Talk:Y combinator

Haskell stateless?

I don't know Haskell, but isn't the definition:

 y f = f (y f)

not stateless as it seems to be defining y by referring to y. Haskell, no doubt has lazy evaluation to make it terminate, but the task does ask for a non-recursive, stateless definition of y. --Paddy3118 09:18, 28 February 2009 (UTC)

I googled this. --Paddy3118 09:22, 28 February 2009 (UTC)