'.' 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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