Using a generic type argument with `typeof T`
typescript
Solution
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
Problem
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?