为Haskell的StateT实现Monad时遇编译错误，求技术解答

我正在通过fp-course学习Haskell，本次练习要求为StateT实现Monad实例，不清楚自己的实现为何触发编译错误，恳请帮忙排查。

我的实现代码 #

instance Monad k => Monad (StateT s k) where
  (=<<) ::
    forall a b. (a -> StateT s k b)
    -> StateT s k a
    -> StateT s k b
  (=<<) f sta = StateT $ \s -> let kas = runStateT sta s
                                   stb = (kas >>= (\(a, s1) -> (f a)))
                               in  runState stb s1

编译错误信息 #

src/Course/StateT.hs:94:17: error:
    • Couldn't match type ‘k’ with ‘(,) b’
      ‘k’ is a rigid type variable bound by
        the instance declaration at src/Course/StateT.hs:89:10
      Expected type: StateT s k b
        Actual type: StateT s ((,) b) b
    • In the expression:
        StateT
        $ \ s
            -> let
                 kas = ...
                 ....
               in runState stb s1
      In an equation for ‘=<<’:
          (=<<) f sta = StateT $ \ s -> let ... in runState stb s1
      In the instance declaration for ‘Monad (StateT s k)’
    • Relevant bindings include
        sta :: StateT s k a (bound at src/Course/StateT.hs:94:11)
        f :: a -> StateT s k b (bound at src/Course/StateT.hs:94:9)
        (=<<) :: (a -> StateT s k b) -> StateT s k a -> StateT s k b
          (bound at src/Course/StateT.hs:94:3)

src/Course/StateT.hs:95:65: error:
    • Couldn't match type ‘k’ with ‘StateT s k’
      ‘k’ is a rigid type variable bound by
        the instance declaration at src/Course/StateT.hs:89:10
      Expected type: k b
        Actual type: StateT s k b
    • In the expression: (f a)
      In the second argument of ‘(>>=)’, namely ‘(\ (a, s1) -> (f a))’
      In the expression: (kas >>= (\ (a, s1) -> (f a)))
    • Relevant bindings include
        s1 :: s (bound at src/Course/StateT.hs:95:57)
        stb :: k b (bound at src/Course/StateT.hs:95:36)
        kas :: k (a, s) (bound at src/Course/StateT.hs:94:36)
        s :: s (bound at src/Course/StateT.hs:94:27)
        sta :: StateT s k a (bound at src/Course/StateT.hs:94:11)
        f :: a -> StateT s k b (bound at src/Course/StateT.hs:94:9)
        (Some bindings suppressed; use -fmax-relevant-binds=N or -fno-max-relevant-binds)

src/Course/StateT.hs:96:49: error:
    • Variable not in scope: s1 :: (b, s)
    • Perhaps you meant ‘s’ (line 94)
Failed, modules loaded: Course.Applicative, Course.Cheque, Course.Comonad, Course.Compose, Course.Contravariant, Course.Core, Course.ExactlyOne, Course.Extend, Course.FastAnagrams, Course.FileIO, Course.Functor, Course.JsonValue, Course.List, Course.Monad, Course.Optional, Course.Parser, Course.Person, Course.State, Course.Validation, Test.Framework.Random.

错误原因分析 #

1. 核心类型不匹配：kas的类型是k (a, s)，使用>>=绑定的时候，回调函数必须返回k类型的值，但你直接返回了f a——它的类型是StateT s k b，和要求的k b完全不匹配，这是最关键的错误。
2. 变量作用域错误：s1是绑定在\(a, s1)这个lambda内部的变量，仅在该lambda的作用域内有效，你在lambda外部的runState stb s1中引用它，自然会提示“变量不在作用域”。
3. 混淆运行函数：你错误使用了runState（这是普通State类型的运行函数），而StateT对应的运行函数应该是runStateT，这导致类型推导混乱，让编译器误以为k是元组类型。

正确实现 #

instance Monad k => Monad (StateT s k) where
  (=<<) ::
    forall a b. (a -> StateT s k b)
    -> StateT s k a
    -> StateT s k b
  (=<<) f sta = StateT $ \s ->
    runStateT sta s >>= \(a, s1) ->
      runStateT (f a) s1

逻辑说明：

• 先用runStateT sta s运行传入的StateT值，得到包裹在k里的(a, s1)。
• 通过>>=绑定这个值，拿到计算结果a和新状态s1。
• 最后用runStateT (f a) s1运行f a生成的StateT值，得到最终的k (b, s2)，整个结果被包裹在StateT构造器中，完全符合类型要求。

内容的提问来源于stack exchange，提问作者zichao liu