My solution to CS50 pset2 - "Parlez-vous français?"

vigenere.c

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

/**
 * Vigenere.c
 * 
 * A program that encrypts messages using Vigenère’s cipher. This program
 * must accept a single command-line argument: a keyword, k, composed entirely
 * of alphabetical characters. If your program is executed without any
 * command-line arguments, with more than one command-line argument, or with one
 * command-line argument that contains any non-alphabetical character, your
 * program should complain and exit immediately, with main returning 1
 * (thereby signifying an error that our own tests can detect). Otherwise, your
 * program must proceed to prompt the user for a string of plaintext, p, which
 * it must then encrypt according to Vigenère’s cipher with k, ultimately
 * printing the result and exiting, with main returning 0.
 * 
 * */
 
 int main(int argc, string argv[])
 {
    // check for 2 arguments only
    if (argc != 2)
    {
        printf("Nope\n");
        return 1;
    }
    
    // check if argument is all alpha char (no punct) - use loop and isalpha
    for (int i = 0; i < strlen(argv[1]); i++)
    {
        if (isalpha(argv[1][i]) == 0)
        {
            printf("Nope\n");
            return 1;
        }
    }
    
    // prompt user for codeword
    string codeword = GetString();
    int j = 0;
    
    // loop through the codeword. If not a letter than print unmodified.
    for (int i = 0, n = strlen(codeword); i < n; i++)
    {
        // to keep looping through the key continously
        j = j % strlen(argv[1]);
        
        // check if the char is alpha
        if (isalpha(codeword[i]))
        {
            // only 4 types of outcomes
            if (islower(codeword[i]) && islower(argv[1][j]))
                printf("%c", (((codeword[i] - 97) + (argv[1][j] - 97)) % 26) + 97);
            else if (isupper(codeword[i]) && islower(argv[1][j]))
                printf("%c", (((codeword[i] - 65) + (argv[1][j] - 97)) % 26) + 65);
            else if (islower(codeword[i]) && isupper(argv[1][j]))
                printf("%c", (((codeword[i] - 97) + (argv[1][j] - 65)) % 26) + 97);
            else if (isupper(codeword[i]) && isupper(argv[1][j]))
                printf("%c", (((codeword[i] - 65) + (argv[1][j] - 65)) % 26) + 65);
            j++;
        }
        else
        {
            printf("%c", codeword[i]);
        }
    }
    printf("\n");
 }

J is because there are two string arrays to loop through: the codeword, which is looped through with strlen on line 46, as well as argv[1], the command line argument word the programmer (Craig) types in. argv[1] is the keyword or k in the comments on line 10.
For each letter in the codeword, codeword[i], the arg[1] also clicks to the next letter in itself, argv[1][j].
If I understand correctly, J functions like an inner loop, a loop within a loop.
Hope this helps.

Now I'm trying to understand line 49 better? Is that like starting a while loop? Anyone?

j = j % strlen(argv[1]);

This is where he modifies the length of J to the length of argv. Once J hits the length of argv it will re-start back over at the first char.

If this wasn't there, j would continue to increase with every char past the strlen of argv. This is a limiter and so needs to be reset prior to going into the if/else if functions.