(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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