Caesar segmentation fault

Question

I prematurely submitted my code without including the if() statements that check if the second command line argument was a digit (if not it would produce an error message). My code was running perfectly fine before I remembered to add this section in, and it got all smileys. But I noticed I hadn't included it and, because I want to actually fulfill all criteria of the project, I added it in afterwards and it seems to be the source of the problem as my code is now giving me Segmentation Fault warnings when I try to run it. If anyone could tell me why this is, I'd be grateful as I'm well and truly stumped.

#include <cs50.h>
#include <string.h>
#include <ctype.h>
#include <stdlib.h> //CAESAR.C, scroll to bottom to see full summary, not the "alphabet" arrays just below as you do

const char upperalphabet[26] = {'A', 'B', 'C', 'D', 'E', 'F', 'G', 'H', 'I', 'J', 'K', 'L', 'M', 'N', 'O', 'P', 'Q', 'R', 'S', 'T', 'U', 'V', 'W', 'X', 'Y', 'Z'};
const  char loweralphabet[26] = {'a', 'b', 'c', 'd', 'e', 'f', 'g', 'h', 'i', 'j', 'k', 'l', 'm', 'n', 'o', 'p', 'q', 'r', 's', 't', 'u', 'v', 'w', 'x', 'y', 'z'};


int main(int argc, string argv[]) //the key will be inputted through a command line argument
{
    //below are terms in which the user's input (for the "key") may be rejected as invalid
    if (argc != 2)
    {
        printf("Error: please limit input to 1 other command-line argument!");
        return 1;
    }
    else if (argv[1] <= 0)
    {
        printf("Error: please provide a positive integer!");
        return 1;
    }
    
    // new code causing problems from here//
    int key;
    if (isdigit(argv[1]))
    {
        key = atoi(argv[1]);
    }
    else
    {
        printf("Usage: ./caesar key");
        return 1;
    }
    //to here//

    string plaintext = get_string("plaintext: ");
    printf("\n ciphertext: ");

    for (int n = 0; n < strlen(plaintext); n++)
    {
        /*Running through the string, if a char isn't an alpha-char, it's printed as is, if it is
        an alpha-char, it'll be sorted by its lettercase*/
        if (isalpha(plaintext[n]))
        {
            if (isupper(plaintext[n]))
            {
                int z;
                for (z = 0; plaintext[n] != upperalphabet[z]; z++)
                    ;
                /*above is a for-loop with nothing to implement, it simply increments the alphabet array through z until plaintext's char matches
                the present element in the alphabet array*/
                int ciphernumber = ((z + key) % 26);
                
                printf("%c", upperalphabet[ciphernumber]);
            }
            else if (islower(plaintext[n])) //code here is same as above, just for lowercase array
            {
                int z;
                for (z = 0; plaintext[n] != loweralphabet[z]; z++)
                    ;

                int ciphernumber = ((z + key) % 26);
                printf("%c", loweralphabet[ciphernumber]);
            }
        }
        else
        {
            printf("%c", plaintext[n]); //non-alphabetical plaintext array elements get printed here
        }
    }
    printf("\n");
    return 0;
}

Answer 1

In other words, the isdigit() function (and all of it's issomething() cousins) takes a single char as input. If you pass it a string, it chokes and generates a seg fault.

Otherwise, @MARS gave a great explanation. ;-)

Answer 2

This is a fairly common error at the beginning of the course, string argv[] is an array of pointers, therefore each of its elements is a pointer, said pointer points to a string, if we now look at the function in the function isdigit () we see that:

isdigit() Prototype function: int isdigit(int arg); The isdigit() function takes a single argument as an integer and returns the value of type int.

Although isdigit() takes an integer as an argument, the character is passed to the function. Internally, the character is converted to its ASCII value for verification.

It is defined in the header file <ctype.h>. Therefore the error is to pass a pointer when we must pass an integer