To gloss over details: it'll reduce x far enough so it knows that it's an Integer, but it won't nessesarally compute that integers value.
No, Integers don't contain any lazy components. It statically knows that it's an integer.
I meant that it would reduce to the outermost constructor but nessesarally evaluate the rest of the object. Ok, I actually looked up the implementation of Integer in GHC.
-- | Arbitrary-precision integers. data Integer = S# Int# -- small integers #ifndef ILX | J# Int# ByteArray# -- large integers #else | J# Void BigInteger -- .NET big ints
You were right and I was wrong, Integers contain no lazy components. Perhaps that just highlights the folly of guessing how much actually gets evaluated in a lazy language.. :) Ben.