On Sat, 22 Dec 2007 17:13:55 +0200, Philippa Cowderoy
function's expansion is ... just like macro expansion.
No, it's not. Expanding variables (swapping f1 for \x->x+1) isn't the evaluation mechanism in haskell, g and h really are semantically equivalent values and we can't do actual computation just by expanding variables. Whereas expanding a macro is equivalent to beta-reduction (evaluating a function application), which isn't required before passing something in Haskell. We pass functions, not just their results.
Here's a trivial example that does so:
(\x -> x) (\x -> x)
A lambda calculus classic that doesn't typecheck in Haskell:
(\x -> x x) (\x -> x x)
Feel free to try evaluating it!
Thank you for your message. I tryed and this is what I've got: ERROR - cannot find "show" function for: *** Expression : (\x -> x) (\x -> x) *** Of type : a -> a