Why does scanf enter a loop when I input a character?

Question

I have to finnish a college project, and a part of my code is acting strangely.

The goal of that part is to get an user input of an integer and store it in a variable so that i can use it later, however if the user inputs a character I have to ask for the number again.

I used the scanf function to get the user input and put it inside a while loop to continuously ask for the input in case it's invalid.

The problem is that when a user inputs a character, the code freaks out and starts running the while loop without stopping in the scanf to get the user input. It makes sense that the loop condition is always true but the strange part is that it doesn't stop to read new inputs.

I deconstructed my code in order to replicate the problem to make it easier to debug. I know that there are some useless variables but in my original code they are useful, I just kept them there to make it look similar to the original. I can only use scanf to get user input, despite knowing them, in this project I am only allowed to use scanf. I can't use scanf's format to get characters, only numerical types are allowed in this project.

C11 is the version of the standart we are using in classes.

I'm sory if the solution for this is a dumb thing, I'm not good at C and I'm having some difficultlies this semester...

Thanks in advance.

    while (!verification) {
        printf(">>>"); //write values in here
        check = scanf("\n%d", &var); //input a number and store the number of valid inputs
        if (check) verification = 1; //if the input is a number then the while condition should turn to false with this statement
        printf("var = %d, check = %d, verification = %d\n", var, check, verification); //printing all variables
    }

Answer 1

If the user does not input an integer there are characters left in the input stream after the call to scanf. Therefor you need to read to end of line before making the next attempt to read an integer. Otherwise scanf will try to read the same non-integer characters again and again. Here is an example:

#include <stdio.h>
#include <stdlib.h>

int main(void)
{
    int ch, i, n;

    n = scanf("%d", &i);
    while (n == 0) {
        fprintf(stderr, "wrong input, integer expected\n");
        do {
            ch = getchar();
        } while ((ch != EOF) && (ch != '\n'));
        n = scanf("%d", &i);
    }
    if (n == 1) {
        printf("%d\n", i);
    } else { /*n == EOF*/
        fprintf(stderr, "reading input failed\n");
        exit(EXIT_FAILURE);
    }
    return 0;
}

Answer 2

Don't use scanf() to read input from the user.

It's really only meant for reading data that's known to be in a particular format, and input from a user... often isn't.

While you do correctly check the return value of scanf("%d"), and could fix the case where the input isn't a number, you'll still have problems if the input is either an empty line, or a number followed by something else (123 foobar).

In the case of an empty line scanf() will continue waiting for non-whitespace characters. This is probably confusing, since users will expect hitting enter to do something.

In the case there's trailing stuff after the number, that stuff stays in the input buffer, and the next time you read something, it gets read. This is again probably confusing, since users seldom expect their input to one question to also act as input to another.

Instead, read a full line with fgets() or getline(), then run sscanf() or strtol() on that. This is much more intuitive, and avoids the disconnect caused by scanf() consuming input lines only partially (or consuming more than one line). See also e.g. scanf() leaves the new line char in the buffer

Here, using getline() (POSIX, even if not in standard C. Use fgets() instead if getline() is not available):

#include <stdio.h>

int main(void)
{
    char *line = NULL;
    size_t len = 0;
    int result;
    printf("Please enter a number: ");
    while (1) {
       if (getline(&line, &len, stdin) == -1) {
           /* eof or error, do whatever is sensible in your case */
           return 1;
       }
       if (sscanf(line, "%d", &result) != 1) {
           printf("That didn't seem like number, please try again: ");
           continue; 
       } 
       break;
    }
    printf("You entered the number %d\n", result);
}

Answer 3

The problem is you must discard offending input when the conversion fails.

Here is a simple solution using only scanf() as instructed:

#include <stdio.h>

int main() {
    int n;

    for (;;) {
        printf("Enter an number: ");
        switch (scanf("%d", &n)) {
        case 1:
            /* successful conversion */
            printf("The number is %d\n", n);
            return 0;
        case 0:
            /* conversion failure: discard the rest of the line */
            scanf("*[^\n]");  // discard characters before the newline if any
            scanf("*1[\n]");  // optional: discard the newline if present
            printf("Invalid input. Try again\n");
            continue;
        case EOF:
            /* input failure */
            printf("Premature end of file\n");
            return 1;
        }
    }
}