GHC / GHCi意外接受的代码

Question

I don't understand why this code should pass type-checking:

foo :: (Maybe a, Maybe b)
foo = let x = Nothing in (x,x)

Since each component is bound to the same variable x , I would expect that the most general type for this expression to be (Maybe a, Maybe a) . I get the same results if I use a where instead of a let . Am I missing something?

Answer 1

Briefly put, the type of x gets generalized by let . This is a key step in the Hindley-Milner type inference algorithm.

Concretely, let x = Nothing initially assigns x the type Maybe t , where t is a fresh type variable. Then, the type gets generalized, universally quantifying all its type variables (technically: except those in use elsewhere, but here we only have t ). This causes x :: forall t. Maybe t x :: forall t. Maybe t . Note that this is exactly the same type as Nothing :: forall t. Maybe t Nothing :: forall t. Maybe t .

Hence, each time we use x in our code, that refers to a potentially different type Maybe t , much like Nothing . Using (x, x) gets the same type as (Nothing, Nothing) for this reason.

Instead, lambdas do not feature the same generalization step. By comparison (\\x -> (x, x)) Nothing "only" has type forall t. (Maybe t, Maybe t) forall t. (Maybe t, Maybe t) , where both components are forced to be of the same type. Here x is again assigned type Maybe t , with t fresh, but it is not generalized. Then (x, x) is assigned type (Maybe t, Maybe t) . Only at the top-level we generalize adding forall t , but at that point is too late to obtain a heterogeneous pair.