你不需要机器代码来理解发生了什么。

根据定义:

赋值操作符求右边表达式的值，并将其存储在临时变量中。 1.1. x的当前值被复制到这个临时变量中 1.2. X现在是递增的。 然后将临时变量复制到表达式的左侧，也就是x !这就是为什么原来的x值又被复制到自身。

这很简单。

从http://download.oracle.com/javase/tutorial/java/nutsandbolts/op1.html

The increment/decrement operators can be applied before (prefix) or after (postfix) the operand. The code result++; and ++result; will both end in result being incremented by one. The only difference is that the prefix version (++result) evaluates to the incremented value, whereas the postfix version (result++) evaluates to the original value. If you are just performing a simple increment/decrement, it doesn't really matter which version you choose. But if you use this operator in part of a larger expression, the one that you choose may make a significant difference.

为了说明这一点，请尝试以下方法:

    int x = 0;
    int y = 0;
    y = x++;
    System.out.println(x);
    System.out.println(y);

它会输出1和0。

    x++
=: (x = x + 1) - 1

So:

   x = x++;
=> x = ((x = x + 1) - 1)
=> x = ((x + 1) - 1)
=> x = x; // Doesn't modify x!

而

   ++x
=: x = x + 1

So:

   x = ++x;
=> x = (x = x + 1)
=> x = x + 1; // Increments x

当然最终结果和x++是一样的;或+ + x;在一条直线上。

这就像你期望的那样。这就是前缀和后缀的区别。

int x = 0; 
while (x < 3)    x = (++x);

自增操作符应用于要赋值的变量。那是自找麻烦。我相信你可以看到你的x变量的值，同时运行这个程序....这应该清楚了为什么循环永远不会结束。

你有效地得到了以下行为。

获取x的值(它是0)作为右边的“结果” 增加x的值(x现在是1) 将右边的结果(保存为0)分配给x (x现在是0)

其思想是，后增量操作符(x++)在返回变量的值后对该变量进行增量，以便在使用该变量的方程中使用。

编辑:由于评论的原因，稍微增加了一点。像下面这样考虑。

x = 1;        // x == 1
x = x++ * 5;
              // First, the right hand side of the equation is evaluated.
  ==>  x = 1 * 5;    
              // x == 2 at this point, as it "gave" the equation its value of 1
              // and then gets incremented by 1 to 2.
  ==>  x = 5;
              // And then that RightHandSide value is assigned to 
              // the LeftHandSide variable, leaving x with the value of 5.