CODEWITHSUNDEEP
javaoopcomputer-science
vincent mathew
vincent mathew

Reputation: 4378

Object Reference passing in Java. Pass by reference working differently in different case?

While playing in Java. I saw different behaviour if an object is modified and given a value and different value if it is assigned a new object. Here is code that I made to show the result.

public class Test {

    int i;

    public Test(int j) {
        this.i = j;
    }

    public static void main(String[] args) {

        Test A = new Test(5);
        Test N = new Test(5);
        add(A);
        makeNew(N);
        System.out.println("Value of A.i= "+A.i);
        System.out.println("Value of N.i= "+N.i);

    }

    private static void add(Test t) {
        t.i+= 3;
        System.out.println("Inside method add() t.i= "+t.i);

    }

    private static void makeNew(Test t) {

        t = new Test(8);
        System.out.println("Inside method makeNew() t.i= "+t.i);

    }

}

Here is the output of the above code.

Inside method add() t.i= 8
Inside method makeNew() t.i= 8
Value of A.i= 8
Value of N.i= 5

In above example object A is modified to value 8. And object B is given a new object itself. But calling them back only object A shows new value. Object B shows the old value itself. Should not they be showing same value because both case are pass by refernce? I was expecting same value for A.i and N.i.

Upvotes: 2

Views: 213

Answers (6)

Jops
Jops

Reputation: 22715

Here's what happens:

Test A = new Test(5);
Test N = new Test(5);

enter image description here

add(A);  // method is add(Test t)

enter image description here

makeNew(N)// method is makeNew(Test t)

enter image description here

t = new Test(8);

enter image description here

System.out.println("Value of A.i= "+A.i);
System.out.println("Value of N.i= "+N.i);

enter image description here

Upvotes: 3

Aniket Thakur
Aniket Thakur

Reputation: 68935

In Java you do not pass the actual object nor do you pass the reference to the object. You pass copy of the reference to that object. Now when you say

makeNew(N);

N which is the reference to new Test(5) is not passed but the copy of it's reference is passed. In the makeNew() function this copy points to some new object and print the value appropriately but the N will still point to the original object.

Upvotes: -1

Stochastically
Stochastically

Reputation: 7846

In your makeNew, you're overwriting the reference to the existing object (that's passed in as the paramter) with your new test(8) object. However, that's local inside makeNew, so the original object sitting inside main(...) is not affected.

Upvotes: 1

Abubakkar
Abubakkar

Reputation: 15644

I think this will make you doubt clear:

Before assigning t inside makeNew to another object

After you assign t to a new object

You see N still point to the first object

Upvotes: 1

LanternMike
LanternMike

Reputation: 664

Whenever you make a variable equal to an object and later use new on that object somewhere else like through another reference, your variable you set to the objects reference no longer points to whatever the new object is, but still holds onto the old. So if multiple variables at different scopes hold a reference, they all need a way to have them made equal to whatever the new object is or they no longer are in synch.

Upvotes: 1

Matthias Van Eeghem
Matthias Van Eeghem

Reputation: 329

Java is pass-by-value. You pass the reference of an object as a value, and you can thus modify that object. However, you cannot modify the actual reference of an object and make it point to something else.

Your question has already been answered here: Is Java "pass-by-reference" or "pass-by-value"?

Upvotes: 0

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