Cannot instantiate Generic item to a list Java

Question

I have a generic class that looks similar to this

public class NetworkItems<T> { private Status status; private T items; public NetworkItems() { this.status = Status.FAILURE; this.items = null; } public NetworkItems(Status status, T listItems) { this.status = status; this.items = listItems; } public NetworkItems(Status status) { this.status = status; this.items = null; } public Status getStatus() { return status; } public void setStatus(Status status) { this.status = status; } public T getItems() { return items; } public void setItems(T items) { this.items = items; } }

All I want to do is check if the parameter T is a List of any kind in the constructors. If T is a List<> then I want to instantiate it into a new List<>();

I tried the following code

if (this.items instanceof List) { this.items = Collections.emptyList(); }

or this

this.items = new ArrayList<>();

But I keep getting an error because I couldn't give a parameter type. How do I make sure that I instantiate the items with a new List if the generic type T is a List?

If you plan on hard-coding a dependency on list, then why not just use NetworkItems<List>? — Tim Biegeleisen
– Tim Biegeleisen, Commented May 2, 2018 at 11:15
Because the container should be able to host both list of items and a standalone item. — Sriram R
– Sriram R, Commented May 2, 2018 at 11:28
@SriramR Because the container should be able to host both list of items and a standalone item - why should it? Why not have two different containers for single item or collection of items? — lexicore
– lexicore, Commented May 2, 2018 at 11:39
I didn't wanna repeat code. The implementation is the same. Except for the T part. So I thought if there's a way to do without repeating code, why not do it? — Sriram R
– Sriram R, Commented May 2, 2018 at 11:49

daniu · Accepted Answer · 2018-05-02 11:19:48Z

Give the class a non-optional constructor argument of type Supplier<T>.

public class NetworkItems<T> { private Status status; private T items; private Supplier<T> defaultCreator; public NetworkItems(Supplier<T> def) { this.status = Status.FAILURE; this.items = null; this.defaultCreator = def; } public T createDefault() { return defaultCreator.get(); } // ...

Now, if you have an instance of NetworkItems, you can always get a default value. Eg,

NetworkItem<List<String>> stringListItems = new NetworkItem<>(ArrayList::new); List<String> defItem = stringListItems.createDefault();

Ehler · Accepted Answer · 2018-05-02 11:20:19Z

0

You can just add a type cast, then it should work like you already said:

if (this.items instanceof List) { this.items = (T) Collections.emptyList(); }

answered May 2, 2018 at 11:20

Ehler

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4 Comments

Sriram R Over a year ago

It gives an UncheckedCastException warning.

Ehler Over a year ago

Yes, that's true but I think you cannot prevent this if you use such generic member variable because the constructor cannot know at compile time which type you want to use there. Maybe you can use another approach, instead using just T items you can use direcly a list List<T> items; where you add your items.

lexicore Over a year ago

Also even if T is List<Something>, this.items may still be null so instanceof won't help much.

Sriram R Over a year ago

Yeah, I did use it earlier as List<T>. The thing is I want this container to be able to hold a List<> and an Object. So I just wanted to know if its possible without writing a new Holder.

Jacob · Accepted Answer · 2018-05-02 13:05:00Z

Based on comments herein and your original post, it seems like items is intended as a collection regardless of your situation. Would you not be better suited to do the following?

 public class NetworkItems<T> { private Status status; private List<T> items; //continue the remainder of your implementation here }

I understand this is not ideal for the scenario where 1 item can exist, but you ensure type safety here and adhere to the concept of this can be one or more items, by making items a collection.

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