JavaScript - Copy Method of Instance

Question

I want a possibility to copy all properties/methods of a class instance:

class A { get prop1() { return 1; } get prop2() { return 2; } doStuff() { return this.prop1 + this.prop2; } } class B extends A { get prop1() { return 5; } }

class AWrapper { constructor(a) { // [1] Copy all methods/propertys of a this.doStuff = () => a.doStuff() + 42; } } const a = new A(); const b = new B(); const wA = new AWrapper(a); const wB = new AWrapper(b); console.log(a.prop1(), wA.prop1(), wB.prop1()); // 1, 1, 5 console.log(a.doStuff(), wA.doStuff()); // 3, 45

I could copy each method/property by hand, but is there a simple command for [1], such that wA has the same signature as a?

You can extend that class (eg: class AWrapper extends A { ... }) — Titus
– Titus, Commented Aug 28, 2019 at 13:30
this? developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/… — Chris Li
– Chris Li, Commented Aug 28, 2019 at 13:30
Well I want to only work with the instance a. I.e. assume a is just an instance of A, not A directly. I've edit the example — random314
– random314, Commented Aug 28, 2019 at 13:33

georg · Accepted Answer · 2019-08-28 13:54:31Z

Usually, Proxy is a tool of choice when working with mixins or decorators:

class A { get prop1() { return 1; } get prop2() { return 2; } doStuff() { return this.prop1 + this.prop2; } } class B extends A { get prop1() { return 5; } } function decorate(target) { let mixin = { doStuff() { return target.doStuff() + 42; } } return new Proxy(target, { get(_, prop) { return (prop in mixin) ? mixin[prop] : target[prop]; } }); } const a = new A(); const b = new B(); const wA = decorate(a) const wB = decorate(b) console.log(a.prop1, wA.prop1, wB.prop1); // 1, 1, 5 console.log(a.doStuff(), wA.doStuff()); // 3, 45

Taki · Accepted Answer · 2019-08-28 13:37:31Z

Use the extends keyword and call the parent's (A) doStuff with : this.doStuff = () => super.doStuff() + 42;

class A { get prop1() { return 1; } get prop2() { return 2; } doStuff() { return this.prop1 + this.prop2; } } class AWrapper extends A { constructor(...args) { super(...args); this.doStuff = () => super.doStuff() + 42; } } const a = new A(); const w = new AWrapper(a); console.log(a.prop1, w.prop1); // 1, 1 console.log(a.doStuff(), w.doStuff()); // 3, 45

Mahendra Pratap · Accepted Answer · 2019-08-28 13:37:32Z

Below should do it.

class A { get prop1() { return 1; } get prop2() { return 2; } doStuff() { return this.prop1 + this.prop2; } } class AWrapper extends A{ constructor(a) { super(a); this.doStuff = () => a.doStuff() + 42; } } const a = new A(); const wA = new AWrapper(a); const wB = new AWrapper(a); console.log(a.prop1, wA.prop1, wB.prop1); // 1, 1, 1 console.log(a.doStuff(), wA.doStuff()); // 3, 45

Jack Bashford · Accepted Answer · 2019-08-28 13:47:48Z

You need to make a BWrapper class too - other than that, extends and super are all you need:

class A { get prop1() { return 1; } get prop2() { return 2; } doStuff() { return this.prop1 + this.prop2; } } class B extends A { get prop1() { return 5; } } class AWrapper extends A { constructor(a) { super(); this.doStuff = () => a.doStuff() + 42; } } class BWrapper extends B { constructor(b) { super(); this.doStuff = () => b.doStuff() + 42; } } const a = new A(); const b = new B(); const wA = new AWrapper(a); const wB = new BWrapper(b); console.log(a.prop1, wA.prop1, wB.prop1); // 1, 1, 5 console.log(a.doStuff(a), wA.doStuff(wA)); // 3, 4

Hm. In my code I have many instances of A. Creating a Wrapper for every one is no option. I just have the concrete instance (a or b) and want to feed it into a function (in here: AWrappers constructur).

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