(g . f) x y = (\z -> g (f z)) x y = g (f x) y, and you need g (f x y), which is definitely not the same thing. 24.09.2010 13:35, Axel Benz пишет:
Hello,
this is maybe a simple question:
cbinary :: a -> b -> (a -> b -> b) -> (b -> c) -> c
-- Version 1 works:
cbinary x y f g = g (f x y)
-- Version 2 should be exactly the same according
-- to my understanding of the "." operator definition,
-- but fails with:
-- Occurs check: cannot construct the infinite type: b = b -> b
-- When generalising the type(s) for `cbinary'
-- cbinary x y f g = (g . f) x y
Can anybody explain why this happens and how I can compose f and g?
Hint: It works fine if f is defined as an unary function.
Thank you very much!
Axel
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