How can I pass an Integer class correctly by reference?

user_251589 0
Viewed 116432

I am hoping that someone can clarify what is happening here for me. I dug around in the integer class for a bit but because integer is overriding the + operator I could not figure out what was going wrong. My problem is with this line:

Integer i = 0;
i = i + 1;  // ← I think that this is somehow creating a new object!

Here is my reasoning: I know that java is pass by value (or pass by value of reference), so I think that in the following example the integer object should be incremented each time. 

public class PassByReference {

    public static Integer inc(Integer i) {
        i = i+1;    // I think that this must be **sneakally** creating a new integer...  
        System.out.println("Inc: "+i);
        return i;
    }

    public static void main(String[] args) {
        Integer integer = new Integer(0);
        for (int i =0; i<10; i++){
            inc(integer);
            System.out.println("main: "+integer);
        }
    }
}

This is my expected output:

Inc: 1
main: 1
Inc: 2
main: 2
Inc: 3
main: 3
Inc: 4
main: 4
Inc: 5
main: 5
Inc: 6
main: 6
...

This is the actual output. 

Inc: 1
main: 0
Inc: 1
main: 0
Inc: 1
main: 0
...

Why is it behaving like this?

pass-by-reference integer java
10 Answers

1 ) Only the copy of reference is sent as a value to the formal parameter. When the formal parameter variable is assigned other value ,the formal parameter's reference changes but the actual parameter's reference remain the same incase of this integer object.

public class UnderstandingObjects {

public static void main(String[] args) {

    Integer actualParam = new Integer(10);

    changeValue(actualParam);

    System.out.println("Output " + actualParam); // o/p =10

    IntObj obj = new IntObj();

    obj.setVal(20);

    changeValue(obj);

    System.out.println(obj.a); // o/p =200

}

private static void changeValue(Integer formalParam) {

    formalParam = 100;

    // Only the copy of reference is set to the formal parameter
    // this is something like => Integer formalParam =new Integer(100);
    // Here we are changing the reference of formalParam itself not just the
    // reference value

}

private static void changeValue(IntObj obj) {
    obj.setVal(200);

    /*
     * obj = new IntObj(); obj.setVal(200);
     */
    // Here we are not changing the reference of obj. we are just changing the
    // reference obj's value

    // we are not doing obj = new IntObj() ; obj.setValue(200); which has happend
    // with the Integer

}

}

class IntObj { Integer a;

public void setVal(int a) {
    this.a = a;
}

}

user_103511210

We can do this using Apache Commons Mutable Int

public static Integer inc(MutableInt i) {
        i.increment();    
        System.out.println("Inc: "+i.getValue());
        return i;
}

public static void main(String[] args) {
        MutableInt integer = new MutableInt(0);
        for (int i =0; i<10; i++){
            inc(integer);
            System.out.println("main: "+integer.getValue());
        }
}

This produces output:

Inc: 1
main: 1
Inc: 2
main: 2
Inc: 3
main: 3
Inc: 4
main: 4
Inc: 5
main: 5
Inc: 6
main: 6
Inc: 7
main: 7
Inc: 8
main: 8
Inc: 9
main: 9
Inc: 10
main: 10
user_24588580
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