C: Scanf Ignored in Type Mismatch

Question

Yes, there are already similar issues on SO on that scanf() does not wait for the user's input before "interpreting" an enter keypress, but my problem is that this problem only occurs when the scan -- which is %ld (with a space in front) -- reads a string instead of another long number. Details are below if this summary seems convoluted.

I have the following program in C:

#include <stdio.h>
#include <stdbool.h>

int main(void)
{
    long int cardnum, tmp;
    int num_of_digits;
    bool validate; // valid value
    LOOP:do
    {
        printf("Number: ");
        scanf(" %ld", &cardnum);

        tmp = cardnum, num_of_digits = 0;
        while (tmp != 0)
        {
            tmp /= 10;
            ++num_of_digits;
        }

        if (num_of_digits != 16)
        {
            printf("INVALID\n");
            goto LOOP;
        }

        validate = false;
        // ... (validate will be processed here, and there will be a case where "validate" will be true)
    }
    while (validate != true);
}

Inputs:

• 4003600000000014: works, length is 16
• 400360000000001: prints INVALID, length is 15
• asdf: infinite loop (notice how not all invalid inputs will result in this bug!)

Number: INVALID
Number: INVALID
Number: INVALID
...

I have read elsewhere that scanf ignores the last line buffer (or something) but in my case, it only happens when scanf did not receive the right input type from the user. Can anyone please help?

Thanks in advance.

Answer 1

how not all invalid inputs will result in this bug!)

If your input is not a number, then scanf will fail. So check if it fails to scan a number and if the scanning failed, then ignore the input until the end of line, for example. Note - you should handle EOF too.

if (scanf(...) != 1) {
    int c;
    // ignore input
    while ((c = getchar()) != EOF) {
        // until a newline!
        if (c == '\n') {
            break;
        }
    }
    // when eof
    if (c == EOF) {
        // then exit our program with a failure!
        printf("ERROR: End of input!");
        exit(-1);
    }
    // if not eof, means user is ready to enter yet another line
    printf("you inputted an invalid line - please input a line consisting of digits only!");
    continue;
}

See return value description in cppreference scanf.

---

LOOP:do { and goto LOOP; looks like a horrible idea - do not use goto there, seems confusing. Instead prefer to restructure your code so that you do not have to use goto. For example:

for (;;) {
    if (some_check) {
         continue;
    }
    if (some_another_check) {
         continue;
    }
    // if we get here, means all checks succeeded - so we need to break
    break;
}

There are nice uses of goto and it happens to be used - see ex. https://www.kernel.org/doc/html/v4.10/process/coding-style.html#centralized-exiting-of-functions .

Answer 2

If you entered something other than an integer number then you need to remove from the input buffer the invalid input.

Here is a demonstrative program.

#include <stdio.h>

int main(void) 
{
    long int x;
    int valid;
    
    do
    {
        if ( !( valid = ( scanf( " %ld", &x ) == 1 ) ) && valid != EOF )
        {
            scanf( "%*[^\n]%*c");
        }
    } while ( !valid );
    
    printf( "x = %ld\n", x );
    
    return 0;
}

The console output might look like

A
12
x = 12

Pay attention to that using the goto statement is a bad programming style. Instead you could use do-while statement as it is shown in my program. That is in fact the goto statement in your program is redundant.