resources to enhance student learning and well-being. [InTASC 9(d), 10(e), 10(l), 10(r); MC1, MC4]
.
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This Dissertation explores the particular circumstances of Mirzapur, a region located in the
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9
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- 3. sel = displayMenu(); switch(sel) { case 1: AddNewStud(studList, &count); break; case 2: PrintStudentList(studList, count); break; case 3: // quit break; default: puts("Invalid Selection!!!"); } }while(sel!=3); WriteStudInfoToFile(studList, count);
- 4. return 0; } int displayMenu(void) { int i; int sel; puts("ttStudent Menu"); for(i=0 ; i<sizeof(menu)/sizeof(menu[0]) ; ++i) { printf("%i. %sn", i+1, menu[i]); } printf("Select: "); scanf("%i", &sel); return sel;
- 5. } //add students to the array void AddNewStud(stStud * studList, int * count) { printf("Enter Student ID:"); scanf("%i", &studList[*count].id); printf("Enter Student name:"); scanf(" %39[^n]", studList[*count].name); (*count)++; } void PrintStudentList(stStud * studList, int count) { int i; for(i=0;i<count;++i) { printf("Stud ID: %i, t Student Name: %sn", studList[i].id, studList[i].name);
- 6. } } void loadStudInfoFromFile(stStud * studList, int * count) { int ret; stream in; unsigned char* buf; in = fopen("stud.bin", "rb"); if(in != NULL) { buf = (unsigned char *)malloc (sizeof(int)); ret = fread(buf,sizeof(int),1,in); if(ret == 1) { fnEncryptDecrypt(buf, sizeof(int), (unsigned
- 7. char *) count); free(buf); buf = (unsigned char *) malloc ((*count) * sizeof(stStud)); fread(buf,(*count) * sizeof(stStud),1,in); fnEncryptDecrypt(buf , (*count) * sizeof(stStud), (unsigned char *) studList); } free(buf); fclose(in); } } void WriteStudInfoToFile(stStud * studList, int count) { stream out; unsigned char * buf; buf = malloc (sizeof(int)); fnEncryptDecrypt((unsigned char *) &count, sizeof(int), buf);
- 8. out = fopen("stud.bin", "wb"); fwrite(buf,sizeof(int),1,out); free(buf); buf =(unsigned char*) malloc (count * sizeof(stStud)); fnEncryptDecrypt((unsigned char *) studList, count * sizeof(stStud), buf); fwrite(buf,count * sizeof(stStud),1,out); free(buf); fclose(out); } void fnEncryptDecrypt(unsigned char * inBuf, int numBytes, unsigned char * outBuf) { int i; for(i=0; i< numBytes;++i) {
- 9. outBuf[i] = inBuf[i] ^ KEY; } } /****************************Group No.07 Assignment**************************** ***************************Group Members: 1.Danda, Praveen 2.Rachamalla, Sandeep 3.Ghaida Almugrin 4.Alghamdi, Hoosam G. ***************************************************** **************************/ #include<stdio.h> /**********************STUDENT functions*****************/ void enterStudentDetails(); void printAllstudentsDetails(); int findStudentByRoleNumber(int);
- 10. void findStudentByName(); int searchGmail(char[]); /************************ BOOK *****************/ void addBook(); void printAllBooksList(); int findBookById(int); void findBookById1(); void findBookByISBN(); void findBookByName(); /********************** Enter Studente Details ******************************/ void enterStudentDetails() { int roll,ch,res; char name[50],email[50]; int flag=0,flag1=0;
- 11. FILE *fp; clrscr(); fp = fopen("student.txt","a"); printf("n ENTER ROLL NUMBER : "); scanf("%d",&roll); flag=findStudentByRoleNumber(roll); if(flag!=1) { printf("ENTER NAME :n"); scanf("%s",&name); printf("ENTER EMAIL:n"); scanf("%s",&email); flag1=searchGmail(email); if(flag1!=1) {
- 12. fprintf(fp,"%d %s %sn",roll,name,email); } else { printf("n email already registered, take a new Emailn"); } } else { printf("This Student Number already registered .If u want add new student take another New RoleNumber :n"); } fclose(fp) ; } /******************* Search gmail ************************/
