Java中的预增量和后增量行为

Question

有人可以解释以下代码的实现：

int j = 1;
System.out.println(j-- + (++j / j++));

我希望输出为3，如下所述： 因为＆＃39; /＆＃39;优先级高于＆＃39; +＆＃39;它首先被评估。

op1 = ++j (Since it is pre-increment, the value is 2, and j is 2)
op2 = j++ (Since it is post-increment, 2 is assigned and j is incremented to 3)

所以＆＃39; /＆＃39;的结果在parantheses中的操作是2/2 = 1。 然后是＆＃39; +＆＃39;操作：

op1 = j-- (Since it is Post-increment, the value is 3, and j is decremented to 2)
op2 = 1 (The result of the '/' was 1)

所以，结果应该是3 + 1 = 4。 但是当我评估这个表达时，我得到2.为什么会发生这种情况？

Answer 1

  

因为＆＃39; /＆＃39;优先级高于＆＃39; +＆＃39;它首先被评估。

不，表达式从左到右计算 - 然后使用优先规则关联每个操作数。

所以你的代码相当于：

int temp1 = j--; //1
int temp2 = ++j; //1
int temp3 = j++; //1
int result = temp1 + (temp2 / temp3); //2