Haskell 函数依赖和类型变量错误

我试图弄清楚为什么在这个代码示例中无法识别返回类型，请参阅

---

当您定义

Foo（StateT s w）

实例时，您没有将

s

约束为

Int

，而是希望从

栏返回它。以后如何使用实例并不重要，因为实例本身必须有效

也许您想定义：

instance (MonadIO w) => Foo (StateT Int w) where
    type Inner (StateT Int w) = w
    bar current = do
        a <- get
        return a

instance（MonadIO w）=>Foo（StateT Int w）其中
类型内部（StateT Int w）=w
巴电流=do
a感谢@Koterpillar的快速回复，效果非常好！
• Couldn't match type ‘s’ with ‘Int’
    arising from a functional dependency between:
      constraint ‘MonadState Int (StateT s w)’
        arising from a use of ‘get’
      instance ‘MonadState s1 (StateT s1 m)’ at <no location info>
  ‘s’ is a rigid type variable bound by
    the instance declaration
    at fsm-try-5-3-q.hs:16:10-55
• In a stmt of a 'do' block: a <- get
  In the expression:
    do a <- get
       return a
  In an equation for ‘bar’:
      bar current
        = do a <- get
             return a

instance (MonadIO w) => Foo (StateT Int w) where
    type Inner (StateT Int w) = w
    bar current = do
        a <- get
        return a