当我引入一个使用单子作为参数的新参数时,为什么函数之间的一元链接会中断?

以下代码:

-- tst2.hs - showing successful monadic linkage.

{-# LANGUAGE NoImplicitPrelude #-}
{-# LANGUAGE FlexibleContexts #-}

import Protolude
import Control.Monad.Extra  (unfoldM)

foo :: Monad m
    => Int
    -> m [[Int]]
foo n = evalStateT (traverse nxt [1..n]) 0
 where nxt _ = do s <- get
                  r <- bar s
                  put $ s + 1
                  return r

bar :: Monad m
    => Int
    -> m [Int]
bar n = unfoldM step n
  where step k = return $ if k > 0 then Just (k, k - 1)
                                   else Nothing

main :: IO ()
main = do xs <- foo 3
          print xs

工作正常,产生以下输出:

Davids-Air-2:haskell-rl dbanas$ stack runghc tst2.hs 
[[],[1],[2,1]]

但是,如果我稍微更改代码:

-- tst3.hs - showing breakage of monadic linkage.

{-# LANGUAGE NoImplicitPrelude #-}
{-# LANGUAGE FlexibleContexts #-}

import Protolude
import Control.Monad.Extra  (unfoldM)

data Dummy m = Dummy (Int -> m Int)

foo :: Monad m
    => Int
    -> Dummy m
    -> m [[Int]]
foo n d = evalStateT (traverse nxt [1..n]) 0
 where nxt _ = do s <- get
                  r <- bar s d
                  put $ s + 1
                  return r

bar :: Monad m
    => Int
    -> Dummy m
    -> m [Int]
bar n d = unfoldM step n
  where step k = return $ if k > 0 then Just (k, k - 1)
                                   else Nothing

main :: IO ()
main = do xs <- foo 3 $ Dummy (\n -> return n)
          print xs

我发现以下编译器错误:

Davids-Air-2:haskell-rl dbanas$ stack runghc tst3.hs 

tst3.hs:15:23: error:
    • Couldn't match type ‘m’ with ‘StateT Integer m’
      ‘m’ is a rigid type variable bound by
        the type signature for:
          foo :: forall (m :: * -> *). Monad m => Int -> Dummy m -> m [[Int]]
        at tst3.hs:(11,1)-(14,16)
      Expected type: StateT Integer m [[Int]]
        Actual type: m [[Int]]
    • In the first argument of ‘evalStateT’, namely
        ‘(traverse nxt [1 .. n])’
      In the expression: evalStateT (traverse nxt [1 .. n]) 0
      In an equation for ‘foo’:
          foo n d
            = evalStateT (traverse nxt [1 .. n]) 0
            where
                nxt _
                  = do s <- get
                       ....
    • Relevant bindings include
        nxt :: forall p. p -> m [Int] (bound at tst3.hs:16:8)
        d :: Dummy m (bound at tst3.hs:15:7)
        foo :: Int -> Dummy m -> m [[Int]] (bound at tst3.hs:15:1)
   |
15 | foo n d = evalStateT (traverse nxt [1..n]) 0
   |                       ^^^^^^^^^^^^^^^^^^^

结果证明,解决方案非常简单,尽管我花了几个小时来推断:

          r <- lift $ bar s d

Davids-Air-2:haskell-rl dbanas$ stack runghc tst3.hs 
[[],[1],[2,1]]

我对此有一个理论,希望得到证实:

Monad在中运行 bar 与在中运行的Monad不同 foo 。 具体来说,Monad在 foo公司 是 已解除 在中运行的Monad版本 酒吧 (提升到a StateT 确切地说,是上下文。)

假使 tst2.hs 如果两者之间没有类型链接 m s(共个) foo公司 和 酒吧 ),此操作非常有效。 然而,在 tst3.hs 我提供了两者之间的类型链接,迫使它们成为同一个Monad。 这就是编译器抱怨的原因。

我的解决方案有效,只是因为Monad在 foo公司 真的是 已解除 在中运行的版本 酒吧 。 如果这两个单子完全不相关,那么我的解决方案 不 工作

一切都对吗?