Haskell - 模式匹配（es）重叠 [英] Haskell - Pattern match(es) are overlapped

问题描述

  test :: String  - >字符串 - > （y：ys）n =错误错误
 test'（x：xs）n =错误
 $ b test'xyn = n 
 test'错误
 test'（x：xs）（y：ys）n = 
 if x == y 
 then test'xs ys n 
 else test'xs ys（n +1）
 test ab = test'ab 0 
  

当我编译这个时，此输出：

 警告：模式匹配重叠
  

答案总是0，这不是我想要的。如何解决这个问题？

解决方案

test'xyn = n 会匹配每个通话，其他模式不会被考虑。我认为这种情况应该是 test'n = n 。如果您在 end 处移动原始行（当所有其他情况都失败时），您会得到相同的结果，但是您应该写 test'_ _ n = n $ b

$ b这表明您故意忽略某些参数。 $ b

较短的解决方案是：

  test ab |长度a ==长度b =总和$ map fromEnum $ zipWith（/ =）a b 
 |否则=错误错误
  

zipWith 表达式会为每个差异生成 Bool 的列表，其中 True 。函数 fromEnum 映射 False 到 0 和 True 至 1 。

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?

解决方案

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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