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

On Mon, 09 Feb 2009 15:10:22 +0100 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?

I believe * is implemented in the normal way and thus is always strict in both arguments.