Higher Order Function (uncurry,curry)

I'm using these functions:

cplus a b = a+b   

ucplus(a, b) = a+b

...to do this experiment:

uncurry cplus=\(a, b) -> cplus a b  

(a,b) is the parameter of the function cplus, so the right side of equatio开发者_StackOverflow中文版n is actually doing the function cplus a b.

Why does uncurry cplus equal the right side of the equation?

uncurry is defined as follows:

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

When you apply the function uncurry to the function f, it returns a function like f, but instead of a 'curried' function, you get a function which takes a tuple.

The type signature of uncurry is (a -> (b -> c)) -> ((a, b) -> c). That is, it takes a function of type a -> (b -> c) and returns a function of type (a, b) -> c.

cplus is a curried function -- that means, it is a function that returns another function. The curried form of cplus could be written:

cplus = (\a -> (\b -> a + b))

But in most functional languages, the syntax cplus a b = a+b is equivalent.