重载增量运算符 [英] Overloading increment operator

问题描述

我只是在重载增量运算符.我希望既有增量又有增量.不幸的是，我看到无论操作员处于前置位置还是后置位置，都调用了相同的方法.因此，我检查了规格并惊讶地发现:

"
请注意，运算符返回通过将1加到操作数而产生的值，就像后缀递增和递减运算符(第7.5.9节) ，以及前缀递增和递减运算符(第7.6.5节)，与C ++不同，此方法不需要，实际上也不必直接修改其操作数的值.
"

没有增量的增量运算符有什么用呢?如果我想做x += 1，我有加法运算符，它工作得很好.我想要x++ 和 ++x，就像内置数字类型一样； x = x++或x = ++x将是荒谬的，但看来我们只允许这样做.


我错过了什么吗?有没有人可以解决?


P.S.还是只是00:34，我应该去睡觉了?它似乎以预期的方式工作了吗? [confused]

I was just working on overloading the increment operator. I was hoping to have both pre-increment and post-increment. Unfortunately I see that the same method is called regardless of whether the operator is in pre- or post-position. So I checked the spec and was surprised to find:

"
Note that the operator returns the value produced by adding 1 to the operand, just like the postfix increment and decrement operators (§7.5.9), and the prefix increment and decrement operators (§7.6.5). Unlike in C++, this method need not, and, in fact, must not, modify the value of its operand directly.
"

What the heck good is an increment operator that doesn''t increment?! If I want to do x += 1 I have the addition operator for that and it works just fine. I want x++ and ++x just like with the built-in numeric types; x = x++ or x = ++x would be ridiculous, but it appears that''s all we''re allowed.


Am I missing something? Does anyone have a work-around?


P.S. Or is just that it''s 00:34 and I should go get some sleep? It seems to be working as expected somehow? [confused]

于2009年2月24日，星期二2:33修改

modified on Tuesday, February 24, 2009 2:33 AM

推荐答案

在调用增量运算符时，将为对象分配一个新的引用，该引用具有新的值.

示例.我们有一个IntWrapper类

When calling the increment operator, the object is assigned a new reference, that reference has the new values.

Example. We have a class IntWrapper

public class IntWrapper
{
    public int IntValue
    {
        get;
        set;
    }
}

而且我们想像这样使用它

and we want to use it like this

IntWrapper a = new IntWrapper();
IntWrapper b = a;
a++;
Console.WriteLine("a = {0}, b = {1}", a.IntValue, b.IntValue);

这是我们类中增量运算符重载的两种可能的实现，第二种是正确的方法

Here are two possible implementations for the increment operator overload in our class, the second is the correct way

public static IntWrapper operator ++(IntWrapper instance)
{
    // WRONG! a and b will be incremented
    instance.IntValue++;
    return instance;
}

public static IntWrapper operator ++(IntWrapper instance)
{
    // CORRECT! Only a will be incremented
    IntWrapper result = new IntWrapper();
    result.IntValue = instance.IntValue + 1;
    return result;
}

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