如何使用bind（>> =）实现功能

Question

我写了一个过滤功能：

f :: (a -> Bool) -> [a] -> [a]
f p xs = case xs of
         [] -> []
         x : xs' -> if p x
                    then x : f p xs'
                    else f p xs'

要了解绑定，我想使用绑定实现此功能。我在想什么：

f p xs = xs >>= (\x xs -> if p x then x : f p xs else f p xs)

但是我得到这个错误：

* Couldn't match expected type `[a]' with actual type `[a] -> [a]'
    * The lambda expression `\ x xs -> ...' has two arguments,
      but its type `a -> [a]' has only one
      In the second argument of `(>>=)', namely
        `(\ x xs -> if p x then x : f p xs else f p xs)'
      In the expression:
        xs >>= (\ x xs -> if p x then x : f p xs else f p xs)
    * Relevant bindings include
        xs :: [a] (bound at <interactive>:104:5)
        p :: a -> Bool (bound at <interactive>:104:3)
        f :: (a -> Bool) -> [a] -> [a] (bound at <interactive>:104:1)

使用foldr以下命令成功完成：

f p xs = foldr (\x xs -> if p x then x : f p xs else f p xs) [] xs

怎么了

Answer 1

要了解绑定，我想将其实现为绑定。

这里没有绑定。如果是do表达式，则添加绑定。上面不是do表达式，因此这里没有绑定。

但是，您可以使用bind编写此代码，例如：

f p xs = xs >>= \x -> if p x then [x] else []

但这不是原始函数的文字映射，我们仅在instance Monad []此处使用实现。不过，您f就filter :: (a -> Bool) -> [a] -> [a]在这里。