Manipulating a List of String in Java

Question

please, I'm having the code shown below:

  public static void main(String[] args) {
    List<String> list1 = new ArrayList<>();
    list1.add("foo");
    System.out.println("\nThis is list 1 : " +list1); //foo

    List<String> list2 = list1;
    System.out.println("\nThis is list 1 : " +list1); // foo
    System.out.println("\nThis is list 2 : " +list2); // foo

    List<String> list3 = new ArrayList<>(list2);
    System.out.println("\nThis is list 3 : " +list3); // foo
    list1.clear();
    System.out.println("\nThis is list 1 : " +list1); // []
    list2.add("bar");
    list3.add("baz");

    System.out.println("\nThis is list 1 : " +list1); // bar
    System.out.println("\nThis is list 2 : " +list2); // bar
    System.out.println("\nThis is list 3 : " +list3); // [foo, baz]

}

output:

**list 1 : [bar], list2 : [bar], list3 : [foo, baz]**

My question is how did list1 get the value bar? because I was thinking it was supposed to be empty at the time. And also, I was thinking that list2 is going to have the values [foo, bar] at this time. Please what is the logic behind this?

Because you have one list, List<String> list2 = list1; does not make a new List. Change it to List<String> list2 = new ArrayList<>(list1); for that. — Elliott Frisch
– Elliott Frisch, Commented Jul 20, 2021 at 3:52

Basil Bourque · Accepted Answer · 2021-07-20 05:40:13Z

1 `ArrayList`, 2 references

Understand that List<String> list1 and List<String> list2 do not hold a list. They hold a reference (pointer) to a list.

Your line:

List<String> list2 = list1;

…copies the contents of list1 (a reference) to list2. You copied the reference (basically a memory address number), not the list.

Your variable list2 points to the same ArrayList object as list1. You have 2 references but only 1 list. So making a change through either reference will be visible through both.

Make a copy

As commented by Elliott Frisch, if you want a copy of a list, pass to the constructor.

List< String > list2 = new ArrayList<>( list1 ) ;  // Make copy of one list by passing to constructor of another.

This is a shallow copy, meaning we are copying the references into a new list the original objects to which the references point are not copied. So the references are being copied, not the referents.

List<String> list1 = new ArrayList<>();
list1.add("foo");
System.out.println("list1: " + list1 );          // foo

List<String> list2 = new ArrayList<>( list1 ) ;  // foo
list1.clear();                                   // empty 
list2.add( "bar" );                              // foo, bar
System.out.println( "list1: " +list1 );          // empty
System.out.println( "list2: " +list2 );          // foo, bar

See this code run live at IdeOne.com.

list1: [foo]
list1: []
list2: [foo, bar]

Immutable copy

Or, for an immutable copy, use List.copyOf. If not already immutable, you get a duplicate list.

List< String > list2 = List.copyOf( list1 ) ;  // Creating an immutable copy of the first list, *if* that list is not already immutable.

If a mutable list underlies the immutable list such as with Collections.unmodifiableList, you can combine these two techniques to ensure a separate fresh immutable list is produced.

List<String> list2 = List.copyOf( new ArrayList<>( list1 ) ) ;

FLAGLORD · Accepted Answer · 2021-07-20 04:15:31Z

1

The following statement lets list2 refer to the same list as list1

List<String> list2 = list1;

However,list3 uses another list to initialize itself.

List<String> list3 = new ArrayList<>(list2);

I think that you should know about the difference between shallow and deep copy...

answered Jul 20, 2021 at 4:15

FLAGLORD

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