Is it possible to create a type that force to return the `typeof` statement in TypeScript?

Issue

For example, I want to create a type that make some like below.

class C1 {}

type TGenericType<T> = typeof T;

let test: TGeneticType<C1>; //=> typeof C1

I don't want to have to pass a typeof explicitly as a generic, like below

let test: TGeneticType<typeof C1>

Solution

There's no way to re-implement or wrap the TypeScript typeof operator as a generic type.

The TypeScript typeof operator takes a value such as a variable or property as an argument, and evaluates to the TypeScript type of that value. If you write typeof C1, then C1 is interpreted as the value named C1, which is the constructor of the C1 class.

But generic type parameters, like the T in the TGenericType<T> type alias, represent types and not values. So typeof T does not make sense because T is itself a type. Hence the compiler error:

type TGenericType<T> = typeof T; // error!
// -------------------------> ~
// 'T' only refers to a type, but is being used as a value here.

---

If you try to fix this, you'll either find yourself writing

type TGenericType<T> = T;
let test: TGenericType<typeof C1> = C1;

which is no better than just writing typeof C1 to start with; or, you'll decide that what you really want is for TGenericType<C1> to produce a constructor type where C1 is the instance, which looks like this:

type TGenericType<T> = new () => T;
let test: TGenericType<C1> = C1; // okay

but that doesn't actually have anything to do with the C1 value, so while you could use this to construct new C1 instances, you would lose track of any static properties:

class C1 { 
    static foo = 1;
    bar = 2;
}
new C1().bar; // okay
new test().bar; // okay
C1.foo; // okay
test.foo // error

It depends on your use case which of these you want to pursue, or whether you want to give up entirely.

Playground link to code

Answered By - jcalz