- 13. int searchGmail(char semail[50]) { int i,ch,roll; char email[50],name[50]; FILE *fm; fm = fopen("student.txt","r"); i=0; while(! feof(fm)) { fscanf(fm,"%d%s%sn",&roll,&name,&email); if(strcmp(email,semail)==0) { i=1; printf("nROLLNO. NAME GMAILn "); printf(" %d %s %s",roll,name,email); break; }
- 14. } fclose(fm); return i; } /********************Print All Students Details**************/ void printAllstudentsDetails() { int rollno[100],x[100],i; char name[100][50],email[100][50]; FILE *fp; clrscr(); fp = fopen("student.txt","r"); i=0; printf("ROLLNO NAME EMAILn"); while(!feof(fp))
- 15. { fscanf(fp,"%d %s %sn",&rollno[i],&name[i],&email[i]); printf(" %d %s %sn",rollno[i],name[i],email[i]); i=i+1; } fclose(fp); printf("nnnPRESS ANY KEY"); getch(); } /************************* Print Student Details using RoleNumber ***************************/ int findStudentByRoleNumber(int roll1) { int i,ch,roll,mark,sroll; char name[50],email[50];
- 16. FILE *fm; fm = fopen("student.txt","r"); i=0; while(! feof(fm)) { fscanf(fm,"%d%s%sn",&roll,&name,&email); if(roll1==roll) { printf("nROLLNO. NAME GMAILn "); printf(" %d %s %sn",roll,name,email); i=1; break; }//endif } if(i==0)
- 17. { printf("n%d",roll1); printf(" Rolenumber does not exit in the recordn"); } fclose(fm); return i; } /************************* To find Student information using student name***************************/ void findStudentByName() { int i=0,roll,ch,mark,roll1; char name[50],email[50],sname[50]; FILE *fm; ch=1;
- 18. while(ch) { fm = fopen("student.txt","r"); printf(" n ENTER STUDENT NAME "); scanf("%s",&sname); while(! feof(fm)) { fscanf(fm,"%d%s%sn",&roll,&name,&email); if(strcmp(sname,name)==0) { i=1; printf("nROLLNO. NAME GMAILn "); printf(" %d %s %s",roll,name,email); break; } }
- 19. // printf("------i----->%d",i); if(i==0) { printf("n%s",sname); printf(" name does not exit in the record"); } printf("nnpress 1 to see Another Student info, otherwise Press 0 to return to main menu: n"); scanf("%d",&ch); fclose(fm); } } /************** BOOKS FUNCTIONS *******************/ void addBook() {
- 20. int i,ch,res,sbookid,bookid,quantity,bookISBN,booksIdPresentEnter [50]; char bookName[50],author[50]; FILE *fp; clrscr(); fp = fopen("book.txt","a"); ch =1; while(ch) { printf("Enter no of books:"); scanf("%d",&quantity); printf("Enter book isbn :"); scanf("%d",&bookISBN); printf("Enter BookNAME :"); scanf("%s",&bookName); printf("Enter Author :"); scanf("%s",&author);
- 21. for(i=quantity;i>=1;i--) { fp = fopen("book.txt","a"); res=0; printf("nEnter book Id : "); scanf("%d",&sbookid); res=findBookById(sbookid); if(res!=1) { fprintf(fp,"%d %s %s %d %d n",sbookid,bookName,author,quantity,bookISBN); booksIdPresentEnter[i]=sbookid; fclose(fp); } else { printf("nBookID what ever you enterd already registered ,Take a new BookID ");
- 22. i++; } } printf("n n press 1 to continue,0 to stop : "); scanf("%d",&ch); } fclose(fp) ; } /* --------------- PRINT ALL BOOKS-------------------------*/ void printAllBooksList() { int bookId[50],quantity[50],bookISBN[50],i; char bookName[100][50],author[50][100],ch; FILE *fp;
- 23. clrscr(); fp = fopen("book.txt","r"); i=0; printf("BookNumber BookNAME Author NoOfBooks ISBNn"); while(! feof(fp)) { fscanf(fp,"%d%s%s%d%dn",&bookId[i],&bookName[i],&autho r[i],&quantity[i],&bookISBN[i]); printf("%d %s %s %d %dn",bookId[i],bookName[i],author[i],quantity[i],bookISBN[i] ); i=i+1; } fclose(fp); printf("nnnPRESS ANY KEY"); getch(); }
- 24. /* ---------------------- find book by ID --------------------*/ void findBookById1() { int i,ch,bid,isbn,quantity,sbid; char bname[50],author[50]; FILE *fm; ch=1; while(ch) { fm = fopen("book.txt","r"); printf(" n ENTER BOOK ID :"); scanf("%d",&sbid); i=0; while(! feof(fm)) {
