Adding to ArrayList in a Constructor

[Bobby Smallman]

Hello Everyone!

I am trying to figure out if there is a way to add a newly created object to an ArrayList in a constructor. It seems that I cannot get it to work the way I would like because really the object is not yet created when it is trying to do it. Is it ever possible and if so is it ever done or is it not proper?

public static void main(String[] args) {
        
        
        
        Animal dog0 = new Dog(dog0, "Fluffy", 1, 1);     // Syntax for new object Animal(array, "name", size, noise)
        Animal dog1 = new Dog(dog1, "Sparky", 5, 5);
        Animal dog2 = new Dog(dog2, "Ellie", 10, 10);
        Animal moose0 = new Moose(moose0, "Moosey", 1, 1);
        Animal moose1 = new Moose(moose1, "Browny", 5, 5);
        Animal moose2 = new Moose(moose2, "Chocolate", 10, 10);
        
        
      
    } 

 
public Animal(Animal aarray, String aname, int asize, int anoise) {
		
		if (asize <= 0 || asize > 10){
			badSize(asize);
		} else {
			setSize(asize);
		}
		if (anoise <= 0 || anoise > 10){
			badNoise(anoise);
		} else {
			numnoise = anoise;
		}	
		setName(aname);
		
		animals.add(aarray);
		
	}

Those are the main() and the constructor which gets called where I would like to add the new objects to the ArrayList.

[Raymond Tong]

Bobby Smallman wrote:

Animal dog0 = new Dog(dog0, "Fluffy", 1, 1);

You are passing dog0 as constructor parameter and assigning the object instance back to dog0.
What you are trying to do?
What is the problem you are trying to solve?
Shouldn't it be

Animal firstDog = new Dog(name, size, noise);
Animal secondDog = new Dog(name2, size2, noise2);
List<Animal> animalList = new ArrayList<Animal>();
animalList.add(firstDog);
animalList.add(secondDog);

Animal shouldn't know about the List, so it shouldn't pass the List as constructor parameter.

[Bobby Smallman]

You can absolutely do that. But I was trying to make the adding of the animal object to the ArrayList in the constructor of the object to streamline the creation of the object. It isn't a big deal to do a animals.add(Animal) for each of the objects in main when there are only a couple of them, but wouldn't running it in the constructor streamline the process for when you have TONS of entries?

[Raymond Tong]

Bobby Smallman wrote:You can absolutely do that. But I was trying to make the adding of the animal object to the ArrayList in the constructor of the object to streamline the creation of the object. It isn't a big deal to do a animals.add(Animal) for each of the objects in main when there are only a couple of them, but wouldn't running it in the constructor streamline the process for when you have TONS of entries?

So you are trying to get rid of typing many animals.add(Animal)?
Let say I have an object called BankAccount, if I pass the bankAccountList to constructor ?
to expose the bankAccount details belong to other bank accounts?
Actually the code could compiles, but this shouldn't be a good design, right ?

If you want to reduce lines of code, you could do

animalList.add(new Dog(name, size, noise));

Or you could do

private void addDog(List<Animal> animalList, String name, int size, int noise) {
  Dog dog = new Dog(name, size, noise);
  animalList.add(dog);
}

[Soniya Ahuja]

HI Bobby,

What are you trying to achieve? I think that what you are trying to do is somehow store all the animals created in an ArrayList? Is that correct?

If yes, then first of all, this ArrayList should belong to the class instead of instances of the class - so it should be marked as static.
Then from the constructor you can call arrayList.add(this)

Note:
A static member belongs to the class rather than a particular object. Each object has it's own copy of non-static objects - however, there is just one copy of a static member which is shared between all instances of that class. Further, a static member is associated with a class. So, existence of a static member depends only on the existence of a class and a static member is accessed as ClassName.staticMemberName. Thus, static members DO NOT need objects to access them.

this keyword refers to the current instance as already pointed by Ernest. You were trying to pass the animal reference but actually it was the object that you needed access to and using "this" you can access that instance.

Eg

public class Animal {
       private static ArrayList<Animal> animals = new ArrayList<Animal>();
       public Animal(String name, int asize, int anoise) {
               // logic for assigning asize and anoise and name will come here
               animals.add(this);
       }
}

Now when you create a new animal, it will automatically be added to the ArrayList animals. You can write more methods to access the ArrayList and you may also mark it as final to make sure that all instances access the same ArrayList and do not create a new one and assign that by mistake.

Again, the code above might not be what you are expecting because I haven't really understood your question or rather your purpose well.

[Campbell Ritchie]

Please explain more about why you think a static List would be a good choice. I don't think you have given enough detail for people "beginning Java" to understand your solution.

[Ernest Friedman-Hill]

Bobby, I think Soniya hit the nail on the head by showing you a bit of Java knowledge you didn't previously have: the this keyword, which refers to the current object.

[Bobby Smallman]

Thank you to everyone who responded. The animals.add(this) solution was what I was looking for. Is it not good practice to allow the constructor to add the new item to an ArrayList? Would a Static Factory (Item 1 of Effective Java) be more efficient?

[Ernest Friedman-Hill]

If the arraylist is supposed to contain all Animals ever created, then yes, that's a perfectly fine way to do it. In a real application there might be issues with garbage collection -- i.e., if every Animal is added to this list, then they can never be collected -- and in that case, something more explicit might be better, so that animals could be added or removed from the list.

[Bobby Smallman]

Perfect, seems all questions have been answered! Thank you to everyone who gave input!

[Campbell Ritchie]

You're welcome

[Campbell Ritchie]

Yesterday, I wrote:Please explain more . . .

Thank you

The edited post now includes that additional explanation.

[Soniya Ahuja]

@Campbell
@Bobby - Welcome

[Bobby Smallman]

Thanks for the welcome! I am excited to be spending a great deal of my time here!