作为我上一个问题的跟进，同时使用makeLenses，类约束和类型同义词，我想了解一个新的类型错误。
类型错误是由type S = (Num n) => State n以下示例中引入类型同义词引起的。
{-# LANGUAGE TemplateHaskell #-}
{-# LANGUAGE RankNTypes #-}
module Foo where
import Control.Lens
data State a = State { _a :: a
} deriving Show
makeLenses ''State -- Requires TemplateHaskell
-- | Smart constructor enforcing class constraint on record field _a.
mkState :: (Num a) => a -> State a
mkState n = State {_a = n}
doStuff1 :: Num a => State a -> State a
doStuff1 s = s & a %~ (*2)
test1a = doStuff1 $ mkState 5 -- results in State {_a = 10.0}
test1b = doStuff1 $ mkState 5.5 -- results in State {_a = 11.0}
type S = (Num n) => State n -- Requires the RankNTypes extensions
doStuff2 :: S -> S
doStuff2 s = s & a %~ (*2)
test2a = doStuff2 $ mkState 5 -- Results in State {_a = 10.0}
--test2b = doStuff2 $ mkState 5.5 -- Type error.
如果我取消注释，则会test2b出现以下错误。
Could not deduce (Fractional n) arising from the literal `5.5'
from the context (Num n)
bound by a type expected by the context: Num n => State n
at Foo.hs:32:10-32
Possible fix:
add (Fractional n) to the context of
a type expected by the context: Num n => State n
In the first argument of `mkState', namely `5.5'
In the second argument of `($)', namely `mkState 5.5'
In the expression: doStuff2 $ mkState 5.5
我希望能够理解为什么引入的类型同义词会导致此错误，以及如何解读错误消息。
S -> S不等同于forall n. Num n => State n -> State n。等同于(forall n. Num n => State n) -> (forall n. Num n => State n)。前者意味着，对于所有数字类型n，我们可以传入aState n并取回a State n（对于同一类型n）。后者意味着我们传递可以State n用于所有数值类型的n东西，并取回可以State n适用于所有类型的东西n。换句话说，参数和结果都是多态的。
这意味着您传入的参数必须具有type Num n => State n，而不是更具体的类型，例如State Int。这是正确的5，具有类型Num n => n，但是没有5.5类型Fractional n => n。
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我来说两句