C compiling Error: non-void function does not return a value in all control paths

I'm doing the CS50 course in C and am working on the problem set from week 2. I'm an absolute beginner so there's probably a lot wrong with my code. For now, I'm trying to create a function that checks if the user has correctly used the command line input and if that input consists of only integers.

this is my code:

#include <cs50.h>
#include <stdio.h>
#include <ctype.h>
#include <string.h>

bool only_digits(string s);

int main(int argc, string argv[])
{
    if (argc == 2 & only_digits > 0)
    {
        printf("test\n");
    }
    else
    {
        printf("usage: ./caesar key\n");
    }
}

bool only_digits(string s)
{
    for (int i = 0, n = strlen(s); i < n; i  )
        {
            if (isdigit(i))
            {
                return 0;
            }

            else
            {
                return 1;
            }
        }
}

This is the error:

caesar/ $ make caesar 
caesar.c:34:1: error: non-void function does not return a value in all control paths [-Werror,-Wreturn-type]
}
^
1 error generated.

Thanks in advance, and excuse me for the many mistakes there probably are in the code. make: *** [: caesar] Error 1

CodePudding user response：

The logic of your function doesn't match its name. Presumably, you want only_digits to check if the passed in string is comprised entirely of digits. However, you return 0 (false) when you find a digit, that makes no sense. You're also not calling the function correctly:

#include <cs50.h>
#include <stdio.h>
#include <ctype.h>
#include <string.h>

int main(int argc, string argv[])
{
    // use && for logical AND, & is bitwise AND
    // Pass argv[1] to only_digits
    if ((argc == 2) && only_digits(argv[1]) == true)
    {
        
        printf("test\n");
    }
    else
    {
        printf("usage: ./caesar key\n");
    }
}

bool only_digits(string s)
{
    // strlen returns a size_t type, use that
    size_t n = strlen(s);
    for (size_t i = 0; i < n; i  )
    {
        // check each digit. If it's _not_ a digit, then we're
        // done, no point in checking the rest of the string
        if (!isdigit(s[i]))
        {
            return false;
        }
    }

    // we've looped thru the entire string. If we make it here
    // and haven't returned false, then we know every char in
    // the string is a digit, return true
    return true;
}

Demo

CodePudding user response：

If you pass empty string "" it will not enter the for loop and all your return statements will be skipped.

bool only_digits(string s)
{
    for (int i = 0, n = strlen(s); i < n; i  )
        {
            if (isdigit((unsigned char)i))
            {
                return false;
            }

            else
            {
                return true;
            }
        }
    return false;
}

bool only_digits(string s)
{
    for (int i = 0, n = strlen(s); i < n || !n; i  )
        {
            if (isdigit((unsigned char)i))
            {
                return false;
            }

            else
            {
                return true;
            }
        }
}