在我尝试理解存在类型时,我已经读过教会编码以及Rank-N类型扩展将足以在没有存在量化的情况下在Haskell中对它们进行编码。我找到了这个直截了当的例子:
type Obj = forall y. (forall x. (Show x) => x -> y) -> y
obj :: Obj
obj f = f "hello"
app :: Obj -> String
app obj = obj (\x -> show x)
在Haskell Wiki中,我偶然发现了以下基于存在类型的异构列表示例:
data Obj = forall a. (Show a) => Obj a
xs :: [Obj]
xs = [Obj 1, Obj "foo", Obj 'c']
doShow :: [Obj] -> String
doShow [] = ""
doShow ((Obj x):xs) = show x ++ doShow xs
现在我尝试用Church表达这个实现,并且因非法多态类型错误而失败:
type Obj = forall y. (forall x. (Show x) => x -> y) -> y
obj1 :: Obj
obj1 f = f 1
obj2 :: Obj
obj2 f = f "foo"
obj3 :: Obj
obj3 f = f 'c'
xs :: [Obj]
xs = [obj1, obj2, obj3]
doShow :: [Obj] -> String
doShow [] = ""
doShow (obj:xs) = obj (\x -> show x ++ doShow xs)
我猜这个翻译很简单,只是错误。存在类型可以用Church / Rank-N编码是正确的吗? 它是如何正确完成的?