构造函数内部和构造函数外部的 forall 量词之间的区别

Question

{-# LANGUAGE DataKinds #-}
{-# LANGUAGE ExplicitForAll #-}
{-# LANGUAGE KindSignatures #-}
{-# LANGUAGE PolyKinds #-}
{-# LANGUAGE RankNTypes #-}
{-# LANGUAGE ExistentialQuantification #-}

import Data.Proxy

data Foo = FooA | FooB

class Bar (a :: k) where
    bar :: Proxy a -> Int

instance Bar FooA where
    bar _ = 1

instance Bar FooB where
    bar _ = 2

foo1 :: forall (a :: Foo). Proxy a -> (Bar a => Proxy a)
foo1 p = p

data BarProxy = BarProxy (forall a. Bar a => Proxy a)

foo2 :: forall (a :: Foo). Proxy a -> BarProxy
foo2 p = BarProxy (foo1 p)

main = print "Hello World"

In this code:

1. Doesn't foo1 , given any Proxy a where a is of kind Foo , return a Proxy a such that a has an instance of Bar ?
2. Doesn't BarProxy constructor accept any Proxy a , where a has an instance of Bar ? How is it different from data BarProxy = forall a. BarProxy (Bar a => Proxy a) data BarProxy = forall a. BarProxy (Bar a => Proxy a) ?
3. Why does foo2 p = BarProxy (foo1 p) fail with the below error?

Test6.hs:27:20: error:
    • Couldn't match type ‘a1’ with ‘a’
      ‘a1’ is a rigid type variable bound by
        a type expected by the context:
          forall (a1 :: Foo). Bar a1 => Proxy a1
        at Test6.hs:27:10-26
      ‘a’ is a rigid type variable bound by
        the type signature for:
          foo2 :: forall (a :: Foo). Proxy a -> BarProxy
        at Test6.hs:26:1-46
      Expected type: Proxy a1
        Actual type: Proxy a
    • In the first argument of ‘BarProxy’, namely ‘(foo1 p)’
      In the expression: BarProxy (foo1 p)
      In an equation for ‘foo2’: foo2 p = BarProxy (foo1 p)
    • Relevant bindings include
        p :: Proxy a (bound at Test6.hs:27:6)
        foo2 :: Proxy a -> BarProxy (bound at Test6.hs:27:1)
   |
27 | foo2 p = BarProxy (foo1 p)
   |                    ^^^^^^

Answer 1

1. No. My understanding is that the signature of foo1 is the same as forall (a :: Foo). Bar a => Proxy a -> Proxy a forall (a :: Foo). Bar a => Proxy a -> Proxy a (though I couldn't find any document). In ghci, :t foo1 gives foo1 :: Bar a => Proxy a -> Proxy a . Given any Proxy a where a is of kind Foo and an instance of Bar , it returns Proxy a .

2. No. The constructor BarProxy has rank-2 polymorphic type (forall a. Bar a => Proxy a) -> BarProxy . This means that an argument p can be passed to BarProxy only if p has the type Proxy a for all type a which is an instance of Bar . If you want existentially quantified one, you may write

   data BarProxy = forall a. Bar a => BarProxy (Proxy a)

   or

   data BarProxy where BarProxy :: forall a. Bar a => Proxy a -> BarProxy

   with -XGADTs enabled.

   Let us call BarProxy of type forall a. Bar a => Proxy a -> BarProxy forall a. Bar a => Proxy a -> BarProxy existential BarProxy and that of type (forall a. Bar a => Proxy a) -> BarProxy universal BarProxy . For existential one, the argument p should have type either Proxy FooA or Proxy FooB (existentially quantified over {a | a is an instance of Bar} = {FooA,FooB} ). For universal one, on the other hand, p should have type both Proxy FooA and Proxy FooB (universally quantified over the same set). Let us consider three proxies below.

    proxyFooA :: Proxy FooA proxyFooA = Proxy proxyFooB :: Proxy FooB proxyFooB = Proxy proxyPoly :: forall a. Proxy a proxyPoly = Proxy

   Existential BarProxy accepts any of the three while universal one accepts only proxyPoly .

3. foo2 p = BarProxy (foo1 p) compiles for existential BarProxy .