存储和检索某个类型类的对象 - 这可能吗？如果有,怎么样？

Question

假设我有一些类型类Foo和一些数据类型FooInst是以下实例Foo:

class Foo a where
  foo :: a -> String

data FooInst = FooInst String

instance Foo FooInst where
    foo (FooInst s) = s

现在,我想定义一种数据类型,该类型存储类型在Foo类型类中的对象,并且能够从该数据类型中提取该对象并使用它.

我找到的唯一方法是使用GADT和Rank2Types语言扩展并定义数据类型,如下所示:

data Container where
  Container :: { content :: Foo a => a } -> Container

但问题是,我无法使用content选择器从Container中获取内容:

cont :: Container
cont = Container{content = FooInst "foo"}

main :: IO ()
main = do
  let fi = content cont
  putStrLn $ foo fi

导致编译错误

Cannot use record selector ‘content’ as a function due to escaped type variables
Probable fix: use pattern-matching syntax instead

但是当我修改let ...线路时

let Conainer fi = cont

我得到一个相当有趣的错误

My brain just exploded
I can't handle pattern bindings for existential or GADT data constructors.
Instead, use a case-expression, or do-notation, to unpack the constructor.

如果我再次尝试修改该let ...行以使用case-expression

let fi = case cont of
           Container x -> x

我得到了一个不同的错误

Couldn't match expected type ‘t’ with actual type ‘a’
  because type variable ‘a’ would escape its scope
This (rigid, skolem) type variable is bound by
  a pattern with constructor
    Container :: forall a. (Foo a => a) -> Container,
  in a case alternative
  at test.hs:23:14-24

那么,我如何存储一个typeclassed的东西并将其取回？

Answer 1

附:

data Container where
    Container :: {content :: Foo a => a} -> Container

甚至没有强制执行类型类约束.那是

void :: Container
void = Container {content = 42 :: Int}

类型检查,即使42 :: Int是没有的一个实例Foo.

但如果你改为:

data Container where
    Container :: Foo a => {content :: a} -> Container

1. 你不再需要Rank2Types语言扩展.
2. 强制执行类型类约束; 因此上面的void例子将不再进行类型检查.

3. 此外,您可以使用模式匹配对内容调用foo(或任何其他带签名的函数Foo a => a -> ...):

   case cont of Container {content = a} -> foo a