Ex 9.9 in Paul Hudak's book
Sun, 31 Mar 2002 22:41:45 +0200
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.
That's ok. But next he asks the reader to rewrite the function remainder
using fix and i cannot find out how.
I tried something like:
remainder2 a b = fix (\x -> x - b)
but there is no base case too.
As soon as the lambda expression wants to evaluate its argument, the interpreter
Can someone help me to solve the problem ?