nic*_*las 3 haskell typeclass quantifiers
我不确定为什么ko不进行类型检查.有没有特别启发性的解释?
{-# LANGUAGE ScopedTypeVariables #-}
{-# LANGUAGE RankNTypes #-}
{-# LANGUAGE NoMonomorphismRestriction, FlexibleInstances #-}
module Wrap where
class ExpSYM repr where
lit :: Int -> repr
newtype Wrapped = Wrapped{unWrap :: forall repr. ExpSYM repr => repr}
a = (lit <$> Just 5) :: ExpSYM expr => Maybe expr
ko :: Maybe Wrapped
ko = do v <- a
return $ Wrapped $ v
ok :: Maybe Wrapped
ok = do v <- Just 5
let e = lit v
return $ Wrapped $ e
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编译器提到
SO.hs:15:14: error:
• No instance for (ExpSYM a0) arising from a use of ‘a’
• In a stmt of a 'do' block: v <- a
In the expression:
do { v <- a;
return $ Wrapped $ v }
In an equation for ‘ko’:
ko
= do { v <- a;
return $ Wrapped $ v }
SO.hs:16:28: error:
• Couldn't match expected type ‘repr’ with actual type ‘a0’
because type variable ‘repr’ would escape its scope
This (rigid, skolem) type variable is bound by
a type expected by the context:
ExpSYM repr => repr
at SO.hs:16:18-28
• In the second argument of ‘($)’, namely ‘v’
In the second argument of ‘($)’, namely ‘Wrapped $ v’
In a stmt of a 'do' block: return $ Wrapped $ v
• Relevant bindings include v :: a0 (bound at SO.hs:15:9)
Failed, modules loaded: none.
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编辑: 在Oleg的注释中找到了一个很好的解决方案来绕过这个,这是专门化类型,以便类型应用程序删除多态,添加实例
instance ExpSYM Wrapped where
lit x = Wrapped $ lit x
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然后我们有
notko :: Maybe Wrapped
notko = do v <- a
return $ v -- note the difference. what's the type of a ?
-- and we get all the usual goodies, no silly impredicative error
alsoOk = lit <$> Just 5 :: Maybe Wrapped
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ko只有在那种类型的a情况下才会起作用
a :: Maybe (? expr . ExpSYM expr => expr)
a = lit <$> Just 5
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...因为只有这样你才能做 - 解开它以获得多态值v :: ? expr . ExpSYM expr => expr.该值必须是多态的,因此它实际上可以用于Wrapped.
但这Maybe (? expr . ExpSYM expr => expr)是一种不可预测的类型.GHC Haskell不支持impredicative类型.
OTOH,在ok,v只是一个无聊的老整数一个引人注意的到来Just 5 :: Maybe Int.只e引入了多态性,但在Maybemonad 之外也是如此.