如何使用Haskell中的函数返回列表中的最后一个元素? [英] How to return last element in the list using function in haskell?

问题描述

我的教授给了我一个使用" laste "函数获取列表中最后一个元素的示例:他说:以" laste xs =…"的形式进行定义是不可接受的，而以" laste =…"的形式进行定义是可以接受的.

My professor gave me an example to get last element in the list using "laste" function: he stated that: definition in the form of "laste xs = …" is not acceptable, whereas definition in the form of "laste = …" is acceptable.

我已经尝试过类似的操作:如果根据问题陈述解决方案不正确，请纠正我.

I have tried something like this: Please correct me if my solution is wrong according to problem statement.

laste :: [a] -> Maybe a 
laste [] = Nothing 
laste (x:[]) = Just x 
laste (x:xs) = laste xs

但这给了我一个答案，例如:

But this gives me answer for example:

ghci>laste[1,2,3,4]
Just 4

我想摆脱这个" Just ".

有什么解决方法可以删除 Just ?

Is there any solution to remove Just?

推荐答案

虽然Charmini2的回答在功能上是正确的，但它不能解决以无点形式检索最后一个元素的问题.考虑

While Charmini2's answer is functionally correct, it doesn't solve the problem of retrieving the last element in pointfree form. Consider

laste :: [a] -> a
laste = foldr1 (\_ a -> a)

它按照规范工作，因为foldr1需要一个非空列表.可以从以下观察得出直觉，为什么它返回列表中的最后一个元素:观察到的是foldr1用上述等式中的lambda替换了列表结构中的每个(:)，基本上选择了两个元素中最右边的一个.重复，您将得到最后一个.

It works according to specs as foldr1 expects a non-empty list. Intuition for why it returns the last element in the list can be gotten from the observation that foldr1 replaces every (:) in the structure of the list with the lambda in the above equation, which basically selects the rightmost of two elements. Repeat, and you get the last.

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