Re: [Haskell-cafe] Composition Operator

I'm not sure what you mean by "abstract function". It's a function like any other function. But otherwise you're right. :) -- Lennart On 9/22/07, PR Stanley wrote:

Ah, I understand now. Let me get this right: The resulting (a -> c) is a kind of abstract function formed by the pairing of b) and (b -> c), hence the lambda \x -> g (f x), correct? Thanks, Paul

At 05:11 22/09/2007, you wrote:

...

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 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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