Is ((f a) b) the same as (f a b) in Haskell?

Question

map2_Maybe :: (a -> b -> c) -> Maybe a -> Maybe b -> Maybe c
map2_Maybe f Nothing _ = Nothing
map2_Maybe f (Just a) Nothing = Nothing
map2_Maybe f (Just a) (Just b) = Just ((f a) b)
-- Or: map2_Maybe f (Just a) mb = fmap (f a) mb

map2_Either :: (a -> b -> c) -> Either e a -> Either e b -> Either e c
map2_Either f (Left e) _ = Left e
map2_Either f (Right a) (Left e) = Left e
map2_Either f (Right a) (Right b) = Right (f a b)
-- Or: map2_Either f (Right a) eb = fmap (f a) eb

In these two examples, Is ((fa) b) the same as (fab) since every function in Haskell can only take one argument?

Answer 1

是的，它们是完全一样的。

Answer 2

Haskell transposes (fab) into ((fa) b). It's called currying. It does that to all functions by default but can be overridden.

 add = (+)
(add 1 2) -- becomes --  ((add 1) 2) -- upon execution.

Both return 3 . The result of a function is its value.

Curried functions are natural.

add1 = add 1
add1 2 -- also returns 3