问题
I'm preparing for my exams but there is something I can't understand.
functions:
tw f x = f (f x)
f x y = (y, x)
I am able to determine the type of 'f' which is
f :: t1 -> t -> (t, t1)
but can't determine the type of 'tw'.
Supposed type of tw:
tw :: (t -> t) -> t -> t
thanks!
回答1:
Let us analyze the function tw:
tw f x = f (f x)
tw takes as parameters f and x. At the moment we dot not know much about these parameters, so we will give these as types f :: a and x :: b.
Now we see a function application with f the function and x the parameter. This thus means that f is a function that takes a value of type b (the type of x), and returns something. We thus specify that f has as type f :: b -> c, with c a new type variable we introduce. We thus know that f x :: c.
We furthermore see, that there is a function application with f :: b -> c the function, and f x :: c the parameter. Since the type of the parameter of f is b, and f x has as type c. We thus come to the conclusion, that b and c must be the same type.
This thus means that we derived as types:
x :: b
f :: b -> b
We can furthermore analyze the type of tw f x by determining the type of f (f x). Since f x has type f x :: b, and f has type f :: b -> b, we know that f (f x) has type f (f x) :: b. So that means that the type for tw is:
tw :: (b -> b) -> b -> b
If we substitute b for t, then we obtain the expected type signature. But since b and t are just variables, that does not matter much.
来源：https://stackoverflow.com/questions/58285479/how-do-i-determine-type-of-haskell-functions