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;

Question

I came across this line of code in java in nutshell book and I would like to know how compiler divide this code

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}

I assume the order of evaluation should be as follows

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

We can prove it by running this snippet

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