A way around my XOR Encryption requiring plaintext size

Question

I looked at some of the XOR examples and am unable to find an answer.

This is the code I'm using for PHP:

 for($i=0;$i<strlen($text);)
 {
     for($j=0;($j<strlen($key) && $i<strlen($text));$j++,$i++)
     {
         $outText .= $text{$i} ^ $key{$j};
         //echo 'i='.$i.', '.'j='.$j.', '.$outText{$i}.'<br />'; //for debugging
     }
 }  

And the resulting code for decryption in C++:

for (int i = 0; i < original.size();)
{
    for (int j = 0; (j < key.size() && i < original.size()); j++, i++)
    {
        decrypted += encrypted[i] ^ key[j];
    }
}

I'm having trouble getting around the need for the plaintext's size (std::string original in this case).

Without, hopefully, drastically changing the way I do the algorithm, is there a way around this?

Thanks in advance!

Answer 1

The ciphertext and plaintext are the same length. So either size will do.

You also do not need two levels of loop nest to do this in PHP or C++.

Your code should look more like this:

 $textlen = strlen($text);
 $keylen  = strlen($key);
 for ($i = 0; $i < $textlen; $i++)
 {
     $outText .= chr( ord( $text{$i} ) ^ ord ( $key{$i % $keylen} ) );
 }  

I answered another question on XOR encryption here, since it seems to be all the rage these days.

BTW, if you try to backport this PHP code back to C++, be aware that (unlike PHP), strlen stops at ASCII NULs, and won't do what you want. You're better off using either std::string or vector<char> for the C++ version. (See link above for my C++ code.)