C Determine if user input is an integer

Question

Hi I need to prompt a user for some input and then validate it. The input must only be validated if it is a positive integer and not greater then 23. The only problem I am having with this is when the user enters a non-numerical input like "hello." The code below does not successfully detect that any input is non-numerical and though I have tried many methods to do this, none of them seem to work. Below is the closest I seem to have gotten by taking the input as a string then converting it to an integer, however it still does not work. Any help would be appreciated.

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


int main(void) {
    int height;
    char input[50];
    int cont = 0;
    while (cont == 0) {
        printf("Please provide a non-negative integer no greater than 23.\n");
        scanf("%s", &input);
        height = atoi(input);
        if (height <= 23 && height >= 0) {
            cont = 1;
        } else {
            //do nothing
        }
    }
    printf("Valid Input.\n");
    return 0;
    }

scanf("%s", &input); is not correct. Hint: what's the type of input? — Yu Hao
– Yu Hao, Commented Aug 5, 2014 at 1:19
Don't use %s for scanf without a field width (and the s specifier takes a char * argument, you passed a char (*)[50] argument): scanf("%49s", input). — mafso
– mafso, Commented Aug 5, 2014 at 1:22

Greg Hewgill · Accepted Answer · 2014-08-05 01:17:47Z

4

The atoi() function has no provision for returning an error indicator. Instead, you can use the strtol() function:

char *end;
height = strtol(input, &end, 10);
if (end == input) {
    // no digits were entered
    puts("Invalid input.");
    continue;
}

answered Aug 5, 2014 at 1:17

Greg Hewgill

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2 Comments

user539810 Over a year ago

And of course you can use end to invalidate strings that start with digits and are followed by characters other than a null terminator or newline instead of accepting them using end == input. It is a matter of preference, but it is worth documenting what happens to "yes" if you allow "12yes\n".

Potatoswatter Over a year ago

… that is, by testing if ( * end != '\0' ) instead. I think that's the better recommendation.

BLUEPIXY · Accepted Answer · 2014-08-05 09:37:37Z

0

#include <stdio.h>

int main(void) {
    int height;

    while(1){
        printf("Please provide a non-negative integer no greater than 23.\n");
        //if(2==scanf("%d%c", &height, &nl) && nl == '\n' && 0<= height && height <= 23)//more limited for "number\n"
        if(1==scanf("%d", &height) && 0<= height && height <= 23)
            break;
        //Clear of invalid input
        while(getchar()!='\n')
            ;
    }
    printf("Valid Input(%d).\n", height);
    return 0;
}

edited Aug 5, 2014 at 9:37

answered Aug 5, 2014 at 9:29

BLUEPIXY

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Comments

Arpit · Accepted Answer · 2014-08-05 18:11:07Z

I am assuming that you have to consider the whole input into the consideration rather than only certain parts like "12jjj" and "23h" as invalid.

In my opinion, since 23 is only 2 char, so there is no harm in checking the length of the string and the individual characters.

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

bool ValidateInput (char * input, int &output)
{
    if (strlen(input) > 2)
        return false;

    for (int index = 0; index < strlen (input); index++)
    {
        if ((input[index] < '0') || input[index] > '9')
            return false;
    }

    output = atoi(input);

    return true;

}


int main(void) {
    int height;
    char input[50];
    int cont = 0;
    while (cont == 0) {
        printf("Please provide a non-negative integer no greater than 23.\n");
        scanf("%s", input);

        if (ValidateInput (input, height))
            break;
    }
    printf("Valid Input.\n");
    return 0;
    }

I hope this helps.

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