JAVA Variable Scope

Question

I am looking through some old questions for a test and I come across a quite easy question however it wasn't what I expected.

public class Exampractice {
  public static void f(int x, int[] y, int[] z) {
    x = 2;
    y[0] = x;
    z = new int[5];
    z[0] = 555;
  }
  public static void main(String[] args) {
    int x = 111;
    int[] y = {222, 333, 444, 555};
    int[] z = {666, 777, 888, 999};
    f(x, y, z);
    System.out.println(x);
    System.out.println(y[0]);
    System.out.println(z[0]);
  }
}

The question asks what is the result of the following code. I get the following results:

111 
2 
666

I understand why x is 111 as local variables override all other variables, y[0] =2 as the code says it equals x which is 2 but I am lost on why z[0] = 666 as it has been rearranged in the f class.

It is fairly easy to understand if you remember few things: 1) in Java all variables are passed by value - i.e. you get a copy of the primitive data types and a copy of the reference to a complex data type. 2) new operator creates completely new object, it does not update the existing object. — Germann Arlington
– Germann Arlington, Commented Apr 3, 2014 at 10:55

Community · Accepted Answer · 2017-05-23 10:31:50Z

6

In Java, the object reference is passed as a value. Therefore, when z = new int[5]; is executed, the local array variable z present in the f() method now refers to the newly created int array and nothing happens to the original array whose reference was passed as a value to it when the method was called.

public static void f(int x, int[] y, int[] z) {
    x = 2;
    y[0] = x;
    // Till here z is referring to the int array passed from main method
    z = new int[5]; // now z is re-assigned with a new reference, the one of the newly created int array
    // thus the reference to the original array is no more being used here
    z[0] = 555; // this modifies the values of the newly created array 
}

Personally, I always suggest reading this answer by Eng.Fouad to understand this concept.

edited May 23, 2017 at 10:31

CommunityBot

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answered Apr 3, 2014 at 10:49

Rahul

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1 Comment

Softey Over a year ago

I see how it works now... Took a little while to get my head around it

ruhungry · Accepted Answer · 2014-04-03 10:49:40Z

1

Because you are using new which creates new object

answered Apr 3, 2014 at 10:49

ruhungry

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Comments

saravanakumar · Accepted Answer · 2014-04-03 11:03:00Z

1

What matters is whether the value or reference is set to variable

x has value

    x=111

and

    y and z
    points two different arrays

y and z are not having value it has a reference of an array

    reference y[0] is edited globally  in f

    z is recreated in the method f locally using "new"

It does not affect original object in method main() so we get that result.

edited Apr 3, 2014 at 11:03

answered Apr 3, 2014 at 10:57

saravanakumar

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Comments

Niels Bech Nielsen · Accepted Answer · 2014-04-03 10:49:03Z

0

The method creates a new int array. The old one is intact after the method exits.

answered Apr 3, 2014 at 10:49

Niels Bech Nielsen

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Ofer · Accepted Answer · 2014-04-03 10:51:33Z

0

Z is a new object, the z in the function is actually this.z

answered Apr 3, 2014 at 10:51

Ofer

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