如何在TypeScript中将函数返回的类用作类型？

Question

I am working on a small library that contains a utility function that generates a Class. This is because the Class has quite a complex signature with a lot of generics. It looks somewhat like this:

function generateClass() {
  class GeneratedClass {
    ...
  }
  return GeneratedClass;
}

The compiler correctly infers the return type of this function. The only problem is, I cannot use the returned value as a type. This only works if I extend the returned GeneratedClass. For a more elaborate example, see this TS Playground snippet .

Is there any way to make the returned class usable as a type? I would rather not require consumers of my library to type class Something extends generateClass() { ... } , because that kind of defeats the purpose of the utility function making it more convenient to create these kinds classes.

Answer 1

You can get the instance type in several ways:

If you have an instance of the class you can use typeof with that variable to type the parameter:

  const FooClass = generateClass();

  const foo = new FooClass(5);
  function processFoo(fooParam: typeof foo) { 

  }

In typescript 2.8 you can use the conditional type InstanceType :

const FooClass = generateClass();

const foo = new FooClass(5);
function processFoo(fooParam: InstanceType<typeof FooClass>) { 

}
processFoo(foo);

Pre 2.8 you can get the instance type without instatiating using a helper function and a dummy variable:

const foo = (function<T>(arg: new (...args: any[])=> T) : T{ return null as any})(FooClass);
function processFoo(fooParam: typeof foo) { 

}
processFoo(foo);

In either case you can store the instance type in a type alias to use in several places:

type GeneratedClass = typeof foo;
type GeneratedClass = InstanceType<typeof FooClass>; // for 2.8