Question

I’m trying to check whether the entered chars are digits or not with pattern checking.

I’ve written following program.

This program not giving me the perfect output for following test cases
can anybody tell me where I’m going wrong with my logic.

/*

 Output test case

 234234 = It's a digit.
 a3434a = It's not a digit.
 33aa3a = It' not a digit.

*/

#define yes 1
#define no 0

#include <stdio.h>

int main(void)
{
    char c[30];
    int arr_size, result, i=0, state;
    printf("Enter your digit:= ");
    scanf("%s",&c);

    arr_size=(sizeof(c)/sizeof(c[0]));

    for(i; i < arr_size; i++)
    {
        if(check_digit(c[i]))
        state = yes;
        else
        state = no;
    }

    if(!state)
        printf("It's not a digit\n");
    else
        printf("It's a digit\n");

    system("\npause");
    return 0;
}

int check_digit(char c)
{
    return (c>='0' && c<='9');
}

Answer 1

1. You’re dealing with a string. Just iterate until you find ‘\0’. No need to get weird with sizeof.
2. Read man isdigit
3. If you really want to pass a pointer to scanf, it needs to be &c[0]. However, c all by its lonesome is a pointer so you can just say scanf("%s", c);
4. And that state variable. Try running some values through by hand on paper and see what happens to it.