Question

我愿履行一项职能,即它的作用。

iterateM :: Monad m => (a -> m a) -> a -> [m a]

然而,我第一次撰写这一职务:

iterateM f x = f x >>= (x  -> return x  : iterateM f x )

我会发现错误:

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,格西兹告诉我,我的职能类型是:

iterateM :: Monad m => (a -> [a]) -> a -> [m a]

我在这里失踪了吗?

Answer 1

我从您的签名中收集的东西:

iterateM :: (Monad m) => (a -> m a) -> a -> [m a]

<代码>nth contentiterateM f x is an action that le fns>m. 这非常接近iterate,我怀疑我们能够按此执行。

iterate :: (b -> b) -> b -> [b]

给我们一份<代码>b清单,我们希望有一份<代码>m as,因此,我怀疑<代码>b = m a。

iterate :: (m a -> m a) -> m a -> [m a]

我们现在需要改变<代码>f a-> m a into some of category m a -> m a. 幸运的是,也就是约束的定义:

(=<<) :: (Monad m) => (a -> m b) -> (m a -> m b)

因此:

f -> iterate (f =<<) :: (a -> m a) -> m a -> [m a]

为了将我们的初始<代码>x * a输入到所希望的m a,我们可以使用return<>code>:



return :: (Monad m) => a -> m a


因此:

iterateM f x = iterate (f =<<) (return x)


免去信道。

Answer 2

您对条约的保留 M正在迫使它被列入僧.名单。你们需要采取行动,并恢复结果。

Try:

iterateM f x = do
      x  <- f x
      xs <- iterateM f x 
      return $ x :xs

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