Is there a way to declare a getter in an interface?

Question

I have a class as follows (just showing part):

export class LinkedListNode<t> extends windward.WrObject implements ILinkedListNode<t>{
   public get next():LinkedListNode<t> {
        return this._next === this._list._first ? null : this._next;
    }
}

So in the interface I want to declare next as a method. However, the following does not work:

export interface ILinkedListNode<t> {
       get next() : LinkedListNode<t>;
}

The following does work, but is inaccurate:

export interface ILinkedListNode<t> {
       next : LinkedListNode<t>;
}

Is there a way to declare a getter in an interface? A getter is different than a member variable, for example it is not passed across to a web worker when posting an object.

thanks - dave

score 3 · Answer 1 · edited May 23 '17 at 12:14

3

No, although this is possible in other languages like C#, it is not currently possible in TypeScript.

If you want that behaviour, it might be better to make it a method:

export interface ILinkedListNode<t> {
    getNext(): LinkedListNode<t>;
}

edited May 23 '17 at 12:14

Community

1
1

answered Aug 12 '15 at 17:26

David Sherret

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The problem is that is not identical. We are passing objects to from web workers and a member variable is passed, but a getter is not. We use the interface to define what we receive in those messages and adding that as a member is inaccurate in that case. – David Thielen Aug 12 '15 at 17:46
@DavidThielen sounds like you could just leave it off altogether where you aren't expecting the getter to be available. – Fenton Aug 12 '15 at 18:51
@SteveFenton The problem is we would really like to use the interface in places for how we want to set up the linked list. But to do so, it needs all the public methods listed. Yet that gets in the way of correctly handling objects passed to/from the web worker. It's a somewhat unusual use case we have but for this all to work, we need it explicitly as a getter in the interface. Oh well... – David Thielen Aug 12 '15 at 19:56

Buzinas · Answer 2 · 2015-08-12T18:22:11.173

At the moment of writing, there is no way to accomplish what you want.

There is an open issue on TypeScript's Github dated since 2014 July, and that feature will be available as soon as they implement it.

The only way of ~simulate~ what you need, is by doing what David Sherret said in his answer: creating a getter method.

export interface ILinkedListNode<t> {
  getNext(): LinkedListNode<t>;
}

Although, I recommend to only create the property inside your interface now, and changing it to the ~readonly~ or ~get~ version when it be available in the future.

Is there a way to declare a getter in an interface?

2 Answers2