CODEWITHSUNDEEP
ccs50
Ayoola
Ayoola

Reputation: 11

Not getting expected result from this code

Please I'm stuck in this code from an exercise on the CS50 course. I was asked to call the only_digits function on argv[1] as stated below but I don't seem to understand how.

Modify main in such a way that it calls only_digits on argv[1]. If that function returns false, then the main should print "Usage: ./caesar key\n" and return 1. Else main should simply return 0.

Here is my code.

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

bool only_digits(string s);

int main(int argc, string argv[])
{
    string s = argv[1];
    if(argc != 2 && only_digits(argv[1]))
    {
        printf("Usage: ./caesar key\n");
        return 1;
    }
    else{
        return 0;
    }
}

bool only_digits(string s)
{
    for(int i = 0; i < strlen(s); i++)
    {
        if(isdigit(s[i]))
        {
            return true;
        }
        else{
            return false;
        }
    }
    return false;
}

Upvotes: 0

Views: 203

Answers (1)

pm100
pm100

Reputation: 50210

2 errors

if (argc != 2 && only_digits(argv[1]))

you will only issue the error message if argc != 2 and its not all digits, that means if they enter the correct numer of args (1) you will never look at it. You mean

if (argc != 2 || only_digits(argv[1]))

in this

for (int i = 0; i < strlen(s); i++)
{
    if (isdigit(s[i]))
    {
        return true;
    }
    else {
        return false;
    }
}

YOu look at the first digit and return true or false, but thats it. YOu have to keep going.

for (int i = 0; i < strlen(s); i++)
{
    if (!isdigit(s[i]))
    {
        return false;
    }
}
// if we get here then all are digits
return true;

Upvotes: 1

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