[Haskell-cafe] Composition Operator

[Tony Morris]

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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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