Can you take a class as an argument and build a type from it in TypeScript?

I have a function that takes a class as an argument, and uses an interface based on that class as a type.

What the function does is not important!. Please don’t make me repost it here (it’s long and 99% of it has nothing whatsoever to do with the question I’m asking, so I’ve made a dummy question here instead).

Everything works great when I take the type as a generic type argument:

class SomeClass{}

interface SomeClassType extends SomeClass {}

// Use them together
doSomething<SomeClassType>(SomeClass, 1234);

// Example doSomething function
const doSomething = <T,> (
  // I found this "type for a class" in another SO post, but I'm
  // not sure it's correct, and it may be the root of my problem
  SomeClass: { new(...args: any[]): any; },
  x: any
) => {
  // Does something with x, SomeClass, and T
}

However, it felt redundant to have to pass T and SomeClass, so I figured, why not pass just the class, and derive its type inside my function? So, I tried to do that:

const doSomething = (
  SomeClass: { new(...args: any[]): any; },
  x: any
) => {
  interface T extends SomeClass {}
  // Does something with x, SomeClass, and T
}

However, that gets me an error:

‘SomeClass’ refers to a value, but is being used as a type here. Did you mean ‘typeof SomeClass’?ts(2749)

My question is simple: in TS, can you pass a class as an argument, and then create a type from it … and if so how?

>Solution :

You still need the generic. You just need to use it in the type of the class – it looks like you want T to be the type of the class instances, which is the return type of the constructor:

const doSomething = <T,> (
  SomeClass: { new(): T; },
//                    ^
  x: unknown
) => {
  // …
}

Then T can be inferred from the class object you’re passing.