【发布时间】:2017-11-07 14:52:32
【问题描述】:
我不确定为什么 ko 不进行类型检查。
有没有特别有启发性的解释?
{-# 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
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.
编辑: 在 Oleg 的笔记中找到了一个很好的解决方案来规避这个问题,即专门化类型,以便通过类型应用程序删除多态性,添加实例
instance ExpSYM Wrapped where
lit x = Wrapped $ lit x
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
【问题讨论】:
-
“不起作用”是什么意思。请提供编译错误/意外输出等。
-
我的意思当然是类型检查。每个人都知道在类型检查之后,一切正常;) - (虽然你是对的)
-
该错误表明
newtype已经出错,而不是ko/ok。
标签: haskell typeclass quantifiers