问题
I'm kind of stuck with an assignement concerning my exams. I want to find out the types of those two functions by applying the unifying algorithm by hand:
map map
(\x -> x >>= (\y -> y))
Could someone point me to the right direction? The only ressource I could find until now was the wikipedia entry which is not really aiding me because of the high level of abstraction.
Greetings and thank you.
回答1:
Let's just do the first.
map :: (a -> b) -> [a] -> [b]
Now we can write it again with two different names, for clarity:
map :: (c -> d) -> [c] -> [d]
Now we substitute the second as the first parameter of the first, getting:
(a -> b) === (c -> d) -> ([c] -> [d]) (recall the associativity of (->))
a === (c -> d)
b === ([c] -> [d])
Now we substitute those type assignments into the remaining portion of the first signature, getting
map map :: [c -> d] -> [[c] -> [d]]
Clear?
回答2:
The type of map is map :: (a -> b) -> [a] -> [b]. Hence the type of map foo is obtained from [a] -> [b] by substituting a and b with what can be derived from foo's type. If, for example, foo :: t -> t, you substitute a = t, b = t and obtain [t] -> [t]. If foo :: [t] -> Int, you obtain [[t]] -> [Int].
In your case, the type of foo (which is map) is (x -> y) -> [x] -> [y]. You have to unify that type with a -> b to find out what a and b have to be substituted with. [Note that the function arrow is right-associative, x -> y -> z = x -> (y -> z).]
To find the type of
\x -> x >>= (\y -> y)
use the known type of (>>=) :: Monad m => m a -> (a -> m b) -> m b. Ignore the constraint (Monad m =>) for now.
As the first argument of (>>=), x must have a type m a for as yet unknown m and a. The second argument of (>>=) is here the identity,
(\y -> y) :: t -> t
so you must unify t -> t with a -> m b. That gives you some information about a, namely a = m b.
That gives x :: m (m b), and (\x -> x >>= (\y -> y)) :: type_of_x -> type_of_rhs.
Finally remember the temporarily forgotten constraint Monad m =>.
来源：https://stackoverflow.com/questions/17148385/unifying-types-in-haskell