Filling array of Java objects in initialization

Question

I have an implemented data structure called List (it's a singly linked list). I have the following code in my Java program:

List[] Adjacent = new List[N]; // Trying to create an array of N List objects

Suppose N = 6. I want Adjacent to be an array of six elements where each one of them is a List object itself. But when I try to access it, the array is actually an array of N null references. What is going on?

Of course I did the following to solve my problem:

for (int p = 0; p < N; p++) 
    Adjacent[p] = new List();

But that's not what I really want. Is there a more efficient/nice way to do it?

Answer 1

When you have

List list;

this is a reference , not an object. To have an object you have to create one with new List(...) . There is no way to work around this. You have to initialise your references to actual objects.

Also, it is far worse to be creating a class which does the same thing with a similar name to something already built in. I suggest you use a List of List like this:

List<List<String>> listofLists = new ArrayList<>();
for(int i = 0; i < N; i++) listOfLists.add(new ArrayList<>());

It is more efficient for you to reuse existing classes.

Answer 2

There will be a “nice” way to to it with Java 8:

MyObject[] array=Stream.generate(MyObject::new).limit(N).toArray(MyObject[]::new);

Stream.generate(MyObject::new) will create an unlimited stream of objects provided by a Supplier which is the default constructor of MyObject in our case, limit(N) obviously limits the number of objects and toArray(MyObject[]::new) will store the objects into an array constructed by an IntFunction<A[]> that constructs an array using a specified int which is set to the equivalent of new MyObject[i] in our case. The stream will provide the limit N for that i parameter.

Though the last part is not the real spirit of streams . Normally, you would direct the stream to the consumer dealing with the objects and omit the array completely.

Answer 3

Nope. After create new array you should fill it manualy. Of course we have Arrays.fill(array, object) method but it will set the same object to all positions in array.

Answer 4

Suppose N is always 6 ...

List[] l = new List[]{new List(),new List(),
       new List(),new List(),new List(),new List()};