-----BEGIN PGP SIGNED MESSAGE----- Hash: SHA1 Apply parentheses from the right. So: (.) :: (b -> c) -> (a -> b) -> a -> c is the same as: (.) :: (b -> c) -> (a -> b) -> (a -> c) is the same as: (.) :: (b -> c) -> ((a -> b) -> (a -> c)) How you read that is up to you, but here is one way of reading it: "accepts a function a to c and returns a function. The function returned takes a function a to b and returns a function a to c" The expression f(g(x)) in C-style languages is similar to (f . g) x Tony Morris http://tmorris.net/ PR Stanley wrote:
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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