我试图将基于变形金刚的monad堆栈中的简单解释器重写为基于更自由的效果，但我遇到了将我的意图传达给GHC类型系统的困难。
我目前只使用State和Fresh效果。我使用了两种状态，我的效果运动员看起来像这样：
runErlish g ls = run . runGlobal g . runGensym 0 . runLexicals ls
where runGlobal = flip runState
runGensym = flip runFresh'
runLexicals = flip runState
除此之外，我已经使用这种类型定义了一个FindMacro函数：
findMacro :: Members [State (Global v w), State [Scope v w]] r
=> Arr r Text (Maybe (Macro (Term v w) v w))
到目前为止所有这些都完美无缺。问题来自于我尝试编写macroexpand2（嗯，macroexpand1，但我简化它以便更容易理解这个问题）：
macroexpand2 s =
do m <- findMacro s
return $ case m of
Just j -> True
Nothing -> False
这会产生以下错误：
Could not deduce (Data.Open.Union.Member'
(State [Scope v0 w0])
r
(Data.Open.Union.FindElem (State [Scope v0 w0]) r))
from the context (Data.Open.Union.Member'
(State [Scope v w])
r
(Data.Open.Union.FindElem (State [Scope v w]) r),
Data.Open.Union.Member'
(State (Global v w))
r
(Data.Open.Union.FindElem (State (Global v w)) r))
bound by the inferred type for `macroexpand2':
(Data.Open.Union.Member'
(State [Scope v w])
r
(Data.Open.Union.FindElem (State [Scope v w]) r),
Data.Open.Union.Member'
(State (Global v w))
r
(Data.Open.Union.FindElem (State (Global v w)) r)) =>
Text -> Eff r Bool
at /tmp/flycheck408QZt/Erlish.hs:(79,1)-(83,23)
The type variables `v0', `w0' are ambiguous
When checking that `macroexpand2' has the inferred type
macroexpand2 :: forall (r :: [* -> *]) v (w :: [* -> *]).
(Data.Open.Union.Member'
(State [Scope v w])
r
(Data.Open.Union.FindElem (State [Scope v w]) r),
Data.Open.Union.Member'
(State (Global v w))
r
(Data.Open.Union.FindElem (State (Global v w)) r)) =>
Text -> Eff r Bool
Probable cause: the inferred type is ambiguous
好的，我可以在类型中添加Members注释：
macroexpand2 :: Members [State (Global v w), State [Scope v w]] r
=> Text -> Eff r Bool
现在我明白了：
Overlapping instances for Member (State [Scope v0 w0]) r
arising from a use of `findMacro'
Matching instances:
instance Data.Open.Union.Member'
t r (Data.Open.Union.FindElem t r) =>
Member t r
-- Defined in `Data.Open.Union'
There exists a (perhaps superclass) match:
from the context (Members
'[State (Global v w), State [Scope v w]] r)
bound by the type signature for
macroexpand2 :: Members
'[State (Global v w), State [Scope v w]] r =>
Text -> Eff r Bool
at /tmp/flycheck408QnV/Erlish.hs:(79,17)-(80,37)
(The choice depends on the instantiation of `r, v0, w0'
To pick the first instance above, use IncoherentInstances
when compiling the other instance declarations)
In a stmt of a 'do' block: m <- findMacro s
In the expression:
do { m <- findMacro s;
return
$ case m of {
Just j -> True
Nothing -> False } }
In an equation for `macroexpand2':
macroexpand2 s
= do { m <- findMacro s;
return
$ case m of {
Just j -> True
Nothing -> False } }
我被告知irc尝试forall r v w.没有任何区别。出于好奇，我在编译这个代码时尝试使用IncoherentInstances（我不喜欢检查更自由和播放的分支），看看它是否会给我一个关于什么是线索的线索继续它没有：
Could not deduce (Data.Open.Union.Member'
(State [Scope v0 w0])
r
(Data.Open.Union.FindElem (State [Scope v0 w0]) r))
arising from a use of `findMacro'
from the context (Members
'[State (Global v w), State [Scope v w]] r)
bound by the type signature for
macroexpand2 :: Members
'[State (Global v w), State [Scope v w]] r =>
Text -> Eff r Bool
at /tmp/flycheck408eru/Erlish.hs:(79,17)-(80,37)
The type variables `v0', `w0' are ambiguous
Relevant bindings include
macroexpand2 :: Text -> Eff r Bool
(bound at /tmp/flycheck408eru/Erlish.hs:81:1)
Note: there are several potential instances:
