Creating a versatile mixin with a generic type - the next step

I'm currently trying to define a generic type for my mixin method that utilizes "extending class creating functions". However, I'm struggling with finding the correct syntax to implement it.

After researching, I came across an implementation in this answer

To illustrate, here's a brief and straightforward example:

// Defining the type for "extending class creating functions"
type Mixable<A extends any[], R> = (
  ctor: new (...args: any[]) => any
) => abstract new (...args: A) => R;

type MixinFunction = {
  <A1 extends any[], R1>(ctor1: Mixable<A1, R1>): new (...args: any) => R1;
  <A1 extends any[], R1, A2 extends any[], R2>
  (ctor1: Mixable<A1, R1>, ctor2: Mixable<A2, R2>): new (...args: any) => R1 & R2;
}

const Mixin: MixinFunction = ((_a: Mixable<any[], any>) => {
  // Example content
  return class {} as (new (...args: any[]) => any);
});

const Base = (superclass: Newable) => class _Base extends superclass {
  foo():number {
    return 1;
  }
};

const Base2 = <T>(superclass: Newable) => class _Base2 extends superclass {
  bar():T {
    return 1 as T;
  }
};

// Attempting to create a class using mixins results in ts(2562) error
class Derived<T> extends Mixin(Base, Base2) {
}

class DerivedString extends Derived<string> {}

(new Derived()).foo();          // Inferred as "number"
(new DerivedString()).bar();    // Desired return type inference is "string"

Answer №1

Regrettably, TypeScript is missing essential features required for you to create this type of intricate composed generic mixin. The capability for higher order type inference from generic functions only functions in very specific scenarios, and utilizing Base2 (a generic function that yields a nongeneric class) as a Mixable (a generic type that yields a nongeneric function that yields a nongeneric class) does not fall into those circumstances.

To enable support for this kind of functionality, TypeScript would likely need, at the least, a relaxation of the restriction on accessing type parameters in base class expressions, as mentioned in microsoft/TypeScript#55972. However, in general, TypeScript struggles to express the higher-kinded types needed to convey the intended operation of Mixin when applied to generic inputs. There is a feature request outlined in microsoft/TypeScript#1213 for this, although it remains unclear whether an implementation of that would automatically resolve your issue.

At present, these capabilities exceed TypeScript's current abilities.


Instead of attempting to coerce TypeScript into tasks beyond its reach, the best approach may be to determine the type you anticipate Mixin(Base, Base2<T>) producing, and then simply assert that the outcome aligns with that type. It may not be ideal, but it is viable:

const Mixed = Mixin(Base, Base2) as new <T>(...args: any) =>
    InstanceType<ReturnType<typeof Base>> &
    InstanceType<ReturnType<typeof Base2<T>>>;
class Derived<T> extends Mixed<T> { }

class DerivedString extends Derived<string> { }
(new Derived()).foo(); // number
(new DerivedString()).bar(); // string

This solution is complex because Base and Base2 have obscured their actual classes within the function body scope, thus lacking an existing name for their instance types. Utilizing the ReturnType and the InstanceType utility types helps address this. The generics are managed by making Mixed a generic class constructor (noted by <T> in

new <T>(...args: any) => ⋯
) and leveraging an instantiation expression to specify T as the type argument for typeof Base2.

Despite its intricacy, this method proves effective; Mixed is recognized as a generic class, enabling you to define

class Derived<T> extends Mixed<T> {}
and consequently assigning a return type of string to bar() in DerivedString, as anticipated.

Playground link to code

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