x += ++x equivalent to x = 2x+1 : Why?

Question

This question is just curiosity : I was wondering what would be the value of some int x after the line x += ++x So I tried that :

int x=10;
x+=++x;
System.out.println(x);

And it printed out :

21

After some tests with other values, it seems to be equivalent to x=2x+1. Why ? Is this line interpreted by the compiler as a byte operation ? (By the way, x += x++ seems to be equivalent to x=2x).

I don't think it's something I'd ever use in a project, but I'm curious to know why I get this result.

Thanks for any explanation or hint

EDIT : First of all, thanks for your answers

I knew how the += operator works, as well as the x++ and ++x , but for some reason the (completely logic and obvious) result seemed strange to me I should probably have thought it through, sorry for your time !

Answer 1

The way it is calculated is

• Step 1: x = x + ++x

• Step 2: It become x = 10 + (incremented x) 11

• Step 3: Final result stored in x ie 21

Here is the proof:

I created a MainClass as below:

public class MainClass{
public static void main(String...s){
int x = 10;
x += ++x;
}
}

and then checked the bytecode using javap -c MainClass

  public static void main(java.lang.String...);
    Code:
       0: bipush        10     // push 10 onto stack
       2: istore_1             // store 10 in local variable 1
       3: iload_1              // load local variable 1 (now 10) back to stack
       4: iinc          1, 1   //increment local variable 1 by 1
       7: iload_1              // load local variable 1  (now 11) back to stack
       8: iadd                 // add top 2 variable on stack ( 10 and 11)
       9: istore_1             // store 21 to local variable 1
      10: return
}

Answer 2

Its about operator precedence and how ++x and x++ are evaluated and used. with ++x , the value of x is incremented and then used so ++x becomes 11 and this x += ++x becomes 21 which is 10 + 11

However x++ says x is used and then its value is incremented

so x+= x++ will mean 10 + 10 ie 20

Answer 3

int x=10;
x+=++x;
System.out.println(x);

x + = ++x is evaluated in the compiler to x = x + ++x => x = 10 + ++x => x = 10 + 11 => x = 21

Answer 4

See here:-

x+=++x; this expression will be executed like x=x+(x+1) so x = 10 + 11

hence x = 21;

You need to understand about pre-increment(++x) and post-increment (x++). see below

 int x = 10;
 if (x++ == 10 )
     System.out.println( "X is equal to 10");// this statement will print

in the above if condition it will execute as true because first it will compare as is 10 == 10 and then x will be incremented by one and x will become 11.

Now see below:-

   if (++x == 10 )
     System.out.println( "X is equal to 10");// this will not print if condition will tern false

In the above if condition x will be pre-incremented so x will become 11 and then a comparison will be done whether 11 == 10 hence if condition will failed.

Hope this will help.

Answer 5

++x will return value of (x+1) , and x value will be increased by one too.

x++ will return value of (x) , and x value will be increased by one too.

so

x+=++x is the same x=x+(x+1) , which is equivalent to x=2*x+1

x+=++x is the same as x=x+(x) , which is equivalent to x=2*x

Answer 6

..... ++ x：首先计算x = x + 1，然后使用x进行比较或计算实际任务..... x ++：首先比较/计算实际任务，然后计算x = x + 1