看看下面code（可作为一个控制台应用程序）：

Take the following code (usable as a Console Application):

static void Main(string[] args)
{
    int i = 0;
    i += i++;
    Console.WriteLine(i);
    Console.ReadLine();
}

的结果，我 0。我预计2（我的一些同事一样）。也许是编译器创建某种结构，导致在我为零。

The result of i is 0. I expected 2 (as some of my colleagues did). Probably the compiler creates some sort of structure that results in i being zero.

我预计2的原因是，在我的思路，右手声明将首先评估，我递增以1比它加入到我。既然我已经是1，则加1比1。所以，1 + 1 = 2。显然，这不是发生了什么。

The reason I expected 2 is that, in my line of thought, the right hand statement would be evaluated first, incrementing i with 1. Than it is added to i. Since i is already 1, it is adding 1 to 1. So 1 + 1 = 2. Obviously this is not what's happening.

您可以解释一下什么是编译器或运行时会发生什么？为什么结果为零？

Can you explain what the compiler does or what happens at runtime? Why is the result zero?

部分-排序的-免责声明：我绝对知道你不会（或许不应该）使用code。我知道我永远也不会。不过，我觉得很有趣的是，知道为什么它的行为以这样的方式和正在发生的事情完全是。的

推荐答案

本：

int i = 0;
i += i++

可以看到，你做什么（以下是过于简单化）：

Can be seen as you doing (the following is a gross oversimplification):

int i = 0;
i = i + i; // i=0 because the ++ is a postfix operator and hasn't been executed
i + 1; // Note that you are discarding the calculation result

而实际上，更多地参与比 - 看看MSDN， 7.5.9后缀增量和减量运算符的：

形式的后缀增量或减量运算的运行时处理X ++或x--包括以下步骤：

The run-time processing of a postfix increment or decrement operation of the form x++ or x-- consists of the following steps:   

如果x属于一个变量：

If x is classified as a variable:   在计算x以产生变量。    x的值被保存。   在所选择的操作符被调用x的保存值作为参数。   由操作者返回的值存储在由x的计算给定的位置。    x的存储的数据变为操作的结果。    x is evaluated to produce the variable. The value of x is saved. The selected operator is invoked with the saved value of x as its argument. The value returned by the operator is stored in the location given by the evaluation of x. The saved value of x becomes the result of the operation.

请注意，由于秩序，后缀 + 出现的在的 + = ，但结果呢？不用的（如previous值我时）。

Note that due to order of precedence, the postfix ++ occurs before +=, but the result ends up being unused (as the previous value of i is used).

的更彻底分解I + = I + 来它是由需要一个知道的部分，这两个 + = 和 + 不是原子（即，没有一个是单人操作），即使它们看起来像他们。这些实施方式包括临时变量，副本我的操作发生前 - 每个操作。 （我将用名称 IADD 和 iAssign 对用于 + 和 + = 分别）。

int i = 0;
int iAdd = i; // Copy of the current value of i, for ++
int iAssign = i; // Copy of the current value of i, for +=

i = i + 1; // i++ - Happens before += due to order of precedence
i = iAdd + iAssign;