我想是因为在java++中有比=(赋值)更高的优先级…不是吗? 看看http://www.cs.uwf.edu/~eelsheik/cop2253/resources/op_precedence.html…

同理，如果x=x+1。+的优先级高于=(赋值)

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++; (increment is overriden by = )

因为上面的表述x永远不会达到3;

这句话:

x = x++;

计算如下:

将x推入堆栈; 增加x; 从堆栈中取出x。

所以这个值是不变的。将其与:

x = ++x;

计算结果为:

增加x; 将x推入堆栈; 从堆栈中取出x。

你想要的是:

while (x < 3) {
  x++;
  System.out.println(x);
}

x++表达式求值为x。++部分影响求值之后的值，而不是语句之后的值。所以x = x++可以有效地转换成

int y = x; // evaluation
x = x + 1; // increment part
x = y; // assignment

其他人已经解释得很好了。我只包含了指向相关Java规范部分的链接。

X = x++是一个表达式。Java将遵循求值顺序。 它将首先计算表达式x++，该表达式将增加x并将结果值设置为x的前一个值。 然后它将表达式结果赋值给变量x。最后，x返回到它之前的值。