haskell翻转仿真问题

Question

flip' :: (a -> b -> c) -> (b -> a -> c)   
flip' f = g   
    where g x y = f y x

I want to ask that in the above flip' function , it seems that flip' will return a function named g and gxy share the same value with fyx

however, in " where gxy = fyx " , fyx is a function call and will return a value, gxy will aslo return a value, so does it make sense that " where v2 = v1 "?

I know that the code will work but I want to know more about the way it make this happen.

does anybody has an idea? thank you so much

Answer 1

In where gxy = fyx , gxy is not a function call g with parameters x and y . It is a declaration of the function g as a function taking 2 arguments x and y and evaluation to fyx .

So it means flip' given a function f taking 2 arguments will evaluate to g . g itself being defined as swapping two arguments to call f .