When exactly does postfix unary operator happen?

Question

I read a few post regarding unary operator: What is the Difference between postfix and unary and additive in java "C - C++" joke about postfix/prefix operation ordering

And a few more.

However, I still don't understand exactly when the value is changed.

For example:

int x = 1;
x = x++;
System.out.print("x = x++ ==> ");
System.out.print(" x = " + x);
System.out.println();

int x = 1;
x = x++ + x++;
System.out.print("x = x++ + x++ ==> ");
System.out.print(" x = " + x);
System.out.println();

The output is:

x = x++ ==>  x = 1
x = x++ + x++ ==>  x = 3

So in the first block x is assigned to x and afterwards incremented, but the value is never used, otherwise the output would have been x = 2.

In the second block, if I understand correctly, the first x++ is evaluated before the assignment and the second x++ is evaluated afterwards but is never used.

If in the second block both x++ would have been evaluated after the assignment but never used, the output would have been x = 2. If both have been used, the output would have been x = 4.

My IDE also indicated that the first x++ is used, but the second is not used:

So to conclude - I'm still confused about when and how exactly the increment is done.

You should read [What is x after “x = x++”?](http://stackoverflow.com/questions/7911776/what-is-x-after-x-x). — Andy Turner, May 23 '16 at 08:00

score 3 · Accepted Answer · edited May 23 '17 at 12:16

At the line

x = x++ + x++;

Assuming x = 1, the first x++ returns "1" as the value, and then it increments x to 2. So basically, it's assigning the old value back to x.

The second x++ does the same; it returns the value of x, which is now 2, and only then increments its value to 3 - that value, is not used.

Your code is equivalent to:

tmp = x;
x = x + 1;

tmp2 = x;
x = x + 1; // not used

x = tmp + tmp2;

Links that may help you:

When exactly does postfix unary operator happen?

1 Answers1