C code that terminates the program

C code that terminates the program

The following code snippet asks for a character which is known for exit. and then enters a loop where it asks for characters one by one and checks for necessary exit value. If found, it exits.

#include<stdio.h> 

#include<conio.h>

void main(){

char exitchar,newchar;

printf("Enter the character for exit.");

scanf("%c",exitchar);       //The character to exit is given as input

while ((newchar=getch())!=exitchar){     //getch() accepts a character and first places in newchar then checks for exit.

printf("%c",newchar);  //print on screen

}

}

You may write the equivalent code 

void main(){

char exitchar,newchar;

printf("Enter the character for exit.");

scanf("%c",exitchar);       //The character to exit is given as input

while (1){     

newchar=getch();

If (newchar==exitchar)

return(0);      //If the exitcharacter is found.

else

printf("%c",newchar);  //print on screen

}

}

C code that terminates the program

I tried this and it worked.

#include<conio.h>

#include<stdio.h>

void main(void) {

    char e,n;

    printf ("Enter the character for exit: ");

    scanf ("%c", e);                    // here e stores the exit character

    while ((n=getch())!=e) {            // the value is stored in n by getch() and checks for the exit value

        printf ("%c",n);                // values are shown on the screen until the e entered

    }

}

This is the code in C++ environment which might be helpful to you …

#include<conio.h>

#include<iostream.h>

#include<stdio.h>

void main(void) {

    char e,n;

    cout << "Enter the character for exit: ";

    cin >> e;                    // here e stores the exit character

    while ((n=getch())!=e) {    // the value is stored in n by getch() and checks for the exit value

        cout << n;                // values are shown on the screen until the e entered

    }

}

THANK YOU FOR THE HELP!!