C: scanf for char not working as expected

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.

Why don't you read the `scanf()` manual? you are using it the wrong way anyway, even if you fix it. — Iharob Al Asimi, Mar 18 '15 at 12:34
You loop over `scanf` and read into he same variable 3 times. — Jared Burrows, Mar 18 '15 at 12:58
Well its just a part of larger program I am doing in which the for loop contain more code. I just posted the problematic bit here — Vivek Sasidharan, Mar 18 '15 at 13:01
@chux no no it is OK. As you can see I have received the answer. Once I added a space it worked correctly. scanf was working correctly but the overall execution made an unusual behavior in that part. — Vivek Sasidharan, Mar 18 '15 at 13:39
@chux Ohh I didn't give this title somebody must have changed it. — Vivek Sasidharan, Mar 18 '15 at 13:46
@Vivek I'm taking down my comments as they only applied to someone else's weak editing and not your work. This comment too - later. — chux - Reinstate Monica, Mar 18 '15 at 13:53

Sourav Ghosh · Accepted Answer · 2015-03-18T13:45:39.787

34

You have to change

scanf("%c", &d);

to

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

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

Note:

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.
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.

edited Mar 18 '15 at 13:45

answered Mar 18 '15 at 12:33

Sourav Ghosh

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Oh thanks for the reply it worked like a charm. why does i need to do that ? i mean why add a space? – Vivek Sasidharan Mar 18 '15 at 12:34
"`\n` is not a valid input for `%d` specifier" is misleading. `\n` _is_ valid input for `"%d"` as that speciifer, unlike `"%c"`, first consumes leading white-space which includes `'\n'` before continuing and scanning for numeric input. – chux - Reinstate Monica Mar 18 '15 at 13:38
@chux Thank you sir for the input. Updated accordingly. Kindly review. – Sourav Ghosh Mar 18 '15 at 13:46

Spikatrix · Answer 2 · 2015-03-18T12:49:50.897

The problem is that when you enter a character for scanf("%c", &d);, you press the enter key. The character is consumed by the scanf and the newline character stays in the standard input stream(stdin). When scanf(with %c) is called the next time, it sees the \n character in the stdin and consumes it, and thus does not wait for further input.

To fix it, change

scanf("%c", &d);

to

scanf(" %c", &d);
//    ^Note the space before %c

The space before the %c instructs scanf to scan any number of whitespace characters including none and stops scanning when it encounters a non-whitespace character. Quoting the standard:

7.21.6.2 The fscanf function

[...]

A directive composed of white-space character(s) is executed by reading input up to the first non-white-space character (which remains unread), or until no more characters can be read. The directive never fails.

The reason that using %d worked is that the %d format specifier automatically skips whitespace characters and since \n is a whitespace character, %d does not scan it. Quoting the standard again:

7.21.6.2 The fscanf function

[...]

Input white-space characters (as specified by the isspace function) are skipped, unless the specification includes a [, c, or n specifier. ²⁸⁴

C: scanf for char not working as expected

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