为什么是 `[1, "a"] :: [forall a. 不允许显示 a => a]`？

Question

In my (might incorrect) understanding, following two lists should be equivalent:

[1, "a"] :: [forall a. Show a => a]

data V = forall a. Show a => V a
[V 1, V "a"] :: [V]

However, the first one is not accepted but the second one works fine (with ExistentialQuantification ).

If the first list doesn't exist, what would be the type in the blank of map V :: ??? -> [V] map V :: ??? -> [V] ? What type mechanism enforces the existence of the wrapper?

Answer 1

Your understanding is not right. A big part of the problem is that the traditional existential quantification syntax you've used is pretty confusing to anyone who isn't thoroughly familiar with it. I therefore strongly recommend that you use GADT syntax instead, which also has the benefit of being strictly more powerful. The easy thing to do is just to enable {-# LANGUAGE GADTs #-} . While we're at it, let's turn on {-# LANGUAGE ScopedTypeVariables #-} , because I hate wondering what forall means in any given spot. Your V definition means exactly the same thing as

data V where
  V :: forall a . Show a => a -> V

We can actually drop the explicit forall if we like:

data V where
  V :: Show a => a -> V

So the V data constructor is a function that takes something of any showable type and produces something of type V . The type of map is pretty restrictive:

map :: (a -> b) -> [a] -> [b]

All the elements of the list passed to map have to have the same type. So the type of map V is just

map V :: Show a => [a] -> [V]

Let's get back to your first expression now:

[1, "a"] :: [forall a. Show a => a]

Now what this actually says is that [1, "a"] is a list, each of whose elements has type forall a . Show a => a forall a . Show a => a . That is, if I provide any a that's an instance of Show , each element of the list should have that type. This is simply not true. "a" does not, for example, have type Bool . There's yet another problem here; the type [forall a . Show a => a] [forall a . Show a => a] is "impredicative". I don't understand the details of what that means, but loosely speaking you've stuck a forall in the argument of a type constructor other than -> , and that's not allowed. GHC might suggest that you enable the ImpredicativeTypes extension, but this really doesn't work right, so you shouldn't. If you want a list of existentially quantified things, you need to wrap them up first in existential datatypes or use a specialized existential list type. If you want a list of universally quantified things, you need to wrap them up first (probably in newtypes).