It's a simple operator problem.

It depends on what you calculate first. In the case of x++, increase or decrease x by 1. For ++x, increase or decrease x by 1. Now, if both cases are the same, we don't have to use two subtly different notation, do we? Maybe if it was Babu who created the programming language. Well, I don't think it's Babu, and it's not really... The order of calculations for z = ++x is (2)z = (1)x+1, i.e. x+1 is calculated first and stored in z. The order of calculations for z = x++ is (1)z = x (2) x = x+1.

public static void main(String[] args) {
    int x = 10; 
    int y = 20;
    int z = (++x) + (y--);
  It's /*y-- right? In the operations of this line, y-1 is the result of all operations
    After it's over. I'm going to be the last one. It's tied up with ()
    I think y-- is going to precede the = operator, but it's not, right?
    I didn't know either. But that's a language rule.
    It's just that the creator of java promised to do so.
    To keep the language consistent! 
   So even the parenthesis operator is pushed back to the last minute!*/
    System.out.println(z);//30


}

The operations differ depending on the increase and decrease operator (++, ---) symbol followed by the latter.

z = (++x) + (y--)    // 31
z = (++x) + (--y)    // 30

You can think of an expression as having a value. So you think that things like "++x" and "x++" are all worth it. The contents are summarized as follows.

x++      // x
++x      // x+1
x--      // x
--x      // x-1