At 16:23 31/03/2002 -0500, Antony Courtney wrote:
Ludovic Kuty wrote:
The exercise is short. First, he defines a "fix" function: fix f = f (fix f) Which has the type (a -> a) -> a Then the function "remainder": remainder :: Integer -> Integer -> Integer remainder a b = if a < b then a else remainder (a - b) b The function fix, has far as i understand the matter, has no base case. So if someone wants to evaluate it, it will indefinitely apply the function f to itself, leading the hugs interpreter to its end.
... Now let's define a generator to test out y:
factGen = \f -> \n -> if n==0 then 1 else n * f (n-1) fact = y factGen
Thanks for the combinator. That's really obscure :) But it works: remainder2 :: Integer -> Integer -> Integer remainder2 = fix remainderGen where remainderGen f a b = if a < b then a else f (a - b) b