Need help in understanding types (based on curry)

Question

 curry f a b = f(a,b)

I thought: the curry function takes function f a b and returns f(a, b), so i thought the type is:

(a -> b -> c) -> (a, b) -> c

so why the type is reversed?:

((a, b) -> c) -> (a -> b -> c)

Answer 1

I believe the source of your confusion is this passage:

the curry function takes function f a b

Not really: curry takes a function, and that function is f. As for a and b, they are arguments that are passed to the curried function. It is easier to see that by either adding a pair of superfluous parentheses, to make the partial application more evident...

(curry f) a b = f (a,b)

... or by shifting a and b to the right-hand side with a lambda:

curry f = \a b -> f (a,b)

f is a function that takes a pair -- note that we give it a pair, (a,b). curry f, on the other hand, takes the two arguments separately. That being so, the type of curry is indeed:

curry :: ((a, b) -> c) -> (a -> b -> c)

Answer 2

Curry takes only one argument which is a function and returns a function. That argument is a function with the signature

((a, b) -> c)

i.e. a function that takes one argument: a pair of things a and b and returns another thing c. Curry does something magical to that function and turns it into a new function:

(a -> b -> c)

i.e. a function that takes two arguments: a thing a and a thing b and returns a thing c.

So the code:

curry f a b

is like:

let newFunction = curry(f)
newFunction a b

Basically curry transforms a function that takes a pair into a function that takes that pair as two arguments instead.