Re: [Haskell-cafe] lazy evaluation is not complete

Peter Padawitz wrote:

A simplied version of Example 5-16 in Manna's classical book "Mathematical Theory of Computation":

foo x = if x == 0 then 0 else foo (x-1)*foo (x+1)

If run with ghci, foo 5 does not terminate, i.e., Haskell does not look for all outermost redices in parallel. Why? For efficiency reasons?

It's a pity because a parallel-outermost strategy would be complete.

(*) is strict in both arguments for Int. If you want to avoid this, you could do newtype X = X Int and write your own implementation of (*) that is nonstrict. -- Jochem Berndsen | jochem@functor.nl GPG: 0xE6FABFAB