How do I set a value to a variable in a method and print it out in the main method?

Question

public class Shape 
{
    public static void main(String args[]){
        int num = 0;
        cng(num);
    }

    public static void cng(int x){
        x = 52;
        System.out.println(x);
    }

}

As you can see, in the cng Method I set the value of x to 52 and then print out the value of x.

Then, back in the main method, the cng method is performed on the num variable.

What I want to do, however is set the value of 52 to x without the System.out.println(x); function in my cng method and print out the value in my main method. How would I go about doing that?

I tried doing

public static void cng(int x){
    x = 52;
}

and then

public static void main(String args[]){
    int num = 0;
    cng(num);
    System.out.println(num);
}

but it only prints out a 0 because num is set to 0. I thought that performing cng on the num variable would change it to 52, but it doesn't.

change return type of `cng()` to int, and return int from `cng()` to main method — Nandkumar Tekale, Nov 23 '12 at 08:27
[Immutable objects](http://docs.oracle.com/javase/tutorial/essential/concurrency/immutable.html) can not be changed like this, Please also read [is java pass by reference](http://stackoverflow.com/questions/40480/is-java-pass-by-reference) — Bo PENG, Nov 23 '12 at 08:34

PermGenError · Answer 1 · 2012-11-23T08:31:54.560

2

make your cng method return an int variable

public static int cng(int num){    
num = 52; 
return num; 
}

In Your Main method, assign the returned variable from cng() method

int num = 0;
 num =cng(num);
 System.out.println(num);

Or:

you could always, make num as a member static variable,

   static int num;

edited Nov 23 '12 at 08:31

answered Nov 23 '12 at 08:26

PermGenError

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Geerten · Answer 2 · 2012-11-23T09:26:04.130

1

You want to pass the data by reference. This is not possible for primitive values in Java (int, double, boolean, etc).

You have the following options:

Make it a member of the class
Create a wrapper object (object references are also passed by value, but you can change the members of that object in a function, just not the object reference itself)
Return the value in the function (as mentioned by other answers)

edited Nov 23 '12 at 09:26

answered Nov 23 '12 at 08:27

Geerten

1,027
1
9
22

partly wrong, Objects are not passed by reference in java . java is always, pass by value in case of both objects and primitives. here's a very good explanation i would suggest. http://stackoverflow.com/questions/40480/is-java-pass-by-reference – PermGenError Nov 23 '12 at 08:37
@GanGnaMStYleOverFlowErroR : oppa gangnam style . You might wany tp refer : http://stackoverflow.com/questions/40480/is-java-pass-by-reference – Mukul Goel Nov 23 '12 at 08:41
1

Actually correct most would be to say : java passes object refferences by value. – Mukul Goel Nov 23 '12 at 08:43
@GanGnaMStYleOverFlowErroR : yea , i think it is neither objects by value nor object as reference it is object references by value. Just for a ref: i posted before you edited ypur initial commemt :-) – Mukul Goel Nov 23 '12 at 08:47
I knew it is like Mukul Goel stated, but I thought that was a bit too complicated for the questioner. In practice it boils down to objects being passed by reference. – Geerten Nov 23 '12 at 09:25

score 1 · Answer 3 · answered Nov 23 '12 at 08:28

1

Try to change your code to this:

public class Shape 
{


public static void main(String args[]){
    int num = 0;
    num = cng(num);
    System.out.println(num);
}

public static int cng(int x){
    x = 52;
    return x;
}

}

answered Nov 23 '12 at 08:28

djakapm

175
7

score 1 · Accepted Answer · answered Nov 23 '12 at 08:30

read along the comments :

public class Shape 
{

    public static void main(String args[]){
    int num = 0;
    num = cng();   //store value returned by cng() in num
    System.out.println("num : " +num); // display num
    }

    public static int cng(){    //change return type to int
    return 52;
    }

}

score 0 · Answer 5 · answered Nov 23 '12 at 08:31

0

The reason for this behaviour is that arguments are passed by value in Java. This implies that it is simply a copy of the variable that you passed to it. Thus the assignment that you do is only local to the method.

answered Nov 23 '12 at 08:31

Karaaie

527
5
15

How do I set a value to a variable in a method and print it out in the main method?

5 Answers5