Typescript generic type constraint from array of types

Question

I have an array of types declared as

const APIS = [NewsApi, CommentsApi, FileApi];

I want to have a method that accepts one of these types from that APIS array, like the following:

myMethod<T extends anyof APIS>(apiType: T): T {
    const api = new apiType(); //pseudo code, does not work off course
    // some common configuration code
    return api;
}

. I cannot do myMethod<T extends NewsApi|CommentsApi>(): T, as the APIS array is auto generated. Is this possible with typescript?

Update:

Aleksey's solution works great for passing in the input. However, the return type cannot be inferred from the usage.

const myApi = myMethod(NewsApi);

In above case, myApi is not of type NewsApi, but NewsApi|CommentsApi|FileApi. Is it possible to set the return type explicitly to instance of input api type?

Aleksey L. · Accepted Answer · 2020-02-15 14:09:22Z

3

If you can change definition of the array to readonly tuple:

const APIS = [NewsApi, CommentsApi, FileApi] as const;

then you can get union of possible array values using lookup type:

type ApiConstructors = typeof APIS[number];

function myMethod<T extends InstanceType<ApiConstructors>>(apiType: new () => T): T {
  return new apiType();
}

const api = myMethod(NewsApi); // api is of type NewsApi

InstanceType utility used to get instance type of a constructor function.

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edited Feb 15, 2020 at 14:09

answered Feb 15, 2020 at 6:05

Aleksey L.

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5 Comments

Victor Mukherjee Over a year ago

Thank you @Aleksey :) . Also, would it be possible to set the return type of the method explicitly to an instance of api type? I understand that the return type will be implicitly inferred from the returned object. Just wondering, if it can also be set explicitly, just in case I am declaring an interface.

Victor Mukherjee Over a year ago

Just noticed that the return type is not exactly inferred from the usage. When I declare const myApi = myMethod(NewsApi); myApi is not inferred as NewsApi, but union of all apis.

Aleksey L. Over a year ago

I'll check this later, but that's not what you've asked originally 🙂

Victor Mukherjee Over a year ago

I agree, I should have been more specific in the first place. Sorry for that. :)

Aleksey L. Over a year ago

@VictorMukherjee I've updated answer and playground

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