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秩-2多态性及其约束

higher-rank-types polymorphism haskell

2

Nikita Volkov · 技术社区 · 12 年前

我正试图重现 ST s rank-2多态性技巧,以确保某些值不能逃脱自定义monad。以下代码代表了我的解决方案的精神:

data STLike s a = STLike a
class Box box where
  runBox :: box a -> a

run :: Box box => (forall s. box (STLike s a)) -> a
run box = let STLike a = runBox box in a

虽然上面的代码编译得很好,但当我试图在 box (STLike s a) :

data STLike s a = STLike a
class Box box where
  runBox :: box a -> a

run :: Box box => (forall s. Eq (box (STLike s a)) => box (STLike s a)) -> a
run box = let STLike a = runBox box in a

第二个代码段编译失败,并显示以下消息:

Could not deduce (Eq (box (STLike t0 a)))
    arising from a use of `box'
  from the context (Box box)
    bound by the type signature for
               run :: Box box =>
                      (forall s. Eq (box (STLike s a)) => box (STLike s a)) -> a
    at src/TPM/GraphDB/Event.hs:36:8-76
  The type variable `t0' is ambiguous

有可能绕过这一点吗?如果没有,为什么?

1 回复 | 直到 12 年前

1

aavogt 12 年前

您需要为ghc提供一种方法来构成该实例。一种方法是:

{-# LANGUAGE FlexibleContexts, RankNTypes #-}
data STLike s a = STLike a deriving (Eq)
class Box box where
  runBox :: box a -> a

newtype Boxed box a = Boxed (box a)

instance (Box box, Eq a) => Eq (Boxed box a) where
    Boxed a == Boxed b = runBox a == runBox b

run :: (Eq a, Box box) => (forall s. Eq (Boxed box (STLike s a)) 
                                            => box (STLike s a))
                       -> a
run box = case runBox box of STLike a -> a

我认为你不能有 instance Eq (box (STLike s a)) ,所以有一个上面不方便的新类型。