Caesar Harvard CS50X

Question

This is a very early version of pset2, caesar. I have figured out how to encrypt, but have found difficulties in figuring out how to convert the ASCII code to the actual letter. Can anyone help?

for (i = 0, n = strlen(plain); i < n; i++)
{
    if (isalpha (plain [i]))
    {
        index = (int) plain [i] - 65;
        cypher = ((index + key) % 26) + 65;
    }

}

Can you specify what you mean with "the actual letter?" Letters are just the way a console/file interprets the ASCII code. — Angew is no longer proud of SO, Jul 08 '18 at 11:54
I want to print a string, but right now the data type is in int. — albertjlay, Jul 08 '18 at 11:56
`isalpha(plain[i])` has undefined behavior. You need `isalpha((unsigned char)plain[i])`. — melpomene, Jul 08 '18 at 11:56
@melpomene How can you know it's UB if we don't even know what `plain` is? — Angew is no longer proud of SO, Jul 08 '18 at 11:57
@melpomene Not to mention the fact that `isalpha` only has UB if the argument does not fit into `unsigned char`, not if it's not typed like that. — Angew is no longer proud of SO, Jul 08 '18 at 11:58
@Angew OK, potentially UB. `plain` better be an array of chars, otherwise `strlen(plain)` won't be valid. — melpomene, Jul 08 '18 at 12:00
I'm such a beginner that I'm not even sure how to do that. @angew — albertjlay, Jul 08 '18 at 12:01
plain is a string, so I guess it is an array of chars? @melpomene — albertjlay, Jul 08 '18 at 12:03
In C, characters *are* their codes, so you can just say `char c = cypher;`. — melpomene, Jul 08 '18 at 12:04
@melpomene I've tried it this way printf("%c", cypher); It kinda works, but it only prints the last character I inputted — albertjlay, Jul 08 '18 at 12:08
@melpomene can I print the string in one go? or do I have to put it in the loop? — albertjlay, Jul 08 '18 at 12:10
@melpomene nevermind, I figured it out. Thanks a lot though! U saved me hours :) — albertjlay, Jul 08 '18 at 12:11

score 1 · Answer 1 · answered Jul 08 '18 at 12:14

Your code doesn't take into account the lowercase part of the alphabet. To fix that, you can do:

//encrypts in place; I've replaced the magic numbers with their char-literal-based derivations
void encrypt(char *plain, int key)
{
    #define cypher plain[i]
    size_t i, n, index;
    for (i = 0, n = strlen(plain); i < n; i++)
    {
        if (isupper (plain [i]))
        {
            index = plain [i] - 'A';
            cypher = ((index + key) % ('Z'-'A'+1) ) + 'A';
        }else if(islower(plain[i])){
            index = plain [i] - 'a';
            cypher = ((index + key) % ('z'-'a'+1)) + 'a';
        }else 
            cypher=plain[i];
    }
}

or, if you're limiting yourself to ASCII, you can speed things up by replacing the locale-based ctype macros with simple char comparisons. If you're further willing to accept as encoding sources/targets the [\]^ _ characters in between the A-Z and a-z range in ascii you could further simplify to

void encrypt(char *plain, int key)
{
    #define cypher plain[i]
    size_t i, n, index;
    for (i = 0, n = strlen(plain); i < n; i++)
    {
        if(plain[i] >= 'A' && plain[i] <='z'){
            index = plain [i] - 'A';
            cypher = ((index + key) % ('z'-'A'+1) ) + 'A';
        } else cypher=plain[i];
    }
}

Example usage (with either version):

int main(int c, char **v)
{
    char s[]="Hello, World!";
    encrypt(s,3);
    puts(s);
    encrypt(s,-3);
    puts(s);
}

Sample run:

$ ./a.out
Khoor, Zruog!
Hello, World!

Thanks, but I haven't really had a chance with tinkering around because of the problem with printing your characters. I also unfortunately cannot read your answer though, I'm trying to solve the problem with my own ability. Thanks though! — albertjlay, Jul 08 '18 at 12:21

Caesar Harvard CS50X

1 Answers1