Using a static class reference as a parameter in a generic type leads to a error

Before I submit a ticket on github, I want to double-check that I'm not making any mistakes. The issue should be clear enough:

class A {}

class B {
  static A = A;
}

function foo<T>(arg: T) {}

// this is valid
const b = new B.A;

// "B" only refers to a type, but it's being used as a namespace.
foo<B.A>(1);

Am I correct in assuming that I can reference a type like this?

While the instantiation deduces the types correctly as:

b = A
B.A = typeof A
B = B

It gives this inference for the last line:

B = any

As a result, VS Code can navigate to the definition for the first one, but struggles to find references for the second.

This appears quite strange to me.

UPDATE: The initial response I received turned out to be incorrect, as this syntax results in a typeof of a typeof:

// arguments of type "A" cannot be assigned to parameters of type "typeof A", property "prototype" is missing.
foo<(typeof B)['A']>(b);

Even if I specify an explicit constructor type, things only get more complicated:

// this will make things worse
class B {
  static A: { new (): A } = A;
}

Answer №1

Is it possible to reference a type in this manner?

In TypeScript, no.

For instance:

class A {}

class B {
  static A = A;
}

You can only utilize B.A as a value and not as a type. It might be misleading since classes in TypeScript can function both as values and types, but this is exclusive to classes, with other entities behaving differently.

As for B.A, it serves as a value where the actual value represents class A, and the type of this value would be typeof A (using a class in a type position results in typeof A, utilized for types and not values).

If you intend on using the type of B.A as a generic type parameter, you can do so by:

a<(typeof B)['A']>(A);

Explanation: A exists within the static aspect of the B class. You reference the static aspect of a class via typeof B, and you refer to a member type using [], which signifies an indexed access type operator.

The function a mentioned earlier accepts a constructor of class A. Should you desire for function a to accept an instance of A, the declaration of B.A should be altered as follows:

class A {}

class B {
  static A: A;
}

This version will compile successfully.

a<(typeof B)['A']>(new A());

UPDATE

If initializing B.A to A's constructor while retaining access to the instance side of A is your goal, you must wait until TypeScript 2.8. Once available, there's a feature that allows extracting the instance type from the constructor type - the following code will compile after running npm i typescript@next:

class A {}

class B {
  static A = A;
}

function a<T>(a: T) { }

const b = new B.A;

type Constructor<T> = { new(...args: any[]): T };

type Instance<C> = C extends Constructor < infer T > ? T : never;

a<Instance<(typeof B)['A']>>(b);

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