Java is Pass by Value right?

Question

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Is Java pass by reference?

How come this code of mine is not working? I am passing an object into a method but it won't modify the original object that I created in main, why is that?

public class TestRun{

        public static void main(String[] args){
            Test t1 = new Test();
            Test t2 = new Test();
            t2.x = 555;
            t2.y = 333;

            System.out.println("The original valuess of X and Y are");
            System.out.println("X =  "+t1.x+"Y = "+t1.y);
            modifyObject(t1,t2);
            System.out.println("After the modification ");
            System.out.println("X =  "+t1.x+"Y = "+t1.y);


        }

        public static void modifyObject(Test arg1,Test arg2){
            arg1 = arg2;
        }
}

public class Test{

        int x = 9999;
        int y = 1;
}

A non-primitive type in C++ is an object but in Java it is implicitly a reference. References are passed by value. — Peter Lawrey, Sep 21 '11 at 07:59

score 2 · Accepted Answer · answered Sep 21 '11 at 01:56

2

You answered it in the title. Java is "pass by value".

This means that the method receives only copies of the object references arg1 and arg2.

You can, however, alter the contents of the objects, by doing

arg1.x=arg2.x
ary1.y=arg2.y

answered Sep 21 '11 at 01:56

Sanjay Manohar

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Soo if I do that, the original object will be modified? – Stacy Olkoha Sep 21 '11 at 01:57
Yep, you got it. And +1 to Sanjay. – Perception Sep 21 '11 at 02:07

score 1 · Answer 2 · answered Sep 21 '11 at 01:56

Its pass by value yes.

But what you need to recognise is that Test are reference types. In C, they would be pointers.

So what you're in fact doing in when calling modifyObject(arg1,arg2) is copying the values of the pointers (i.e. copying the memory address of the pointer, or copying the value of the variable with is the reference).

score 0 · Answer 3 · answered Sep 21 '11 at 01:58

0

in function modifyObject:

arg1 receives an address of an object of type Test, let's say x1000 arg2 receives another address of an object, let's say x2000

arg1 contains a reference to an object, but if you do arg1 = arg2, you are just telling arg1 to point to another address.

this way, you are not touching the content of the object pointed by arg1

arg1 and arg2 are local variables in the function. to make changes to the object, you need to address to the memory where it is stored, but arg1.x or arg1.y

notice that the object is somewhere in the memory, and arg1/arg2/t1/t2 are just saying where they are located, they are not the "actual" objects

answered Sep 21 '11 at 01:58

galchen

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soo in order for me to change their values, all I have to do is this? arg1.x = arg2.x am I right? – Stacy Olkoha Sep 21 '11 at 02:03
yes. this will copy the values in the memory from the instance pointed by arg2 to the instance pointed by arg1. arg1.x = arg2.x; arg1.y = arg2.y; will do the work. – galchen Sep 21 '11 at 02:08
soo thus the orignal objects value will change? – Stacy Olkoha Sep 21 '11 at 02:23
yes, it will change the values of the original object – galchen Sep 21 '11 at 02:26

score 0 · Answer 4 · answered Sep 21 '11 at 02:04

Uh?!!

What was your expectations? So, you create and print object t1 whose x,y values are 999,1 respectively. If you print it, that's what you'll get

$ java TestRun
The original valuess of X and Y are
X =  9999, Y = 1
After the modification 
X =  9999, Y = 1

So... what is your question again?

Q Is Java pass by value? A: Yes
Q It won't modify the original object? A: That's correct, it won't be modified.
Q Why is that? A: Because Java is pass by value

after the modification the value will be x = 555 and y = 333; — Stacy Olkoha, Sep 21 '11 at 02:58

Java is Pass by Value right?

4 Answers4