I'm trying to migrate some legacy PHP code to ruby, and I've encountered a problem with some 3DES encryption. This is the PHP implementation using mcrypt:
function encrypt_3DES($message, $key){
$bytes = array(0,0,0,0,0,0,0,0); //byte [] IV = {0, 0, 0, 0, 0, 0, 0, 0}
$iv = implode(array_map("chr", $bytes)); //PHP 4 >= 4.0.2
$ciphertext = mcrypt_encrypt(MCRYPT_3DES, $key, $message, MCRYPT_MODE_CBC, $iv);
return $ciphertext;
}
and this is my ruby code:
def encrypt_3DES(message, key)
des=OpenSSL::Cipher.new('des3')
des.encrypt
des.key = key
des.update(message)+des.final
end
However results are slightly different (base64 encoded):
//PHP
ZpgH7NWpRx+Mi6tDBZ9q2Q==
# Ruby
ZpgH7NWpRx/usGDIsQ+A8A==
As you can see it's the lowest portion of the string's bytes that differs. Any pointers are much appreciated.
I answer my own question.
It was an issue about how openssl and mcrypt implementations use padding. My cryptography knowledge isn't too deep, but I found a usable code sample here http://opensourcetester.co.uk/2012/11/29/zeros-padding-3des-ruby-openssl/
#ENCRYPTION
block_length = 8
des.padding = 0 #Tell Openssl not to pad
des.encrypt
json = '{"somekey":"somevalue"}'
json += "\0" until json.bytesize % block_length == 0 #Pad with zeros
edata = des.update(json) + des.final
b64data = Base64.encode64(edata).gsub("\n",'')
Basically, ruby openssl will use PKCS padding, while mcrypt uses 0 padding. So in our code I had to tell openssl not to pad the string with des.padding = 0
and then do the padding manually: json += "\0" until json.bytesize % block_length == 0
.
Those are the important bits that were missing in my original implementation.