- 25. //printf("------- find book by ID while 1-----n"); fscanf(fm,"%d%s%s%d%dn",&bid,&bname,&author,&quantity, &isbn); if(sbid==bid) { printf("n bookID BookName Author quantity ISBNnumbern "); printf(" %d %s %s %d %dn",bid,bname,author,quantity,isbn); break; } else i=i+1; } printf("nnpress 1 to see another book info, 0 to return to main menun"); scanf("%d",&ch); fclose(fm); }
- 26. } /*-------int findBookById(int sbid)------*/ int findBookById(int sbid) { int i=0,ch,bid,isbn,quantity; char bname[50],author[50]; FILE *fm; fm = fopen("book.txt","r"); while(! feof(fm)) { fscanf(fm,"%d%s%s%d%dn",&bid,&bname,&author,&quantity, &isbn); if(sbid==bid) { printf("n bookID BookName Author quantity ISBNnumbern");
- 27. printf(" %d %s %s %d %dn",bid,bname,author,quantity,isbn); i=1; break; } } fclose(fm); return i; } /*--------------------------- findBookByISBN-----------------*/ void findBookByISBN() { int i,ch,bid,isbn,quantity,sisbn; char bname[50],author[50]; FILE *fm; ch=1;
- 28. while(ch) { fm = fopen("book.txt","r"); printf(" n ENTER BOOK ISBN :"); scanf("%d",&sisbn); i=0; while(! feof(fm)) { fscanf(fm,"%d%s%s%d%dn",&bid,&bname,&author,&quantity, &isbn); if(isbn==sisbn) { printf("n bookID BookName Author quantity ISBNnumbern "); printf(" %d %s %s %d %dn",bid,bname,author,quantity,isbn); break; } else
- 29. i=i+1; } printf("nnpress 1 to see another book info, 0 to return to main menu :"); scanf("%d",&ch); fclose(fm); } } /*--------------------------- find Book By its name----------------- */ void findBookByName() { int ch,bid,isbn,quantity; char bname[50],author[50],sbname[50]; FILE *fm; ch=1; while(ch)
- 30. { fm = fopen("book.txt","r"); printf(" n ENTER BOOK NAME :"); scanf("%s",&sbname); while(!feof(fm)) { fscanf(fm,"%d %s %s %d %sn",&bid,&bname,&author,&quantity,&isbn); if(strcmp(sbname,bname) == 0) { printf("n bookID BookName Author quantity ISBNnumbern "); printf(" %d %s %s %d %dn",bid,bname,author,quantity,isbn); break; } } printf("nnpress 1 to see another Book info, 0 to return to
- 31. main menu :"); scanf("%d",&ch); fclose(fm); } } /**************** FUNC. ENDS************************/ void main() { int i,rollno[100],x[100],n,j,roll,c,roll1,sroll,res,sbid; char ch,name[100][10],email; while(c!=10) { printf("GIVE CHOICE--n"); printf(" 1 TO ENTER STUDENT INFORMATIONn"); printf(" 2 TO SEE ALL STUDENTS INFORMATIONn");
- 32. printf(" 3 TO PRINT STUDENT INFOORMATION USING ROLL NOn"); printf(" 4 TO PRINT STUDENT INFO. USING HIS NAMEn"); printf(" 5 TO ENTER New Book INFO.n"); printf(" 6 TO PRINT ALL BOOKS INFO.n"); printf(" 7 TO Find BOOK INFO by BookById.n"); printf(" 8 TO Find Book Info By ISBN number.n"); printf(" 9 TO Find Book Info By its Name.n"); printf(" 10 QUIT nn--"); scanf("%d",&c); clrscr(); switch(c) { case 1: enterStudentDetails(); break; case 2:
- 33. printAllstudentsDetails(); break; case 3: do { res=0; printf("Enter Student Roll Number-----> "); scanf("%d",&roll1); res=findStudentByRoleNumber(roll1); printf("nDo u want search again if yes PRESS 1 else PRESS 0 :"); scanf("%d",&i); }while(i!=0); if(res==0) { printf("nThis RoleNumberis not yet registered");
- 34. } break; case 4: findStudentByName(); break; case 5: addBook(); break; case 6: printAllBooksList(); break; case 7: findBookById1();
- 35. break; case 8: findBookByISBN(); break; case 9: findBookByName(); break; case 10: exit(); break; default: break; } }
- 36. } S14_CSCI 6610 Assignment 3 Enhance your library system with the following set of requirements: 1) Use dynamic memory allocation for the book array and student array. This will help your program to expand without any limitation on fixed number of books or students. 2) Use binary files to store the data instead of using text files. You’ll be using binary modes instead of a text mode. 3) Data has to be encrypted before it gets stored in file. Use 8 bits key encryption using “exclusive or” operator (^) as we discussed in the class. Data has to be decrypted before loading the arrays again.