什么时候确实发生postfix一元运算符？

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.

Answer 1

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:

• JSL - 15.14.2. Postfix Increment Operator ++
• JLS - 15.15.1. Prefix Increment Operator ++
• What is x after “x = x++”?