java code why println 10?

Question

public class Exa {
    public static void main(String[] args) {
        Integer b = new Integer(10);
        add(b);
        System.out.println(b.intValue());
   }

    public static void add(Integer b){
        int i = b.intValue();
        i += 3;
        b = new Integer(i);
        System.out.println("b="+b+",i="+i);
    }
}

I wrote the above code, and run output 10. Why did it not change?

Why output 10, please give detailed instructions，thanks！

You made a new `Integer`. Did you really expect the old one to change when you did that? — Dawood ibn Kareem, May 22 '18 at 06:47

score 2 · Accepted Answer · edited Jun 20 '20 at 09:12

The method local variable b in your main method is different from the method local variable in your add() method. Changing one won't affect the other.

b = new Integer(i);

Here you are referring to the variable in the add(Integer b) method not the variable in the main method.

Java is always pass-by-value. Unfortunately, they decided to call the location of an object a "reference". When we pass the value of an object, we are passing the reference to it. This is confusing to beginners.

(Extracted from this answer by erlando)

score 0 · Answer 2 · answered May 22 '18 at 06:58

0

The value does not change because you only assign b locally inside the add method. In Java, you can not change the value of an Integer, you only change the reference. To have the intended side effect, you could wrap b in an object.

answered May 22 '18 at 06:58

Lars Gendner

1,816
2
14
24

java code why println 10?

2 Answers2