Question

我试图允许在我的免费monad中嵌入州monad;这是我的简单尝试：

{-# language FlexibleInstances, MultiParamTypeClasses #-}
module Main where

import Control.Monad.Free
import Control.Monad.State
import Data.Bifunctor

data Toy state next =
      Output String next
    | LiftState (state -> (next, state))
    | Done

instance Functor (Toy s) where
  fmap f (Output str next) = Output str $ f next
  fmap f (LiftState stateF) = LiftState (first f . stateF)
  fmap f Done = Done

instance MonadState s (Free (Toy s)) where
  state = overState

overState :: (s -> (a, s)) -> Free (Toy s) a
overState = liftF . LiftState

output :: Show a => a -> Free (Toy s) ()
output x = liftF $ Output (show x) ()

done :: Free (Toy s) r
done = liftF Done

program :: Free (Toy Int) ()
program = do
  start <- get
  output start
  modify ((+10) :: (Int -> Int))
  end <- get
  output end
  done

interpret :: (Show r) => Free (Toy s) r -> s -> IO ()
interpret (Free (LiftState stateF)) s = let (next, newS) = stateF s
                                         in interpret next newS
interpret (Free (Output str next)) s = print str >> interpret next s
interpret (Free Done) s = return ()
interpret (Pure x) s = print x

main :: IO ()
main = interpret program (5 :: Int)

我收到错误：

• Overlapping instances for MonadState Int (Free (Toy Int))
    arising from a use of ‘get’
  Matching instances:
    instance [safe] (Functor m, MonadState s m) =>
                    MonadState s (Free m)
      -- Defined in ‘Control.Monad.Free’
    instance MonadState s (Free (Toy s))
      -- Defined at app/Main.hs:18:10
• In a stmt of a 'do' block: start <- get
  In the expression:
    do { start <- get;
         output start;
         modify ((+ 10) :: Int -> Int);
         end <- get;
         .... }
  In an equation for ‘program’:
      program
        = do { start <- get;
               output start;
               modify ((+ 10) :: Int -> Int);
               .... }

我可以收集;它正在尝试应用此实例：

(Functor m, MonadState s m) => MonadState s (Free m)

来自 free package here;但是在这种情况下，它必须匹配Free (Toy s)并且根据需要没有MonadState s (Toy s)，所以我不明白为什么它认为它适用。

如果我删除了我的实例定义，我会得到：

• No instance for (MonadState Int (Toy Int))
    arising from a use of ‘modify’

这支持了我的想法，其他实例并没有真正适用;如何使用我指定的实例进行编译？你能解释为什么会这样吗？是因为FlexibleInstances被使用了吗？

谢谢！

Answer 1

选择实例时，将忽略实例上下文（(Functor m, MonadState s m)位）。这是为了防止编译器不必进行可能代价高昂的回溯搜索来选择实例。因此，如果两个实例适用并且仅因为实例上下文而排除了一个实例，就像在您的情况下那样，它是重叠的。

这是mtl设计的一个不幸的部分，我认为每个Haskell程序员在某些方面都遇到了反对。解决方案没有太多选择;通常你添加一个newtype并给你的实例，如

newtype FreeToy s a = FreeToy (Free (Toy s) a)
instance MonadState s (FreeToy s) where -- ...

为什么我会得到＆＃39;重叠实例＆＃39;当一个人不匹配时出错？

1 个答案: