Haskell中的整数与模式匹配给出错误的结果 [英] Pattern matching with integer in haskell gives wrong result

问题描述

我注意到，尝试在带有参数的haskell中执行模式匹配并不总是可行.这是一个示例:

I noticed that trying to perform pattern matching in haskell with the arguments doesn't quite always work. Here's an example:

test :: Integer -> Integer -> String
test num1 num2 = case num1 of
                    num2 -> "foo"
                    otherwise -> "bar"

当我将其加载到解释器中时，它会警告我重叠的模式匹配.此外，对于任何两个整数 a 和 b 的 test ab 都会返回"foo" ，无论它们是否是否相等.看来，模式匹配中的 num2 与参数中的 num2 不同.

When I load this in the interpreter, it warns me of overlapping pattern matches. Further more, test a b for any two integers a and b returns "foo", regardless of whether they're equal or not. It appears that the num2 in the pattern match is not the same as the one in the arguments.

我想知道为什么会这样.我非常感谢您对此情况有任何见识.

I want to know why exactly this is happening. I'd really appreciate any insight into the situation.

推荐答案

num2 是可与任何值匹配的模式.总是.它与范围内的现有变量 num2 无关.(在大小写替代项的右侧，这里为"foo" ，匹配的值将绑定到名称 num2 ，而现有名称为 num2.但这在这里无关紧要，因为您始终不使用 num2 .)

num2 is a pattern that matches any value. Always. It has nothing to do with an existing variable num2 that is in scope. (In the right-hand side of the case alternative, here "foo", the value matched will be bound to the name num2, shadowing the existing name num2. But that's not relevant here since you don't use num2 anyways.)

想象一下，如果您所有的 map f(x:xs)= fx:map f xs 模式都匹配，则意味着有人定义了一个名为 x 的顶级变量！

Imagine if all your map f (x : xs) = f x : map f xs pattern matches changed meaning just because someone defined a top-level variable called x!

类似地， otherwise 是一种匹配所有内容的模式，与顶级值 otherwise 无关.否则应该用于类似 |的防护中否则= ... ;它的定义是 otherwise = True .

Similarly, otherwise is a pattern that matches everything, and has nothing to do with the top-level value otherwise. otherwise is supposed to be used in guards like | otherwise = ...; its definition is otherwise = True.

在这种情况下，修复程序的最简单方法就是

In this case, the easiest way to fix your program is just

test num1 num2 = if num1 == num2 then "foo" else "bar"

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