ProgrammingCryptol zipWith HMAC

[weaversa]

On page 25, HMAC is defined using zipWith twice. What is zipWith buying us here? Doesn't xor work pairwise already?

Cryptol> zipWith (^) (join [0..7]) (join (repeat 0x5c))
Showing a specific instance of polymorphic result:
  * Using '3' for type argument 'n' of 'Cryptol::repeat'
0x59652b
Cryptol> (join [0..7]) ^ (join (repeat 0x5c))
Showing a specific instance of polymorphic result:
  * Using '3' for type argument 'n' of 'Cryptol::repeat'
0x59652b

[yav]

Not sure about the reasoning, might be just an oversight. I agree that using ^ without zipWith is more direct, and xor-ing words together is really not that exotic.

[robdockins]

I think that (^) is equal to zipWith (^) for all the types where that makes sense.

[brianhuffman]

We might add some text explaining that (^) is equivalent to zipWith (^); that would be a useful lesson to communicate in the book.

[brianhuffman]

That particular part of the book is explaining how modules work; it is not trying to explain the code itself. So I think it probably makes sense to just change the code without changing any of the surrounding text.