instance (r ~ (t' : r'), Data.Open.Union.Member' t r' n) =>
Data.Open.Union.Member' t r ('Data.Open.Union.S n)
-- Defined in `Data.Open.Union'
instance (r ~ (t : r')) =>
Data.Open.Union.Member' t r 'Data.Open.Union.Z
-- Defined in `Data.Open.Union'
In a stmt of a 'do' block: m <- findMacro s
In the expression:
do { m <- findMacro s;
return
$ case m of {
Just j -> True
Nothing -> False } }
In an equation for `macroexpand2':
macroexpand2 s
= do { m <- findMacro s;
return
$ case m of {
Just j -> True
Nothing -> False } }
所以，这是我对更自由的内部结构的理解用完了，我有疑问：
干杯！
答案 0 :(得分：17)
可扩展效果的类型推断在历史上一直很糟糕。让我们看一些例子：
{-# language TypeApplications #-}
-- mtl
import qualified Control.Monad.State as M
-- freer
import qualified Control.Monad.Freer as F
import qualified Control.Monad.Freer.State as F
-- mtl works as usual
test1 = M.runState M.get 0
-- this doesn't check
test2 = F.run $ F.runState F.get 0
-- this doesn't check either, although we have a known
-- monomorphic state type
test3 = F.run $ F.runState F.get True
-- this finally checks
test4 = F.run $ F.runState (F.get @Bool) True
-- (the same without TypeApplication)
test5 = F.run $ F.runState (F.get :: F.Eff '[F.State Bool] Bool) True
我将尝试解释一般问题并提供最少的代码说明。代码的自包含版本可以是found here。
在最基本的级别（忽略优化的表示），Eff的定义如下：
{-# language
GADTs, DataKinds, TypeOperators, RankNTypes, ScopedTypeVariables,
TypeFamilies, DeriveFunctor, EmptyCase, TypeApplications,
UndecidableInstances, StandaloneDeriving, ConstraintKinds,
MultiParamTypeClasses, FlexibleInstances, FlexibleContexts,
AllowAmbiguousTypes, PolyKinds
#-}
import Control.Monad
data Union (fs :: [* -> *]) (a :: *) where
Here :: f a -> Union (f ': fs) a
There :: Union fs a -> Union (f ': fs) a
data Eff (fs :: [* -> *]) (a :: *) =
Pure a | Free (Union fs (Eff fs a))
deriving instance Functor (Union fs) => Functor (Eff fs)
换句话说，Eff是来自仿函数列表的联合的免费monad。 Union fs a表现得像n-ary Coproduct。对于两个仿函数，二进制Coproduct类似于Either：
data Coproduct f g a = InL (f a) | InR (g a)
相比之下，Union fs a让我们从仿函数列表中选择一个仿函数：
type MyUnion = Union [[], Maybe, (,) Bool] Int
-- choose the first functor, which is []
myUnion1 :: MyUnion
myUnion1 = Here [0..10]
-- choose the second one, which is Maybe
myUnion2 :: MyUnion
myUnion2 = There (Here (Just 0))
-- choose the third one
myUnion3 :: MyUnion
myUnion3 = There (There (Here (False, 0)))
让我们以State效果为例。首先，我们需要为Functor设置Union fs个实例，因为Eff只有Monad实例Functor (Union fs)。
Functor (Union '[])是微不足道的，因为空联盟没有值：
instance Functor (Union '[]) where
fmap f fs = case fs of {} -- using EmptyCase
否则我们会在联盟中添加一个仿函数：
instance (Functor f, Functor (Union fs)) => Functor (Union (f ': fs)) where
fmap f (Here fa) = Here (fmap f fa)
fmap f (There u) = There (fmap f u)
现在是State定义和跑步者：
run :: Eff '[] a -> a
run (Pure a) = a
data State s k = Get (s -> k) | Put s k deriving Functor
runState :: forall s fs a. Functor (Union fs) => Eff (State s ': fs) a -> s -> Eff fs (a, s)
runState (Pure a) s = Pure (a, s)
runState (Free (Here (Get k))) s = runState (k s) s
runState (Free (Here (Put s' k))) s = runState k s'
runState (Free (There u)) s = Free (fmap (`runState` s) u)
我们已经可以开始编写和运行我们的Eff程序了，尽管我们缺乏所有的糖和便利：
action1 :: Eff '[State Int] Int
action1 =
Free $ Here $ Get $ \s ->
Free $ Here $ Put (s + 10) $
Pure s
-- multiple state
action2 :: Eff '[State Int, State Bool] ()
action2 =
Free $ Here $ Get $ \n -> -- pick the first effect
Free $ There $ Here $ Get $ \b -> -- pick the second effect
Free $ There $ Here $ Put (n < 10) $ -- the second again
