Why doesn't scanf with character set format read input in a for loop?

Question

While trying to get newline character as input using scanf and display it as output as this discussion suggests, I tried the following code,

#include <stdio.h>
void main()
{
    int n;
    printf("Loop:\n");

    scanf("%d",&n);
    for (int i = 0; i < n; ++i)
    {
        char in[28];
        scanf("%27[^\n]%*c",&in);
        printf("%s\n",in);
    }
}

During execution, inside the for loop the input stream doesn't accept any inputs and instead displays n smiley faces. Is this because of the trailing newline character after reading n?

You must check return value of `scanf`. I assume it returns `0` because there is a pending `\n` in the input buffer from first `scanf` that does not match your format specifier. As a result `in` is not filled with any new input. — Gerhardh, Jun 22 '22 at 10:25
Please stop trying to kludge away trailing newlines: deal with them as they arrive. With `%d` and `%f` and `%s` you don't need to. With `%c` and `%[]` just place a space before: such as `" %27[^\n]"` This will also deal the the newline from the previous scanf before the loop. Please see [scanf() leaves the newline char in the buffer](https://stackoverflow.com/questions/5240789/scanf-leaves-the-new-line-char-in-the-buffer). — Weather Vane, Jun 22 '22 at 10:26
What is the input you are giving it, and what is the expected output? — Weather Vane, Jun 22 '22 at 10:28
If you want to write portable C programs, don't make `main` "return" `void`. Make it `int main` — Ted Lyngmo, Jun 22 '22 at 10:34

chqrlie · Accepted Answer · 2022-06-22T11:44:05.990

Beyond the type mismatch, the reason scanf("%27[^\n]%*c",&in); does not read extra input from the user is there is a pending newline left by scanf("%d",&n);. scanf() fails because there is no match for the character set and the newline is not consumed by %*c because the previous mismatch stops the scan.

Here is a modified version:

#include <stdio.h>

int flush_input(void) {
    int c;
    while ((c = getchar()) != EOF && c != '\n')
        continue;
    return c;
}

int main() {
    int n;
    printf("Loop:\n");

    if (scanf("%d", &n) != 1)
        return 1;

    flush_input();
    for (int i = 0; i < n; ++i) {
        char in[28];
        if (scanf("%27[^\n]", in) != 1) {
            // blank line or EOF
            *in = '\0';
        }
        printf("%s\n", in);
        if (flush_input() == EOF)
            break;
    }
    return 0;
}

score 2 · Answer 2 · answered Jun 22 '22 at 10:48

How to compile any C program as a beginner: What compiler options are recommended for beginners learning C?

Following this advise and compiling with gcc gives 2 problems:

<source>:2:6: error: return type of 'main' is not 'int' [-Wmain]
    2 | void main()
      |      ^~~~
<source>: In function 'main':
<source>:11:21: error: format '%[^
   ' expects argument of type 'char *', but argument 2 has type 'char (*)[28]' [-Werror=format=]
   11 |         scanf("%27[^\n]%*c",&in);
      |                ~~~~~^~      ~~~
      |                     |       |
      |                     char *  char (*)[28]

The first reported error is because void main() is an implementation-defined form of main() which isn't suitable for gcc unless you explicitly compile for embedded systems or the like. Switch to int main (void).
The second reported error says that the conversion %c expected a parameter of type char*. You gave it a parameter of type char (*)[28]. Huh, what is that? Well, it is a pointer to an array of 28 char. Not the same thing and not what you want, but what you get if you do &in instead of in.

Luckily, viewing multiple lines of the gcc output gives you the exact location of the bug, after which you will find the bug in seconds:

   11 |         scanf("%27[^\n]%*c",&in);
      |                ~~~~~^~      ~~~
      |                     |       |
      |                     expect  BUG HERE FIX ME

Now if you follow the above guidelines, you should be able fix the next trivial bug that the compiler has already found.

Why doesn't scanf with character set format read input in a for loop?

2 Answers2