Haskell << loop >> Exception

Question

我正在为我的State Monad实例获得一个<< loop >> Exception,我想这是指一个无限循环,但我看不出我的代码在使用时会如何导致一个:

instance Monad (State' s) where

    -- return :: a -> State' s a
    return x = State' (\(s,c) -> (x, s, (c <> oneReturn) ))

    -- (>>=) :: State' s a -> (a -> State' s b) -> State' s b
    st >>= k = State' $ \(s,c) -> let (a, s', c) = runState' st (s,c)
                                  in runState' (k a) (s',(c <> oneBind) )

instance MonadState (State' s) s where

    -- get :: State' s s
    get = State' $ \(s,c) -> (s,s, (c <> oneGet))

    -- put :: s -> State' s ()
    put s = State' $ \(_,c) -> ((),s, (c <> onePut))

如果有人有线索,可以帮助我,非常感谢!

最好的问候,Skyfe.

编辑:作为参考,一个< SomeMonadicFunc >用于在绑定当前Counts值时正确增加计数器 -

oneBind   = Counts 1 0 0 0
oneReturn = Counts 0 1 0 0
oneGet    = Counts 0 0 1 0
onePut    = Counts 0 0 0 1

Answer 1

        let (a, s', c) = runState' st (s,c)

这是一个递归定义:c结果中的结果用于计算结果,用于...

你可能意味着let (a,s',c') = runState' st (s,c) in runState' (k a) (s',(c' <> oneBind) ),没有阴影c变量.