我想实现一个函数iterateM,它的类型是这样的:
iterateM :: Monad m => (a -> m a) -> a -> [m a]
然而,我第一次写这个函数:
iterateM f x = f x >>= (x' -> return x' : iterateM f x')
给出错误:
Could not deduce (m ~ [])
from the context (Monad m)
bound by the type signature for
iterateM :: Monad m => (a -> m a) -> a -> [m a]
at main.hs:3:1-57
`m' is a rigid type variable bound by
the type signature for
iterateM :: Monad m => (a -> m a) -> a -> [m a]
at main.hs:3:1
Expected type: [a]
Actual type: m a
In the return type of a call of `f'
In the first argument of `(>>=)', namely `f x'
In the expression: f x >>= ( x' -> return x' : iterateM f x')
如果我删除我的类型签名,ghci告诉我函数的类型是:
iterateM :: Monad m => (a -> [a]) -> a -> [m a]
我在这里错过了什么?
我从你的签名中得知:
iterateM :: (Monad m) => (a -> m a) -> a -> [m a]
表示n
元素iterateM f x
将是运行f
n
次的动作。这非常接近iterate
,我怀疑我们可以在这方面实现它。
iterate :: (b -> b) -> b -> [b]
iterate
给了我们一个b
s的列表,我们想要一个m a
s的列表,所以我怀疑是b = m a
。
iterate :: (m a -> m a) -> m a -> [m a]
现在我们需要一种方法将f :: a -> m a
转换为m a -> m a
类型的东西。幸运的是,这正是bind:
(=<<) :: (Monad m) => (a -> m b) -> (m a -> m b)
:
f -> iterate (f =<<) :: (a -> m a) -> m a -> [m a]
要将初始的x :: a
转换为所需的m a
,我们可以使用return
:
return :: (Monad m) => a -> m a
:
iterateM f x = iterate (f =<<) (return x)
pointfreeze to taste.
你递归地使用iterateM是在强迫它在列表单子中。您需要运行iterateM操作并返回其结果。
试题:
iterateM f x = do
x' <- f x
xs <- iterateM f x'
return $ x':xs