C: scanf for char not working as expected [duplicate]

Question

I was recently running a c program in my PC. It have a for loop in which some char d is scanned. The for loop runs for 3 times. During each running it prints the the count of running and then scans the value of char d. The program is as follows

#include<stdio.h>

int main(){
    int f;
    char d;
    for(f=0;f<3;f++){
        printf("Choice %d\n", f);
        scanf("%c", &d);
    }
    return 0;
}

Now the trouble is that when I run the program, the for skips the scanf part when f is 1. Now if i changed the code as follows

#include<stdio.h>

int main(){
    int f;
    int d;
    for(f=0;f<3;f++){
        printf("Choice %d\n", f);
        scanf("%d", &d);
    }
    return 0;
}

Now the program works fine. and scanf is executed for every iteration of for loop.

What does seem to be the problem here? I mean when d is of type int it works fine, but when d is of type char it does not work correctly.

Answer 1

You have to change

scanf("%c", &d);

to

scanf(" %c", &d);
       ^
       |

Otherwise, scanf() will consider the previously entered ENTER key press.

Note:

1. ENTER key press generates a \n, which is a vaild input for %c format specifier. Adding a space before %c tells scanf() to ignore all leading whitespace-like inputs (including that previously stored \n) and read the first non-whitespace character from stdin.

2. As for the case with %d format specifier, it consumes (and ignores) any leading whitespace-like inputs before scanning for numeric inputs, so the second case does not suffer any issues.