How to implement type declaration with generics?

Question

How can I define a function which is an implementation of a type declaration that includes generics?

Here's what I mean:

abstract class Base {}
class Sub extends Base {}

declare type BaseFactory = <T extends Base>() => T;
let subFactory: BaseFactory = <Sub>():Sub => new Sub();

console.debug(subFactory()); // expect instance of Sub

Unfortunately, this produces the following error:

Sub is not assignable to parameter of type Base

Can this be done?

/edit nevermind, it actually works in Playground. Guess it's a compiler version issue

Nitzan Tomer · Accepted Answer · 2017-06-05 13:44:37Z

2

This works:

declare type BaseFactory = <T extends Base>() => T;
let subFactory: BaseFactory = (): Sub => new Sub();

let a = subFactory<Sub>(); // type of a is Sub

(code in playground)

But I'd do it like this:

declare type BaseFactory<T extends Base> = () => T;
let subFactory: BaseFactory<Sub> = (): Sub => new Sub();

let a = subFactory(); // type of a is Sub

(code in playground)

Edit

If you're using typescript >= 2.3 then you can use default generics:

declare type BaseFactory<T extends Base = Base> = () => T;

And then you don't need to specify T everywhere.

edited Jun 5, 2017 at 13:44

answered Jun 5, 2017 at 13:21

Nitzan Tomer

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2 Comments

Benny Bottema Over a year ago

The problem with the second version is now you have to declare T everywhere BaseFactory is used as property or parameter, where as with the first version, T is only needed when actually invoking the function.

Nitzan Tomer Over a year ago

Check my revised answer

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