为什么这些模式匹配重叠？

本文介绍了为什么这些模式匹配重叠？的处理方法，对大家解决问题具有一定的参考价值，需要的朋友们下面随着小编来一起学习吧！ 问题描述 （q，m）（q4，m4）= case $ m2 m - > deleteRec键q m2 _ - > （q，m2） 这个警告符合 警告：模式匹配重叠在另一种情况下：_ - > ... 并且不能按我的想法工作。它似乎是（q4，m4）它始终返回 []，fromList [] 无视m2和m的值。是否有任何我不指望他们的局部变量？ 我想用文字来实现：如果m2和m相等，那么（q4，m4）应该评估为 deleteRec key q m2 ，否则为（q，m2）。 > m ）匹配所有内容并将其分配给 m 。第二个匹配所有内容并丢弃它（ _ ），但没有任何匹配，因为 m 将获得所有内容。 我认为你的意思是为了像 switch 语句一样工作，但它实际上可以作为一组模式，很像函数声明。因此，您的案例与以下类似： check m2 其中check m = deleteRec key q m2 check _ =（q，m2） 在这段代码中，如果最好使用： if m == m2 then deleteRec key q m2 else（q，m2） 您如果语句不同，也可以考虑缩进： if m == m2 然后deleteRec key q m2 else（q，m2） 应该也可以。 然而，一般来说，你可以在 case 语句中使用警卫，所以这可以起作用太： case m2 of val | m2 == m - > deleteRec键q m2 |否则 - > （q，m2） 如果，但是如果你有更多的分支或者需要做一些实际的模式匹配，这是有道理的。 Why do my pattern matches inside a do block overlap? (q, m) <- newRq let m2 = appendMsg "first" key m (q4, m4) = case m2 of m -> deleteRec key q m2 _ -> (q, m2)This compiles with the warningWarning: Pattern match(es) are overlapped In a case alternative: _ -> ...and does not work as I want to. It just seems that for (q4, m4) it always returns [], fromList []disregarding what the values of m2 and m are. Is there any local variables where I do not expect them?What I want to achieve in words: If m2 and m are equal then (q4, m4) should evaluate to deleteRec key q m2, otherwise to (q, m2). 解决方案 The first pattern in the case (m) matches everything and assigns it to m. The second one matches everything and discards it (_), but has nothing left to match because m will get everything.I think you meant for the case to work like a switch statement, but it actually works as a set of patterns, much like a function declaration. So your case is the same as something like:check m2 where check m = deleteRec key q m2 check _ = (q, m2)In this code, you're probably best off just using an if:if m == m2 then deleteRec key q m2 else (q, m2)You might also consider indenting the if statement differently:if m == m2then deleteRec key q m2else (q, m2)should also work.However, in general, you can actually use guards in a case statement, so this would work too:case m2 of val | m2 == m -> deleteRec key q m2 | otherwise -> (q, m2)This is obviously harder to read than an if, but if you had more branches or needed to do some actual pattern matching, it would make sense. 这篇关于为什么这些模式匹配重叠？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！