How to input just character in scanf("%d%c");

Question

#define _CRT_SECURE_NO_WARNINGS
#include <stdio.h>
int main(void) {
    int input;
    char ch;
    scanf("%d%c", &input, &ch);
    printf("%d %c", input, ch);
}

I just want ch has some character and input has trash value.

So i typed [q] and [enter].

But output is [trash value ?].

I don't know why ch has '?' not 'q'.

What happens in my buffer?

You need to give the input buffer something to feed the `%d` format specifier ... a number! — Adrian Mole, Aug 19 '21 at 12:27
You should check the return value of `scanf` to see how many inputs were actually matched. — mediocrevegetable1, Aug 19 '21 at 12:28
ALWAYS check the return value of `scanf()`. (Just to understand mediocrevegetable1 comment correctly, it is not only here where you need to check it). — 12431234123412341234123, Aug 19 '21 at 13:42

pmg · Accepted Answer · 2021-08-19T12:45:56.470

scanf("%d%c", &input, &ch);

With input "q", scanf will try to interpret 'q' as the first character of an integer (for the "%d" conversion) and fail... scanf() returns immediately with value 0 and without changing the value of ch (the value of input is indeterminate).
"q" is still in the input buffer ready for the next input operayion.

Always check the return value of scanf()

int chk = scanf("%d%c", &value, &ch);
switch (chk) {
    default: /* this didn't happen */;
             /* let the compiler writers know about this */
             break;
    case EOF: /* error */;
              /* value unchanged */
              /* ch unchanged */
              break;
    case 0: /* no input found */;
            /* value indeterminate */
            /* ch unchanged */
            break;
    case 1: /* no input for ch found */;
            /* value ok; (possibly) changed */
            /* ch indeterminate */
            break;
    case 2: /* all ok */;
            /* value ok; (possibly) changed */
            /* ch ok; (possibly) changed */
            break;
}

Adrian Mole · Answer 2 · 2021-08-19T12:48:19.040

This is an excellent example of why you should always check the return value of scanf (and related functions).

The format string you give to scanf ("%d%c") expects, as input, a valid decimal number followed (immediately) by a character. If you input q by itself, the function will fail to read a decimal value and stop at that point – returning 0, because none of the arguments was successfully assigned a value. In that case, neither input or ch will have predictable values, as you have not initialized either.

Also note that, if you type in just a number (i.e. 12), the terminating newline (from the Enter key) will be assigned to ch.

Here's your code with an input-check added and default values assigned to your two variables:

#define _CRT_SECURE_NO_WARNINGS 1
#include <stdio.h>

int main(void)
{
    int input = 42;
    char ch = '$';
    if (scanf("%d%c", &input, &ch) != 2) {
        printf("Error: Please input a number and a character!\n");
    }
    printf("%d %c", input, ch);
    return 0;
}

Alternatively, if (as I think you suggest in your question), you want to still read the ch value but leave input as its original ('trash') value, you can re-read the input if/when the first call to scanf fails. On that second call, remove the %d format specifier:

int main(void)
{
    int input = -137; // Let's use this as a 'trash' value.
    char ch = '$';
    if (scanf("%d%c", &input, &ch) != 2) {
        if (scanf("%c", &ch) != 1) printf("Error: Failed to read anything!");
    }
    printf("%d %c", input, ch);
    return 0;
}

score 0 · Answer 3 · answered Aug 19 '21 at 12:42

0

#include <stdio.h>
int main(void) {


int input;
char ch;
scanf("%c%d", &input, &ch);
printf("%c %d",ch, input);

}

This code gives just the trash value as output nothing else.

answered Aug 19 '21 at 12:42

Prathamesh Jagale

1
1

How to input just character in scanf("%d%c");

3 Answers3