[Haskell-cafe] pair (f,g) x = (f x, g x)?

[Marc A. Ziegert]

'.' 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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