[Haskell-cafe] Composition Operator
Peter Verswyvelen
bf3 at telenet.be
Sat Sep 22 06:45:07 EDT 2007
Tony Morris wrote:
> is the same as:
> (.) :: (b -> c) -> ((a -> b) -> (a -> c))
> ..
> "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"
IMO, that should be
"accepts a function b to c and returns a function. The function returned
takes a function a to b and returns function a to c"
Personally I also find the following a good explanation, since it does
not introduce lambdas or other scary things for newbies.
f . g = composite
where composite x = f (g x)
for example
printLn = putStrLn . show
so
printLn x
becomes
putStrLn (show x)
Cheers,
Peter
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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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>>
>>
>>
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