cs50 Pset2 *Spoiler*

Question

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

int main(int argc, char *argv[])

{
    //create key
    string key = argv[1];


    //check if number of argument is not 2
    if (argc != 2)
    {
        printf("Usage: %s key\n", argv[0]);
        return 1;
    }



    //convert string to int

    int k = atoi(key);

    //prompt user for plaintext
    string p = get_string("plaintext: ");
    printf("ciphertext:");

    int n = strlen(p);

    //shift letters only
    //preserve case and iterate over each character
    //ASCII value then encipher back to ASCII
    for (int i = 0; i < n; i++)
    {
        if (isalpha(p[i]))
        {

          if (isupper(p[i]))
          {
          printf("%c", (((p[i] + k) - 65) % 26) + 65);
          }

          if (islower(p[i]))
          {
          printf("%c", (((p[i] + k) - 97) % 26) + 97);
          }
        }
        else //print character as is
        {
        printf("%c", p[i]);
        }



    }
    //print newline
    printf("\n");

    //program exit
    return 0;

}

This is almost done but not quite. I keep getting ":( handles non-numeric key; timed out while waiting for program to exit. I was thinking isdigit but I'm not sure that would work with atoi. Help please.

Answer 1

There are two ways to handle the non-numeric key. The first is using the isdigit function from ctype.h library. the isdigit function can be used to checked each character in the string. The second one is a very simple method.

# include<string.h>
...
for (int i = 0; i < strlen(argv[1]); i++){
    if ((argv[1][i] < '0')||(argv[1][i] > '9')){
        printf("Usage: ./caesar key\n");
        return 1;
    }
}
...

The for loop check if all the keys are numeric. If it finds a non-numeric key then it will end the program and exit.

Hope this helps. If it does, then please check the tickmark.