Question

我正在尝试编写一个类似于zip但不会丢弃额外元素的函数。我觉得我在某个地方犯了一个非常愚蠢的错误。

示例输入：

zipMaybe [1,2,3] [1,2]

期望的输出：

[(Just 1, Just 1), (Just 2, Just 2), (Just 3, Nothing)]

zipMaybe :: [a] -> [b] -> [(Maybe a, Maybe b)]
zipMaybe (a:as) (b:bs) = (Just a, Just b) : zip as bs -- line with error
zipMaybe (a:as) [] = (Just a, Nothing) : zip as []
zipMaybe [] (b:bs) = (Nothing, Just b) : zip [] bs
zipMaybe _ _ = []

然而，这不会编译。

Test.hs:2:49:
    Couldn't match type `a' with `Maybe a'
      `a' is a rigid type variable bound by
          the type signature for
            zipMaybe :: [a] -> [b] -> [(Maybe a, Maybe b)]
          at Test.hs:1:13
    Expected type: [Maybe a]
      Actual type: [a]
    In the first argument of `zip', namely `as'
    In the second argument of `(:)', namely `zip as bs'
    In the expression: (Just a, Just b) : zip as bs

Answer 1

你应该递归调用zipMaybe而不是退回到类型错误的香草zip。

zipMaybe :: [a] -> [b] -> [(Maybe a, Maybe b)]
zipMaybe (a:as) (b:bs) = (Just a, Just b) : zipMaybe as bs
zipMaybe (a:as) [] = (Just a, Nothing) : zipMaybe as []
zipMaybe [] (b:bs) = (Nothing, Just b) : zipMaybe [] bs
zipMaybe _ _ = []

顺便说一下，这个函数的定义较短：

zipMaybe (x:xs) (y:ys)  =  (Just x, Just y) : zipMaybe xs ys
zipMaybe xs     []      =  [(Just x, Nothing) | x <- xs]
zipMaybe []     ys      =  [(Nothing, Just y) | y <- ys]

我的改版拉链有什么问题？

1 个答案: