我的zip版本有什么问题？

Question

我试图编写一个类似于zip的函数，但不会丢弃额外的元素。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] 

希望的输出：

[(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 _ _ = [] 

然而这将不编译。

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

你应该叫zipMaybe递归，而不是釜底抽薪香草zip，其中有错误的类型。 。

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 _ _ = [] 

顺便说一句，有此功能的更短的定义：

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]