Typescript mixin function with optional superClass

Question

I'm trying to build a mixin function that optionally expects a super class. The rationale is that we often just build intermediate classes to start off with our mixins. I am quite confident with the following declarations, but they don't work, however:

interface Test {
  readonly __TEST: "test";
  new (...args: any[]): {
    readonly test: "test";
  };
}

function TestMixin<SuperClass extends new (...args: any[]) => any>(
  superClass?: SuperClass
) {
  const defaultClass = class {};
  /* Error: Type 'new (...args: any[]) => any' is not assignable to type 'SuperClass extends undefined ? typeof defaultClass : undefined'.ts(2322) */
  const sc: typeof superClass extends undefined ? typeof defaultClass : undefined = superClass === undefined ? defaultClass : superClass;

  /* Error: Type 'SuperClass extends undefined ? typeof defaultClass : undefined' is not a constructor function type. */
  class T extends sc implements InstanceType<Test> {
    public static readonly __TEST = "test";
    public readonly test = "test";
  }

  return T;
}

You can't extend typeof defaultClass Nothing extends typeof — AlexOwl
– AlexOwl, Commented Mar 12, 2019 at 12:34

Titian Cernicova-Dragomir · Accepted Answer · 2019-03-12 15:10:05Z

1

You can't extend a conditional type, Typescripts expects the clause in the extends clause to be a constructor, not any other more complicated type.

I think the simplest solution in this case would be to lie to the compiler with a type assertion:

interface Test {
    readonly __TEST: "test";
    new(...args: any[]): {
        readonly test: "test";
    };
}

function TestMixin<SuperClass extends new (...args: any[]) => any = new () => {}>(
    superClass?: SuperClass
) {
    const defaultClass = class { };
    /* ok */
    const sc = (superClass === undefined ? defaultClass : superClass) as SuperClass;

    /* Ok now */
    class T extends sc implements InstanceType<Test> {
        public static readonly __TEST = "test";
        public readonly test = "test";
    }

    return T;
}

let a = TestMixin();
new a().test;

let b = TestMixin(class {
    constructor(n: number) { console.log("Hi") }
    m() { }
});
new b(1).test;
new b(1).m();

As long as nobody specifies an explicit type parameter to TestMixin and omits the parameter, it should work fine.

edited Mar 12, 2019 at 15:10

answered Mar 12, 2019 at 13:18

Titian Cernicova-Dragomir

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1 Comment

Titian Cernicova-Dragomir Over a year ago

@FlorianBachmann glad to hear, upvotes and mark as answered are also appreciated ;)

Karol Majewski · Accepted Answer · 2019-03-12 13:55:10Z

0

Find a common denominator — the Constructor type.

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

interface Test {
    readonly __TEST: "test";
    new(...args: any[]): {
        readonly test: "test";
    };
}

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

function TestMixin<SuperClass extends new (...args: any[]) => any>(
    superClass?: SuperClass
) {
    const defaultClass: Constructor = superClass || class { };

    return class T extends defaultClass implements InstanceType<Test> {
        public static readonly __TEST = "test";
        public readonly test = "test";
    }
}

answered Mar 12, 2019 at 13:55

Karol Majewski

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3 Comments

Titian Cernicova-Dragomir Over a year ago

Minor problem, this looses constructor arguments of the passed in class. Not that you can't pass them in they are just not checked

Karol Majewski Over a year ago

You, sir, are correct! I got too focused on getting rid of compiler errors and missed it.

Florian Bachmann Over a year ago

The result of this is no instance of superClass anymore. We'd loose all its member.

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