Pure ()
现在：
> run $ runState action1 0
(0,10)
> run $ (`runState` False) $ (`runState` 0) action2
(((),0),True)
这里只缺少两件必不可少的东西。
第一个是Eff的monad实例，它允许我们使用do - 符号代替Free和Pure，还让我们使用许多多态monadic功能。我们将在此处跳过它，因为它写得很直接。
第二个是推理/重载，用于从效果列表中选择效果。以前我们需要编写Here x以选择第一个效果，There (Here x)选择第二个效果，依此类推。相反，我们想在效果列表中编写多态的代码，因此我们必须指定的是，某些效果是列表的元素，而某些隐藏的类型类魔法将插入适当的{ {1}} - ：S
当There是Member f fs的元素时，我们需要一个f a类，可以将Union fs a - s注入f - s。从历史上看，人们已经以两种方式实施了它。
首先，直接使用fs：
OverlappingInstances其次，通过首先使用类型系列计算class Member (f :: * -> *) (fs :: [* -> *]) where
inj :: f a -> Union fs a
instance Member f (f ': fs) where
inj = Here
instance {-# overlaps #-} Member f fs => Member f (g ': fs) where
inj = There . inj
-- it works
injTest1 :: Union [[], Maybe, (,) Bool] Int
injTest1 = inj [0]
injTest2 :: Union [[], Maybe, (,) Bool] Int
injTest2 = inj (Just 0)
中f的索引，然后使用不重叠的类实现fs，由inj引导-s计算索引。这通常被认为更好，因为人们往往不喜欢重叠的实例。
f data Nat = Z | S Nat
type family Lookup f fs where
Lookup f (f ': fs) = Z
Lookup f (g ': fs) = S (Lookup f fs)
class Member' (n :: Nat) (f :: * -> *) (fs :: [* -> *]) where
inj' :: f a -> Union fs a
instance fs ~ (f ': gs) => Member' Z f fs where
inj' = Here
instance (Member' n f gs, fs ~ (g ': gs)) => Member' (S n) f fs where
inj' = There . inj' @n
type Member f fs = Member' (Lookup f fs) f fs
inj :: forall fs f a. Member f fs => f a -> Union fs a
inj = inj' @(Lookup f fs)
-- yay
injTest1 :: Union [[], Maybe, (,) Bool] Int
injTest1 = inj [0]
库使用第二种解决方案，而freer使用第一种方法用于早于7.8的GHC版本，第二种用于较新的GHC-s。
无论如何，两个解决方案都有相同的推理限制，即我们几乎总是extensible-effects只有具体的单态类型，而不是包含类型变量的类型。 ghci中的示例：
Lookup这是有效的，因为> :kind! Lookup Maybe [Maybe, []]
Lookup Maybe [Maybe, []] :: Nat
= 'Z
或Maybe中没有类型变量。
[Maybe, []]这个因为> :kind! forall a. Lookup (Either a) [Either Int, Maybe]
forall a. Lookup (Either a) [Either Int, Maybe] :: Nat
= Lookup (Either a) '[Either Int, Maybe]
类型变量阻止减少而被卡住了。
a这是有效的，因为我们对任意类型变量的唯一了解是它们等于它们自己，而> :kind! forall a. Lookup (Maybe a) '[Maybe a]
forall a. Lookup (Maybe a) '[Maybe a] :: Nat
= Z
等于a。
一般来说，类型族减少会受到变量的影响，因为约束求解可能会在以后将它们精炼到不同类型，因此GHC不能对它们做任何假设（除了等于它们自己）。基本上同样的问题出现在a实现中（尽管没有任何类型的系列）。
让我们根据这一点重新审视OverlappingInstances。
freer GHC知道我们有一个具有单一效果的堆栈，因为import Control.Monad.Freer
import Control.Monad.Freer.State
test1 = run $ runState get 0 -- error
适用于run。它也知道该效果必须是Eff '[] a。但是当我们写State s时，GHC只知道它对某些新的get变量有State t效果，并且t必须保持，所以当它尝试计算相当于Num t的{{1}}，它会卡在类型变量上，并且任何进一步的实例解析都会在卡住类型族表达式上跳起来。另一个例子：
freer这也失败了，因为GHC需要计算Lookup (State t) '[State s]，虽然我们知道状态必须是foo = run $ runState get False -- error
，但由于Lookup (State s) '[State Bool]变量，这仍然会卡住。
Bool这是有效的，因为s的状态类型可以解析为foo = run $ runState (modify not) False -- this works
，modify not会减少。
现在，经过这次大迂回之后，我将在你的帖子结尾处提出你的问题。
Bool并不表示任何可能的解决方案，只是一种类型错误工件。我需要更多的代码上下文来确定它究竟是如何产生的，但我确定它不相关，因为只要Lookup (State Bool) '[State Bool]被卡住，案件就变得无望了。
Overlapping instances也无关紧要，没有帮助。我们需要一个具体的效果位置索引，以便能够为程序生成代码，如果Lookup卡住，我们就无法通过索引来提取索引。
IncoherentInstances的问题在于它对状态内的类型变量有Lookup个效果。只要您想使用findMacro，就必须确保State和findMacro的{{1}}和v参数是已知的具体类型。您可以通过输入注释来执行此操作，或者更方便地使用TypeApplications，并编写w以指定Scope和Global。如果您在多态函数中有findMacro @Int @Int，则需要为该函数使用v = Int注释启用w = Int，绑定findMacro和ScopedTypeVariables，并写{ {1}}当你使用它时。您还需要为多态v或w启用forall v w.（如评论中所述）。我认为虽然在GHC 8中它是一个合理的扩展，与findMacro @v @w一起启用。
附录：
然而，幸运的是，新的GHC 8功能允许我们修改可扩展效果的类型推断，我们可以推断{-# language AllowAmbiguousTypes #-}可以做的所有事情，以及v无法处理的一些事情。关于效果的排序，新类型推断也是不变的。
我有一个minimal implementation here以及一些例子。但是，它还没有用在我所知道的任何效果库中。我可能会对其进行一次注释，并执行拉取请求以将其添加到w。