CODEWITHSUNDEEP
typescripttypescript-typingstypescript-genericsstatic-typingtypescript-class
Stefan Parker
Stefan Parker

Reputation: 31

Typescript: define a class type from the return of a function (and use it as a type)

Is there a way to define a class from the return of a function, and then use that as a property type in an interface?

class Foo {};
function makeFoo<T extends Foo>(classType: T): T {
  return classType;
}
const Foo2 = makeFoo(Foo);
new Foo(); // OK
new Foo2(); // OK
interface IFoo {
  foo1: Foo; // OK
  foo2: Foo2; // 'Foo2' refers to a value, but is being used as a type here.
}

The real world use case is I have a function that adds some static methods to classes using the class's generic types. So,

class DataModel {}
class Foo<T extends DataModel> {}
class Bar extends Foo<DataModel> {
  getInstance() {
    return 'I am an instance';
  }
}

function addStatics<T extends DataModel, TF extends Foo<T>>(
  classType: TF,
): TF & { getStatic: () => T } {
  return Object.defineProperty(classType, 'getStatic', {
    value: () => {},
  });
}

// This is: typeof Bar & { getStatic: () => typeof DataModel; }
const BarWithStatics = addStatics(Bar);
console.log(BarWithStatics.getStatic(), new BarWithStatics().getInstance());

In reality it's doing some things with a singleton instance, but yeah, this is the gist of it. I was hoping there was some way to do class BarWithStatics = addStatics(Bar) but no such luck.

Upvotes: 1

Views: 210

Answers (1)

Stefan Parker
Stefan Parker

Reputation: 31

Of course as soon as I ask this I figure out the solution. You can extend the return of a function:

class BarWithStatics extends addStatics(Bar) {}
interface IFoo {
  bar: BarWithStatics;
}

Upvotes: 1

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