Question

我试图在一个Reader中传递一个函数，该函数将从与调用函数相同的monad中调用，但是我得到了一个无限类型的错误。

简化代码是：

{-# LANGUAGE FlexibleContexts #-}

module G2 where

import Control.Monad
import Control.Monad.Reader

data Foo m = Foo { bar :: m () }

runFoo :: MonadReader (Foo m) m => m ()
runFoo = do
    b <- asks bar
    b

main :: Monad m => m ()
main = do
    let bar = return () :: m ()
        foo = Foo bar
    runReaderT runFoo foo

错误是：

    • Occurs check: cannot construct the infinite type:
        m0 ~ ReaderT (Foo m0) m
        arising from a functional dependency between:
          constraint ‘MonadReader
                        (Foo (ReaderT (Foo m0) m)) (ReaderT (Foo m0) m)’
            arising from a use of ‘runFoo’
          instance ‘MonadReader r (ReaderT r m1)’ at <no location info>
    • In the first argument of ‘runReaderT’, namely ‘runFoo’
      In a stmt of a 'do' block: runReaderT runFoo foo
      In the expression:
        do let bar = ...
               foo = Foo bar
           runReaderT runFoo foo
    • Relevant bindings include main :: m () (bound at G2.hs:16:1)
   |
19 |     runReaderT runFoo foo
   |                ^^

非常感谢任何帮助，谢谢！

Answer 1

runFoo :: MonadReader (Foo m) m => m ()

让我们忘记课程，并假设MonadReader env mon表示mon ~ ((->) env)。这相当于简单地使用(->)作为我们的monad而不是鸽友ReaderT。然后你得到m ~ ((->) m) => m ()。您看到m需要包含自身（具体而言，m的参数为m）。这对于值来说是可以的，但如果类型检查器必须处理无限大的类型，那将是非常糟糕的。 ReaderT也是如此（您需要使用ReaderT，因为您调用了runReaderT runFoo）。您需要定义另一个newtype来编码此递归：

data RecReader c a = RecReader { runRecReader :: c (RecReader c) -> a }
instance Functor (RecReader c) where
  fmap f (RecReader r) = RecReader $ f . r
instance Applicative (RecReader c) where
 pure = RecReader . const
 RecReader f <*> RecReader g = RecReader $ \e -> f e (g e)
instance Monad (RecReader c) where
  return = pure
  RecReader x >>= f = RecReader $ \e -> runRecReader (f (x e)) e
instance MonadReader (c (RecReader c)) (RecReader c) where
  ask = RecReader id
  local f (RecReader x) = RecReader $ x . f

它有效：

runRecReader runFoo (Foo $ return ())
-- ==>
()

＆＃34; MonadReader（Foo m）m＆＃34;从函数依赖导致无限类型

1 个答案: