\n

[vgoel38]

this code works only if there is \n before %c in scanf statement. please explain me why?

#include<stdio.h>
#include<conio.h>

int main()
{
char ch;
do{
printf("k");
printf("do u want to stop?(y/n)");
scanf("\n%c",&ch);
}while(ch=='n');
getch();
}

[Cubbi]

When you enter n on keyboard, you hit two keys: n and Enter. Two characters are sent to the program: 'n' and '\n'

On the first loop iteration, scanf processed 'n', but not '\n' (there are no more directives after %c, so once it reads a char, it's done)

On the second loop iteration, scanf('%c',&ch); processed '\n' without waiting for more input, and proceeded to check if ch == 'n', which it wasn't. So the loop ended.

To consume all endlines, spaces, and tabs, and make sure the char you read is a non-whitespace char, you can use scanf("\n%c",&ch);, although scanf(" %c",&ch); (with a space before %c) is a more idiomatic.

[vlad from moscow]

Then there is specification %c then white spaces are not skipped. So when you enter a symbol and after that you press ENTER the input buffer will contain two characters: the symbol and the new line character. Next time when you read a a symbol again if there is no \n in the format string then ch will get this new line symbol instead of the character you was going to read.