Using a generic type argument with `typeof T`

Question

I have a factory-like function that is meant to return an sub-class instance of BaseApi. It currently looks something like this (trimmed out irrelevant parts):

function getApi<T extends BaseApi>(apiClass: typeof BaseApi): T {     return new apiClass(); } 

And I use it like this:

const someApi = getApi<SomeApi>(SomeApi); 

This works, but I would like <SomeApi> to be inferred by virtue of the fact that I'm passing the SomeApi constructor in. If I omit <SomeApi> then someApi is inferred to be of type BaseApi, not SomeApi. Worse, there's really no compiler correlation between <T extends BaseApi> and typeof BaseApi being the same thing, so you can incorrectly do something like getApi<SecondApi>(FirstApi) without a compiler error.

So I tried defining the apiClass as typeof T:

function getApi<T extends BaseApi>(apiClass: typeof T): T {     return new apiClass(); } 

And I found that TS did not understand this usage of T. Is there any way to do this?

Answer 1

What you actually want is new() => T, since you intend to use the argument as a constructor and produce a T. Even if you could write typeof T, that wouldn't be what you want, since T might not have a zero-argument constructor.

Remember that the typeof operator takes a value and produces a value. T is already a type; it is not a value.

Naturally, this is addressed in the TypeScript FAQ https://github.com/Microsoft/TypeScript/wiki/FAQ#why-cant-i-write-typeof-t-new-t-or-instanceof-t-in-my-generic-function