答案是:(a -> b) -> (c -> d -> a) -> c -> d -> b
(a -> b) -> (c -> d -> a) -> c -> d -> b
但我不知道怎么去那里。
haskell types type-inference currying parametric-polymorphism
currying ×1
haskell ×1
parametric-polymorphism ×1
type-inference ×1
types ×1