Typescript - type to represent any class?

Question

What type should I use in typescript to represent any class?

I'm trying to write a function that takes an array of classes and returns an array with different order.

function shuffle(classes: typeof Object[]) : typeof Object[] {
    return ...;
}

class A { }
class B extends A { }
class C extends B { }
class D extends B { }
shuffle([A, B, C, D]);

Argument of type 'typeof A[]' is not assignable to parameter of type 'ObjectConstructor[]'.

Then I've tried:

shuffle([typeof A, typeof B, typeof C, typeof D]);

error TS2345: Argument of type 'string[]' is not assignable to parameter of type 'ObjectConstructor[]'. Type 'string' is not assignable to type 'ObjectConstructor'.

What's the right way? Generics? How? This doesn't work:

export function <T extends typeof Object> shuffle(classes: T[]) : T[]

This neither:

export function <T extends Object> sortClassesBySpeciality(classes: typeof T[]) : typeof T[]

~~Also why typeof (typeof A) is "string" and "" + typeof A is function?~~ Ok, got this, typeof has two very different meanings context of type definition and expression.

(The ultimate goal is to sort the classes by level of extends from Object.)

Here's the result - sortClassesBySpeciality(classes: Array<AnyClass>) : Array<AnyClass> — Ondra Žižka
– Ondra Žižka, Commented Nov 4, 2016 at 14:01

Nitzan Tomer · Accepted Answer · 2016-10-11 12:34:28Z

12

You should avoid using the type Object in typescript, you better use any as the docs say:

You might expect Object to play a similar role, as it does in other languages. But variables of type Object only allow you to assign any value to them - you can’t call arbitrary methods on them, even ones that actually exist

But if you want to represent classes then you need to have the following form:

{ new (): CLASS_TYPE }

Or in your case:

function shuffle(classes: Array<{ new (): any }>): Array<{ new (): any }> {
    return [];
}

class A { }
class B extends A { }
class C extends B { }
class D extends B { }
shuffle([A, B, C, D]);

(code in playground)

If all of your classes are based on a super class (as your example implies) then you can simply do:

function shuffle(classes: Array<{ new (): A }>): Array<{ new (): A }> {
    return [];
}

Edit

Just saw that you want to

sort the classes by level of extends from Object

To answer that:

function shuffle(classes: Array<{ new (): any }>): Array<{ new (): any }> {
    return classes.sort((a, b) => getInheritanceLevel(a) - getInheritanceLevel(b));
}

function getInheritanceLevel(cls: { new (): any }): number {
    let level = 0;

    while (Object.getPrototypeOf(cls.prototype) !== Object.prototype) {
        level++;
        cls = Object.getPrototypeOf(cls.prototype).constructor;
    }

    return level;
}

shuffle([D, A, C, B]); // returns [A, B, D, C]

(code in playground)

edited Oct 11, 2016 at 12:34

answered Oct 11, 2016 at 11:48

Nitzan Tomer

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5 Comments

Ondra Žižka Over a year ago

Am I right that the type syntax means "type(s) whose constructor returns A"? So there's no direct support like Class<T extends A> like in Java? How would this work with interfaces?

Nitzan Tomer Over a year ago

It's more like a "type who has a constructor for A", you can also define static class functions there. I don't think that there are plans to make it more like the java way, but what are you missing from the typescript way? It suits what you're after.

Ondra Žižka Over a year ago

Also, it could also be { new (): any }[], is that equivalent?

Ondra Žižka Over a year ago

I've introduced an alias: type AnyClass = { new (): any };

Nitzan Tomer Over a year ago

Yeah, should be fine. The convention is to call those constructors, so type AnyConstructor, just like the ArrayConstructor

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