重载后增量运算符会导致参数值为 0

发布于 2024-11-09 10:16:25 字数 964 浏览 3 评论 0原文

可能的重复：
是否允许在后缀运算符 ++ 中命名参数？

我创建了一个对象来保存在内部维护当前位置的对象列表，因此我认为这是重载前增量运算符和后增量运算符的好地方，这些运算符通过边界检查有效地增量此内部位置。

我注意到，当您在对象上调用 ++ 时，参数为 0。

测试代码：

#include <stdio.h>
class A {
public:
   A& operator++(int n) { printf("%d  ", n); return *this; }
};
int main() {
   A a;
   a++;
   a.operator++(0);
   a.operator++(1);
   a.operator++(10);
   return 0;
}

返回0 0 1 10。据我了解，这是正常行为。所以，它让我重新思考 operator++ 应该如何工作。以前，如果边界检查通过，我只是在内部位置变量上调用 ++ 。但无论输入参数是什么，都会增加 1。接下来，我虽然使用 += 并使用参数 n 作为右侧，但您会注意到，只需调用 ++ > 没有运算符（按照惯例），给出零并且位置不递增。

基本上，这是我应该担心的事情吗？如果是这样，我如何检测用户是否真的想要 0，或者默认行为 (a++) 是否是有意的并且我应该增加 1？

原文

Possible Duplicate:
Is it allowed to name the parameter in postfix operator ++?

I created an object to hold a list of objects that maintains the current position internally, so I thought this was a great place to overload the pre and post increment operators, which effectively increment this internal position with bounds checking.

What I noticed is, when you call ++ on the object, the argument is 0.

Test code:

#include <stdio.h>
class A {
public:
   A& operator++(int n) { printf("%d  ", n); return *this; }
};
int main() {
   A a;
   a++;
   a.operator++(0);
   a.operator++(1);
   a.operator++(10);
   return 0;
}

This returns 0 0 1 10. From what I understand, this is normal behavior. So, it has made me rethink how operator++ should work. Previously, I was simply calling ++ on my internal position variable if bounds checking passed. But this has the affect of incrementing by 1 no matter what the input argument is. Next, I though of using the += using the argument n as the right hand side, but as you'll notice, simply calling ++ with no operators (as is customary), gives a zero and the position is not incremented.

Basically, is this something I should even worry about? If so, how do I detect if the user really wanted 0, or if the default behavior (a++) was intended and I should increment by 1?

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