list - 如何使用 bind (>>=) 实现一个函数

Question

我写了一个过滤函数:

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'

为了理解绑定(bind)，我想使用绑定(bind)来实现它。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成功做到了:

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

出了什么问题？

Answer 1

To understand bind, i want to implement this as bind.

这里没有绑定(bind)。在 do expression 的情况下添加绑定(bind).上面不是do-expression，所以这里没有bind。

但是你可以用 bind 写这个，比如:

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

但这不是原始函数的文字映射，我们在这里简单地使用instance Monad [] 实现。然而，您的 f 只是 filter :: (a -> Bool) -> [a] -> [a]在这里。