Hello,
It's probably easiest to think of composition as a function which
takes two arguments (both functions), (g :: b -> c) and (f :: a -> b),
and returns a new function of type a -> c. We could write this
explicitly as
composition :: (b -> c, a -> b) -> a -> c
composition (g,f) = \x -> g (f x)
then (.) is the currying of composition:
(.) = curry composition
or
(.) g f = \x -> g (f x)
-Jeff
On 9/21/07, PR Stanley
Hi (.) :: (b -> c) -> (a -> b) -> (a -> c) While I understand the purpose and the semantics of the (.) operator I'm not sure about the above definition. Is the definition interpreted sequentially - (.) is a fun taht takes a fun of type (b -> c) and returns another fun of type (a -> b) etc? Any ideas? Thanks, Paul
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