Caesar cipher help!

[GillianG2203]

Body

I am working on the cs50 problem set 2 and have run into an issue with my caesar ciper problem. Everything is working, as in it will make the program without errors, but after imputing a number on the command line, the user should be prompted for text. My code is not prompting for text. Please help! I'm not sure what I'm missing.

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

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

for (int i = 0; i < strlen(argv[1]); i++)
{
    if (!isdigit(argv[1][i]))
    {
        printf("Usage: ./caesar key\n");
    }
    return 1;
}
int k = atoi(argv[1]);

string plaintext = get_string("Plaintext: ");

printf("Ciphertext: \n");

for (int j = 0; j < strlen(plaintext); j++)
{
    if (isupper(plaintext[j]))
    {
        printf("%c", (plaintext[j] - 65 + k) % 26 + 65);
    }

    else if (islower(plaintext[j]))
    {
        printf("%c", (plaintext[j] - 97 + k) % 26 + 97);
    }

    else
    {
        printf("%c", plaintext[j]);
    }
}
printf("\n");

}

Guidelines

• I have read and understood this category's guidelines before making this post.

[rangarajanpraneeth]

Both your "return 1;" statements are outside their respective if statements. So regardless of whether or not your if condition is met, the return will execute. Move them both inside and it should work. Also add a "return 0;" at the end of the main function since a successful program should do that.

[varunpats]

Hey @GillianG2203,

You should only return 1; inside the error cases, not after. Only exit if the input is invalid.
Below is a fixed version of your code, if this is the solution, then please mark this as an answer. Thanks!

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

int main(int argc, string argv[])
{
// Check if user provided exactly 1 argument
if (argc != 2)
{
printf("Usage: ./caesar key\n");
return 1; // only exit here if wrong number of args
}

// Ensure every character in argv[1] is a digit
for (int i = 0; i < strlen(argv[1]); i++)
{
    if (!isdigit(argv[1][i]))
    {
        printf("Usage: ./caesar key\n");
        return 1;   // exit only if invalid
    }
}

// Convert string to int
int k = atoi(argv[1]);

// Prompt user for plaintext
string plaintext = get_string("Plaintext: ");

printf("Ciphertext: ");

// Encrypt and print ciphertext
for (int j = 0; j < strlen(plaintext); j++)
{
    if (isupper(plaintext[j]))
    {
        printf("%c", (plaintext[j] - 'A' + k) % 26 + 'A');
    }
    else if (islower(plaintext[j]))
    {
        printf("%c", (plaintext[j] - 'a' + k) % 26 + 'a');
    }
    else
    {
        printf("%c", plaintext[j]);
    }
}
printf("\n");

}

[jfullstackdev]

here is my take :

---

Move the return 1; inside the conditional blocks:

if (argc != 2)
{
    printf("Usage: ./caesar key\n");
    return 1;
}

for (int i = 0; i < strlen(argv[1]); i++)
{
    if (!isdigit(argv[1][i]))
    {
        printf("Usage: ./caesar key\n");
        return 1;
    }
}

With this change, the program will only exit when there’s an error and otherwise will proceed to:

int k = atoi(argv[1]);
string plaintext = get_string("Plaintext: ");

---

after this fix, the program should correctly prompt for plaintext input and then perform the Caesar cipher encryption

[educhile1]

Code Comparison: Caesar Cipher
This document highlights the key differences between the original and the corrected version of the caesar.c program. Although the changes are small, they are fundamental for the program to work correctly.

1. The Most Important Change: The Position of return 1;
   The main error in the original code was the location of the return 1; statements. This caused the program to terminate immediately after the first check, regardless of whether the input was correct or not.

Original Code (Incorrect)

if (argc != 2)
{
    printf("Usage: ./caesar key\n");
}
return 1; // ERROR! This line ALWAYS runs.

In this snippet, the return 1; is outside the if block. This means that, regardless of the value of argc, the program will print the message (if necessary) and then stop when it reaches return 1;. It never had the chance to ask the user for the text.

Corrected Code

if (argc != 2)
{
    printf("Usage: ./caesar key\n");
    return 1; // CORRECT! This line only runs if the condition is true.
}

In the corrected version, the return 1; is inside the if block. Now, the program will only stop if the condition (argc != 2) is true. If the condition is false (meaning the number of arguments is correct), the program ignores the if block and continues its execution normally.

The same error and the same correction were applied in the validation to check if the key contains only digits.

2. Successful Exit Indicator
   It is good practice in C programming to indicate when a program has finished successfully.

Original Code (Missing)
The original code did not have a return statement at the end if everything went well.

Corrected Code

// ... all the encryption code ...

printf("\n");
return 0; // Added to indicate that the program finished successfully.

return 0; was added at the end of the main function. By convention:

return 0; means the program executed without errors.

return 1; (or any other non-zero number) means an error occurred.

3. Header Cleanup (#include)
   This is a minor change, but it is a good programming practice.

Original Code
#include <math.h>

Corrected Code
The line #include <math.h> was removed.

The math.h library was not being used anywhere in the program, so it was not necessary to include it. It is advisable to include only the libraries that are actually needed.

Summary
In summary, the fundamental difference was not in the encryption logic (which was correct), but in the program's control flow. By moving the return statements to their correct place, we allow the program to stop only when it finds an error and continue to the end when the user's input is valid.