Question

I read from the official tutorial of Java that prefix and postfix ++ — have different precedences:

postfix: expr++ expr–

unary: ++expr –expr +expr -expr ~ !

Operators

According to the tutorial, shouldn’t this

d = 1; System.out.println(d++ + ++d);

print out 6 (d++ makes d 2, ++d makes it 3) instead of 4?

I know the explanation of ++d being evaluated beforehand, but if d++ has higher precedence then ++d, why isn’t d++ being first evaluated? And what is more, in what case should d++ shows that it has higher precedence?

EDIT:

I tried the following:

d = 1; System.out.println(++d * d++);

It returns 4. It seems that it should be 2*2, instead of 1*3.

Answer 1

The inside of the println statement is this operation
(d++) + (++d)

1. It is as follows, the value of d is read (d = 1)
2. current value of d (1) is put into the addition function

3. value of d is incremented (d = 2).

4. Then, on the right side, the value of d is read (2)

5. The value of d is incremented (now d = 3)

6. Finally, the value of d (3) is put into the addition function

   thus 1 + 3 results in the 4

edit: sorry for the format, I’m rather bad at using the list haha