Haskell - 解析错误/使用多个 where 子句
当尝试定义一个函数来删除集合 m 的最大子集（同时也是集合 a 中集合 a 的子集）时，我遇到了以下错误：
filename.hs:7:33:parse error (possibly incorrect indentation)
对于以下代码：
exclude :: Integral t => [t] -> [t] -> [t]
a `exclude` m
| m == [] = a
| a == (b ++ c) = b
| otherwise = []
where b /= []
where c = [z | z <- m]
如何实现多个条件/定义（使用 where 或其他方式），或更正函数以以不同的方式正常工作？
when trying to define a function that would remove the largest subset of set m that is also a subset of set a from set a, I encountered the following error:
filename.hs:7:33:parse error (possibly incorrect indentation)
for the following code:
exclude :: Integral t => [t] -> [t] -> [t]
a `exclude` m
| m == [] = a
| a == (b ++ c) = b
| otherwise = []
where b /= []
where c = [z | z <- m]
how do I implement multiple conditions/definitions (using where or otherwise), or correct the function to properly work in a different way?
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评论（3）
你的问题的一部分很容易回答。您可以在一个
where子句中拥有多个定义，例如，您可以嵌套
where子句。但在where子句中，只能有定义。条件将进入守卫（或右侧的if then else表达式），并且可以与布尔运算符(&&)、组合>(||)，not...至于你的代码，到目前为止我还没有弄清楚你打算做什么。
One part of your question is easily answerable. You can have multiple definitions in one
whereclause, as inand you can have nested
where-clauses. But in awhere-clause, you can only have definitions. Conditions would go into the guards (orif then elseexpressions on the right hand side) and can be combined with the boolean operators,(&&),(||),not...As for your code, so far I haven't figured out what you intended it to do.
说“集合
m的最大子集也是集合a的子集”与“
m的所有元素也是a的元素”相同。那么问题的解决方案简单地表述为：
当应用于
m时，将为您提供m对任何元素取模的子集它们也是
a的成员。也就是说，它将“删除最大的子集m也是a的子集”。Saying "the largest subset of set
mthat is also a subset of seta"is the same as saying "all elements of
mthat are also elements ofa".Then the solution to your problem is stated simply as:
which when applied to
mwill give you a subset ofmmodulo any elementsthat are also members of
a. That is, it will "remove the largest subset ofmthat is also a subset ofa".事实上，Data.List和Data.Set中都有一个函数叫做“\”。我将展示 Data.List 的 '\' 函数。
In fact，there is a function in Data.List and Data.Set called '\'. I'll show '\' function of Data.List .