haskell类型签名问题

Question

Can someone explain me, why do these functions have different number of arguments and behavior , but the same type signature , yet they are both correct?

comp1 :: (a -> b) -> (b -> c) -> a -> c
comp1 f g = g.f

comp2 :: (a -> b) -> (b -> c) -> a -> c
comp2 f g x = g (f x)

also, why does comp2 has

comp2 :: (a -> b) -> (b -> c) -> a -> c

instead of something like

comp2 :: a -> (a -> b) -> (b -> c) -> a -> c

?

Thank you.

Answer 1

comp1 fg = gf is written in point-free style (not referring to points, but to values ). When you call comp1 , there is implicitly a third parameter being passed to gf , which is the composition of the two functions g and f : (gf) x equals g (fx) , ie g is passed the result of fx . No parameter x exists in comp1 because it's implicitly passed to the function. (You could think of comp1 as a partially applied or curried function if it makes you feel better.)

comp2 's type asks for two functions, one from (a->b) and another (b->c) , as well as a parameter of type a . There is no need to put an a -> in its signature.

The two functions are really equivalent; one simply uses some Haskell tricks to be more concise.

Answer 2

comp2 f g x = g (f x)

is syntactic sugar for

comp2 = \f -> \g -> \x -> g (f x)

Similarly

comp1 f g = g.f

is sugar for

comp1 = \f -> \g -> g.f

The definition of . is:

f1 . f2 = \x -> f1 (f2 x) -- Names of arguments have been changed to avoid confusion

So if we insert the definition into the desugared form of comp1 , we get:

comp1 = \f -> \g -> \x -> g (f x)

This is exactly the same as the desugared form of comp2 , so clearly the definitions are equivalent.

Answer 3

Currying . A multi-argument function in ML and Haskell, is just syntactic sugar for a one-argument function that returns a function; the function that it returns takes the remaining arguments.