等式=> Haskell中的功能 [英] Eq => function in Haskell

问题描述

我试图弄清楚如何用Haskell中的Eq函数编写此函数.

I am trying to figure out how to write this function with the Eq function in Haskell.

我要实现的一个简单功能是:

An easy function that I am trying to implement is:

f :: Eq a => [a] -> [[a]]

其中f将在单独的子列表下收集每个重复的连续元素，例如:

Where f will gather each repeated consecutive elements under separate sub-lists, For example:

f [3] = [[3]]
f [1,1,1,3,2,2,1,1,1,1] = [[1,1,1],[3],[2,2],[1,1,1,1]]

我考虑过此功能:

f :: Eq a => [a] -> [[a]]
f [] = []
f (x:[]) = [[x]]
f (x:x':xs) = if x == x' then [[x, x']] ++ (f (xs))
                    else [[x]] ++ (bundler (xs))

它似乎不能很好地工作，因为当它到达最后一个元素时，它想将其与连续的元素进行比较，而后者显然是不存在的.

It seems to not work well since when it arrives to the last element, it wants to compare it to its consecutive, which clearly does not exist.

此外，我觉得我对Eq =>函数不使用任何东西.

Moreover, I feel like I do not use anything with the Eq => function.

我想收到一个答案，该答案将说明如何在我的情况下正确使用Eq.

I would like to receive an answer that will show how to use Eq properly in my case.

谢谢.

推荐答案

您的类型中出现的Eq类型类是一个红色鲱鱼:它与您报告的错误无关.发生此错误的原因是，您已定义了当列表为空(f [] =)且列表中至少包含两个元素(f (x:x':xs) =)时函数的行为方式，但定义了列表中仅包含一个元素时函数的行为.解决方案是添加一个开始的案例*

The presence of the Eq typeclass in your type is a red herring: it is not related to the error you report. The error you report happens because you have defined how the function should behave when the list is empty (f [] =) and when the list has at least two elements (f (x:x':xs) =), but not when the list has exactly one element. The solution will be to add a case that begins*

f [x] = ????

并决定如何处理一个元素列表.或者找到其他方式编写处理所有情况的函数.

and decide how to deal with one-element lists. Or to find some other way to write the function that deals with all of its cases.

*请注意，f [x]与f (x:[])是同一件事.

* Note that f [x] is the same thing as f (x:[]).

这篇关于等式=> Haskell中的功能的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！