假设 i=0 并且数组的所有元素都初始化为 0 a[i++] = a[i++] + 2；

Question

我在 java in nutshell 书中看到了这行代码，我想知道编译器是如何划分这段代码的

a[i++] += 2;
a[i++] = a[i++] + 2;

Answer 1

15.26.1. Simple Assignment Operator =

If the left-hand operand is an array access expression (§15.10.3), possibly enclosed in one or more pairs of parentheses, then:

First, the array reference subexpression of the left-hand operand array access expression is evaluated. If this evaluation completes abruptly, then the assignment expression completes abruptly for the same reason; the index subexpression (of the left-hand operand array access expression) and the right-hand operand are not evaluated and no assignment occurs.

Otherwise, the index subexpression of the left-hand operand array access expression is evaluated. If this evaluation completes abruptly, then the assignment expression completes abruptly for the same reason and the right-hand operand is not evaluated and no assignment occurs.

^{https://docs.oracle.com/javase/specs/jls/se12/html/jls-15.html#jls-15.26.1}

我假设评估的顺序应该如下

a[i++] = a[i++] + 2;
  ^      ^ ^
  1      3 2
         ----------
             ^
             4
------
  ^
  5
--------------------
         ^
         6

我们可以通过运行这个片段

来证明

int[] a = {0, 10, 0, 0};
int i = 0;
a[i++] = a[i++] + 2;

System.out.println(Arrays.toString(a)); // [12, 10, 0, 0]
System.out.println(i);                  // 2

假设 i=0 并且数组的所有元素都初始化为 0 a[i++] = a[i++] + 2；

What's the order of evaluation for this line assuming i=0 and all elements of array are initialized to 0 a[i++] = a[i++] + 2;

java

assignment-operator

post-increment

15.26.1. Simple Assignment Operator =