Why x!=3? In Java

Question

int x=3;x+=x++; // In the end you have x ==2. Why?
https://code.sololearn.com/cPFwT69FFOvM/?ref=app

Answer 1

If I'm correct this behaviour is similiar to this one:
https://stackoverflow.com/questions/7911776/what-is-x-after-x-x

post increment means, it returns the value and then increment it.

x is 1 -> x += 1 -> x is 2 -> increment x but 2 is already assigned to x. So I would say the old value (2) overrides the new value (3)
So in the end you get x = 2

Another example:

int x = 1
int y = 1
x += y++

x uses the old value, so x gets 2, increment y, so y gets also 2

x = 1
x += ++x

In this case it increments x and then the value is added to x, so you get 3.

Answer 2

Java operates a stack in addition to values in memory.

int x = 1;
x += x++;

After the first instruction, the value of x at its referenced location is 1.

The second instruction is an addition to x. So, throw x as summand onto the stack. What is added to x is x, throw x as second summand onto the stack, then increment x at its memory location. The increment happens AFTER the value of x is added onto the stack because it is a postincrement operator.

At this time, two 1s are on the stack, x in memory is 2 because of the increment.

Then perform the addition on the stack: 1+1=2. And assign the result to x at its memory location. Hence, x gets the value 2.

Answer 3

Yes, I think you are right. Tanks a lot!!!