Java增量和赋值运算符

问题描述：

我对post ++和pre ++运算符感到困惑，例如在以下代码中

I am confused about the post ++ and pre ++ operator , for example in the following code

int x = 10;
x = x++;

sysout(x);

将打印10？

打印10，但我预计它应该打印11

It prints 10,but I expected it should print 11

但是当我这样做时

x = ++x; instead of x = x++;

它会按照我的预期打印11，所以为什么x = x ++;不改变x的值？

it will print eleven as I expected , so why does x = x++; doesn't change the the value of x ?

答

不，打印输出10是正确的。理解结果背后原因的关键是预增量 ++ x 和后增量 x ++ 之间的差异复合作业。使用预增量时，表达式的值在执行增量后获取。但是，当您使用后递增时，表达式的值在递增之前取，并在将递增结果写回变量之后存储供以后使用。

No, the printout of 10 is correct. The key to understanding the reason behind the result is the difference between pre-increment ++x and post-increment x++ compound assignments. When you use pre-increment, the value of the expression is taken after performing the increment. When you use post-increment, though, the value of the expression is taken before incrementing, and stored for later use, after the result of incrementing is written back into the variable.

以下是导致您所看到的事件序列：

Here is the sequence of events that leads to what you see:

x 被分配 10

由于 ++ 在后增量位置，存储 x 的当前值（即 10 ）供以后使用

新值 11 存储在 x

10 的临时值存储回 x ，写在上11 已存储在那里。

x is assigned 10
Because of ++ in post-increment position, the current value of x (i.e. 10) is stored for later use
New value of 11 is stored into x
The temporary value of 10 is stored back into x, writing right over 11 that has been stored there.

Java增量和赋值运算符

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