Why does (a == (a = b)) not same as (a = b) == a?

The result of ‘(a == (a = b))’ is not same as ‘(a = b) == a’. It is because, as per Java specification, left-hand operand of a binary operator appears to be fully evaluated before any part of the right-hand operand is evaluated.

Let’s see an example.

For int i = 2; int j = i * (i = 3);

Above snippet is evaluated like below.

int j = i * (i = 3)

       = 2 * (i = 3)

       = 2 * 3

       = 6

For int i = 2; int k = (i = 3) * i;

Above snippet evaluated like below.

int k = (i = 3) * i

        = 3 * i

        = 3 * 3

        = 9

Find the below working application.

App.java

package com.sample.app;

public class App {

 public static void main(String args[]) {
  int i = 2;
  int j = i * (i = 3);
  int k = (i = 3) * i;

  System.out.println("j : " + j);
  System.out.println("k : " + k);
 }

}

Output

j : 6

k : 9

Similarly let’s evaluate the expression 'a == (a = b)' for a = 1 and b = 3.

expression = a == (a = b)

           = 1 == (a = b)

           = 1 == (a = 3)

           = 1 == 3

           = false

          

Let's evaluate the expression '(a = b) == a' for a = 1 and b = 3.

expression = (a = b) == a

           = (a = 3) == a

           = 3 == a

           = 3 == 3

           = true

Find the below working application.

App.java

package com.sample.app;

public class App {

    public static void main(String args[]) {
        int a = 1;
        int b = 3;

        System.out.println("a == (a = b) : " + (a == (a = b)));
        a = 123;
        System.out.println("(a = b) == a : " + ((a = b) == a));
    }

}

Output

a == (a = b) : false

(a = b) == a : true

Reference

https://docs.oracle.com/javase/specs/jls/se11/html/jls-15.html#jls-15.7.1

You may like

Miscellaneous

Interview Questions

Programming Questions

Thread.start() vs Runnable.run()

Pass ArrayList to a method with varargs?

Can I define synchronized methods in interface?

How to define global variables in Java?