[Haskell-cafe] pair (f,g) x = (f x, g x)?
Marc A. Ziegert
coeus at gmx.de
Sat Jul 2 14:59:08 EDT 2005
'.' is not always a namespace-separator like '::','.','->' in c++ or '.' in java.
it is used as an operator, too.
(.) :: (b->c) -> (a->b) -> (a->c)
(f . g) x = f (g x)
remember the types of fst and snd:
fst :: (a,b)->a
snd :: (a,b)->b
so the function (.) combines
square :: Int -> Int
with fst to
(square . fst) :: (Int,b) -> Int
the same with toUpper:
(Char.toUpper . snd) :: (a,Char) -> Char
so you have with 'pair (f,g) x = (f x,g x)':
pair (square . fst,Char.toUpper . snd) (2,'a')
==>
((square . fst) (2,'a'), (Char.toUpper . snd) (2,'a'))
==>
( square (fst(2,'a')), Char.toUpper (snd(2,'a')) )
==>
( square 2 , Char.toUpper 'a' )
==>
(4,'A')
- marc
Am Samstag, 2. Juli 2005 08:32 schrieb wenduan:
> I came across a haskell function on a book defined as following:
>
> pair :: (a -> b,a -> c) -> a -> (b,c)
> pair (f,g) x = (f x,g x)
>
> I thought x would only math a single argument like 'a', 1, etc....,but
> it turned out that it would match something else, for example, a pair as
> below:
>
> square x = x*x
>
> pair (square.fst,Char.toUpper.snd) (2,'a')
> (4,'A')
>
> The type declaration of pair is what confused me,
> pair :: (a -> b,a -> c) -> a -> (b,c),it says this function will take a
> pair of functions which have types of a->b,a->c,which I would take as
> these two functions must have argument of the same type, which is a,and
> I didn't think it would work on pairs as in the above instance,but
> surprisingly it did,can anybody enlighten me?
>
> --
> X.W.D
>
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