In addition to what others have said, you could use pointfree[1] to do this automagically!
pointfree "h x y = (g 0 (f x y))" h = (g 0 .) . f
[1] http://hackage.haskell.org/package/pointfree
On 11 April 2011 10:22, Adam Krauze
Hello, as I am newbie to Haskell and my introductory question is:
given functions say f and g with type signatures
f :: (Num a) => [a] -> [a] -> [(a,a)] // f takes two lists and zips them into one in some special way g :: (Num a) => a -> [(a,a)] -> [a] // g using some Num value calculates list of singletons from list of pairs
of course g 0 :: (Num a) => [(a,a)] ->[a]
now I want to create function h :: (Num a) => [a] -> [a] -> [a] in such way
that (g 0) consumes output of f.
But when I try
Prelude> :t (g 0).f
I get an error:
<interactive>:1:9: Couldn't match expected type `[(a0, a0)]' with actual type `[a1] -> [(a1, a1)]' Expected type: [a1] -> [(a0, a0)] Actual type: [a1] -> [a1] -> [(a1, a1)] In the second argument of `(.)', namely `f' In the expression: (g 0) . f
In pointfull representation it works well
Prelude> let h x y = (g 0 (f x y))
How to do pointfree definition of h?
Ajschylos.
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