Haskell - 模式匹配重叠

发布于 2024-12-08 05:28:29 字数 416 浏览 0 评论 0原文

test :: String -> String -> Int

test' x y n = n
test' "" (y:ys) n = error "error"
test' (x:xs) "" n = error "error"
test' (x:xs) (y:ys) n =
        if      x == y
        then    test'  xs ys n
        else    test'  xs ys (n+1)
test a b = test' a b 0

当我编译这个时，我得到这个输出：

Warning: Pattern match(es) are overlapped

答案总是“0”，这不是我想要的。代码有什么问题以及如何修复它？

原文

test :: String -> String -> Int

test' x y n = n
test' "" (y:ys) n = error "error"
test' (x:xs) "" n = error "error"
test' (x:xs) (y:ys) n =
        if      x == y
        then    test'  xs ys n
        else    test'  xs ys (n+1)
test a b = test' a b 0

When I compile this, I get this output:

Warning: Pattern match(es) are overlapped

And the answer is always "0", which is not what I intended. What is the problem with the code and how to fix it?

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空城之時有危險 2024-12-15 05:28:29

test' xyn = n 将匹配每次调用，其他模式将不被考虑。我认为这种情况应该是test' "" "" n = n。如果将原始行移到末尾（当所有其他情况都失败时），您会得到相同的结果，但是您应该编写 test' _ _ n = n ，其中显示你故意忽略一些论点。

[编辑]

一个较短的解决方案是：

test a b | length a == length b = sum $ map fromEnum $ zipWith (/=) a b
         | otherwise = error "error"

zipWith 表达式生成一个 Bool 列表，对于每个不同之处。函数 fromEnum 将 False 映射到 0，将 True 映射到 1。

test' x y n = n will match for every call, the other patterns won't be considered. I think this case should be test' "" "" n = n. You get the same result if you move your original line at the end (when all other cases fail), but then you should write test' _ _ n = n which shows that you deliberately ignore some of the arguments.

[Edit]

A shorter solution would be:

test a b | length a == length b = sum $ map fromEnum $ zipWith (/=) a b
         | otherwise = error "error"

The zipWith expression generates a list of Bool which is True for every difference. The function fromEnum maps False to 0 and True to 1.

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月光色 2024-12-15 05:28:29

按顺序尝试这些模式。 test' 的第一个模式始终匹配，因此始终使用该大小写。第一种情况可能应该是

test' "" "" n = n

相反。

The patterns are tried in order. The first of your patterns for test' always matches, so that case is always used. The first case should probably be