Marc A. Ziegert wrote:
'.' 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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you are correct,but as in the following,
(square . fst) :: (Int,b) -> Int
(Char.toUpper . snd) :: (a,Char) -> Char
you get a Int and Char out of the two composed functions, namely square.fst,
Char.toUpper.snd.But in the type declaration of
pair, which appeared to me,it meant its arguments must be two functions which
are of the same type namely a,whereas Int and
Char passed to as arguments are of different types here, and that's the reason
I thought it wouldn't work.
Thank you,
regards.
--
X.W.D
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