CODEWITHSUNDEEP
javaoutput
skrtbhtngr
skrtbhtngr

Reputation: 2251

Cannot understand why this Java program produces such output

Here is the code :

class Value
{
    public int i = 15;
}
public class Test
{
    public static void main(String argv[])
    {
        Test t = new Test();
        t.first();
    }
    public void first()
    {
        int i = 5;
        Value v = new Value();
        v.i = 25;
        second(v, i);
        System.out.println(v.i);
    }
    public void second(Value v, int i)
    {
        i = 0;
        v.i = 20;
        Value val = new Value();
        v =  val;
        System.out.println(v.i + " " + i);
    }
}

I cannot understand why this code prints

15 0
20

on the console.

Why is it not

15 0
15

?

Upvotes: 1

Views: 244

Answers (1)

Sotirios Delimanolis
Sotirios Delimanolis

Reputation: 279880

Everything in Java is passed by value. In this first method

public void first()
{
    int i = 5;
    Value v = new Value();
    v.i = 25;
    second(v, i);
    System.out.println(v.i);
}

you pass the value of the reference stored in v (which points to a Value object with an i field with value 15) to second.

public void second(Value v, int i)
{
    i = 0;
    v.i = 20;
    Value val = new Value();
    v =  val;
    System.out.println(v.i + " " + i);
}

It dereferences the reference value to find a Value object and changes its i field value to 20. You then create a new Value object with its i field value initialized to 15. That's what you're printing

15 0

The method returns and first prints the value of the object the local variable v is referencing, ie.

20

Upvotes: 3

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