CODEWITHSUNDEEP
javaarraysarraylist
Sahil Chaudhary
Sahil Chaudhary

Reputation: 503

Passing ArrayList as a parameter

Let me start by saying that I can't put any code here because Internet on my laptop is not working so I am posting this through my phone. Okay the problem is that say I have two classes: class one and two. Class one has an ArrayList as one of its attributes and it calls a void method from class two and passes that ArrayList as a parameter. Now that method initializes another ArrayList and makes it equal to the parameter passed by me and makes changes to that new ArrayList. Funny thing is that even my original ArrayList which was passed as parameter is also changing. What could be the possible reason?

Upvotes: 12

Views: 125165

Answers (6)

Stephen
Stephen

Reputation: 10059

Sample Code:

public class MethodArguments {

public static void main(String args[]) {

    ArrayList<String> a = new ArrayList<String>();

    a.add("Steve");

    a.add("Daniel");
    a.add("John");
    a.add("Maxi");
    a.add("Jeni");

    System.out.println(a);

    display(a);

    getSize(a);

}

static void display(ArrayList<String> arrayList1) {

    arrayList1.add("Pollard");

    System.out.println(arrayList1); // passing the arraylist values and
                                    // adding the element

}

static void getSize(ArrayList<String> arrayList1) {

    System.out.println(arrayList1.size()); // getting the size of arraylist
                                            // by passing arguments to
                                            // method
 }

}

Output:

[Steve, Daniel, John, Maxi, Jeni]

[Steve, Daniel, John, Maxi, Jeni, Pollard]

6

Upvotes: 3

Chris Dargis
Chris Dargis

Reputation: 6043

The = operator in Java will just copy the ArrayList reference (the same is for all objects). See this answer for making a deep copy of an ArrayList.

Upvotes: 0

PCM
PCM

Reputation: 873

This is because the new array list points to the same old array when you make it equal to.

This small example should clarify it.

import java.util.ArrayList;
import java.util.List;


public class JavaApplication1 {


    public static void main(String[] args) {

        List <String> origList = new ArrayList<>();
        origList.add("a");
        origList.add("b");

        JavaApplication1 app = new JavaApplication1();
        app.addToList(origList);

        for(String str:origList){            
            System.out.println(str);
        }

    }

    private void addToList(List<String> strList){
        System.out.println("inside addToList");
        List <String> newList = new ArrayList<>();
        // newList = strList; //This is how you are doing it
        newList.addAll(strList); //This is how you should do it.

        newList.add("x");
    }
}

Upvotes: 0

AlexWien
AlexWien

Reputation: 28727

The reason is that when you pass an ArrayList as argument, the called method can change the content of the array. The ArrayList contains references to Objects. If you want to avoid that some class will change the content of your ArrayList you have to give back Copy of your ArrayList where all objects inside are clone()s of the objects in your list.

Use object.clone() 

ArrayList clonedCopy = new ArrayList(list1.size());
for (Object obj : list1) {
  clonedCopy.add(obj.clone());
}

Now give this clonedCopy back. But make sure obj is cloneable!

Upvotes: 3

mauro.dec
mauro.dec

Reputation: 556

The problem is that when you use = to make the new ArrayList a copy of the original, you're just creating a new reference to the same ArrayList. Think of it as two variables pointing at the same object.

Check this out, it might help you understand what's happening: Is Java "pass-by-reference" or "pass-by-value"?

In order to solve your problem, you need to create a new ArrayList by using the "new" keyword and then adding all of the objects, or use the clone() method.

Upvotes: 6

DarthVader
DarthVader

Reputation: 55022

Because they point to the same reference.

Upvotes: 1

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