CODEWITHSUNDEEP
typescript
Kyle Fahringer
Kyle Fahringer

Reputation: 294

Type checking class (not instance of class) extends another class

Given the class below:

class BaseClass {}

I want to create another class which extends this class, i.e.:

class ExtendedClass extends BaseClass {}

I then want to pass the ExtendedClass as an argument to another function (not an instance of an ExtendedClass, but the class itself!), and type check this by ensuring the ExtendedClass extends BaseClass.

This doesn't work:

function helloClass (c: BaseClass) {}

because c: BaseClass denotes that c should be an instance of a BaseClass (or something which extends a BaseClass) - but not a class which extends BaseClass itself.

Is this at all possible within TypeScript?

Upvotes: 9

Views: 4078

Answers (1)

Nitzan Tomer
Nitzan Tomer

Reputation: 164367

You can represent classes like this: { new(): BaseClass }.

Or in an example:

interface BaseClassConstructor<T extends BaseClass> {
    new (): T;
}

class BaseClass {
    str: string;
}

class ExtendedClass extends BaseClass { }

function fn<T extends BaseClass>(ctor: BaseClassConstructor<T>) {

}

fn(BaseClass); // fine
fn(ExtendedClass); // fine

class A {
    prop: any;
}

fn(A);  // error

(code in playground)

Upvotes: 11

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