Use of .apply() with 'new' operator. Is th

Question

In JavaScript, I want to create an object instance (via the new operator), but pass an arbitrary number of arguments to the constructor. Is this possible?

What I want to do is something like this (but the code below does not work):

function Something(){
    // init stuff
}
function createSomething(){
    return new Something.apply(null, arguments);
}
var s = createSomething(a,b,c); // 's' is an instance of Something

---

The Answer

From the responses here, it became clear that there's no built-in way to call .apply() with the new operator. However, people suggested a number of really interesting solutions to the problem.

My preferred solution was this one from Matthew Crumley (I've modified it to pass the arguments property):

var createSomething = (function() {
    function F(args) {
        return Something.apply(this, args);
    }
    F.prototype = Something.prototype;

    return function() {
        return new F(arguments);
    }
})();

Answer 1

I just came across this problem, and I solved it like this:

function instantiate(ctor) {
    switch (arguments.length) {
        case 1: return new ctor();
        case 2: return new ctor(arguments[1]);
        case 3: return new ctor(arguments[1], arguments[2]);
        case 4: return new ctor(arguments[1], arguments[2], arguments[3]);
        //...
        default: throw new Error('instantiate: too many parameters');
    }
}

function Thing(a, b, c) {
    console.log(a);
    console.log(b);
    console.log(c);
}

var thing = instantiate(Thing, 'abc', 123, {x:5});

Yeah, it's a bit ugly, but it solves the problem, and it's dead simple.

Answer 2

Solution without ES6 or polyfills:

var obj = _new(Demo).apply(["X", "Y", "Z"]);


function _new(constr)
{
    function createNamedFunction(name)
    {
        return (new Function("return function " + name + "() { };"))();
    }

    var func = createNamedFunction(constr.name);
    func.prototype = constr.prototype;
    var self = new func();

    return { apply: function(args) {
        constr.apply(self, args);
        return self;
    } };
}

function Demo()
{
    for(var index in arguments)
    {
        this['arg' + (parseInt(index) + 1)] = arguments[index];
    }
}
Demo.prototype.tagged = true;


console.log(obj);
console.log(obj.tagged);

output

Demo {arg1: "X", arg2: "Y", arg3: "Z"}


... or "shorter" way:

var func = new Function("return function " + Demo.name + "() { };")();
func.prototype = Demo.prototype;
var obj = new func();

Demo.apply(obj, ["X", "Y", "Z"]);

edit:
I think this might be a good solution:

this.forConstructor = function(constr)
{
    return { apply: function(args)
    {
        let name = constr.name.replace('-', '_');

        let func = (new Function('args', name + '_', " return function " + name + "() { " + name + "_.apply(this, args); }"))(args, constr);
        func.constructor = constr;
        func.prototype = constr.prototype;

        return new func(args);
    }};
}

Answer 3

This one-liner should do it:

new (Function.prototype.bind.apply(Something, [null].concat(arguments)));

Answer 4

Yes we can, javascript is more of prototype inheritance in nature.

function Actor(name, age){
  this.name = name;
  this.age = age;
}

Actor.prototype.name = "unknown";
Actor.prototype.age = "unknown";

Actor.prototype.getName = function() {
    return this.name;
};

Actor.prototype.getAge = function() {
    return this.age;
};

when we create an object with "new" then our created object INHERITS getAge(), But if we used apply(...) or call(...) to call Actor, then we are passing an object for "this" but the object we pass WON'T inherit from Actor.prototype

unless, we directly pass apply or call Actor.prototype but then.... "this" would point to "Actor.prototype" and this.name would write to: Actor.prototype.name. Thus affecting all other objects created with Actor...since we overwrite the prototype rather than the instance

var rajini = new Actor('Rajinikanth', 31);
console.log(rajini);
console.log(rajini.getName());
console.log(rajini.getAge());

var kamal = new Actor('kamal', 18);
console.log(kamal);
console.log(kamal.getName());
console.log(kamal.getAge());

Let's try with apply

var vijay = Actor.apply(null, ["pandaram", 33]);
if (vijay === undefined) {
    console.log("Actor(....) didn't return anything 
           since we didn't call it with new");
}

var ajith = {};
Actor.apply(ajith, ['ajith', 25]);
console.log(ajith); //Object {name: "ajith", age: 25}
try {
    ajith.getName();
} catch (E) {
    console.log("Error since we didn't inherit ajith.prototype");
}
console.log(Actor.prototype.age); //Unknown
console.log(Actor.prototype.name); //Unknown

By passing Actor.prototype to Actor.call() as the first argument, when the Actor() function is ran, it executes this.name=name, Since "this" will point to Actor.prototype, this.name=name; means Actor.prototype.name=name;

var simbhu = Actor.apply(Actor.prototype, ['simbhu', 28]);
if (simbhu === undefined) {
    console.log("Still undefined since the function didn't return anything.");
}
console.log(Actor.prototype.age); //simbhu
console.log(Actor.prototype.name); //28

var copy = Actor.prototype;
var dhanush = Actor.apply(copy, ["dhanush", 11]);
console.log(dhanush);
console.log("But now we've corrupted Parent.prototype in order to inherit");
console.log(Actor.prototype.age); //11
console.log(Actor.prototype.name); //dhanush

Coming back to orginal question how to use new operator with apply, here is my take....

Function.prototype.new = function(){
    var constructor = this;
    function fn() {return constructor.apply(this, args)}
    var args = Array.prototype.slice.call(arguments);
    fn.prototype = this.prototype;
    return new fn
};

var thalaivar = Actor.new.apply(Parent, ["Thalaivar", 30]);
console.log(thalaivar);

Answer 5

modified @Matthew answer. Here I can pass any number of parameters to function as usual (not array). Also 'Something' is not hardcoded into:

function createObject( constr ) {   
  var args =  arguments;
  var wrapper =  function() {  
    return constr.apply( this, Array.prototype.slice.call(args, 1) );
  }

  wrapper.prototype =  constr.prototype;
  return  new wrapper();
}


function Something() {
    // init stuff
};

var obj1 =     createObject( Something, 1, 2, 3 );
var same =     new Something( 1, 2, 3 );

Answer 6

Here's a generalized solution that can call any constructor (except native constructors that behave differently when called as functions, like String, Number, Date, etc.) with an array of arguments:

function construct(constructor, args) {
    function F() {
        return constructor.apply(this, args);
    }
    F.prototype = constructor.prototype;
    return new F();
}

An object created by calling construct(Class, [1, 2, 3]) would be identical to an object created with new Class(1, 2, 3).

You could also make a more specific version so you don't have to pass the constructor every time. This is also slightly more efficient, since it doesn't need to create a new instance of the inner function every time you call it.

var createSomething = (function() {
    function F(args) {
        return Something.apply(this, args);
    }
    F.prototype = Something.prototype;

    return function(args) {
        return new F(args);
    }
})();

The reason for creating and calling the outer anonymous function like that is to keep function F from polluting the global namespace. It's sometimes called the module pattern.

[UPDATE]

For those who want to use this in TypeScript, since TS gives an error if F returns anything:

function construct(constructor, args) {
    function F() : void {
        constructor.apply(this, args);
    }
    F.prototype = constructor.prototype;
    return new F();
}