JavaScript constructor alternative

Question

Since it is not possible to use new operator with an arguments array (.call or .apply), Then I wondering if I can do it by hand.

In the following code, are obj1 and obj2 equivalent (in ES3) ?

function MyClass(a, b, c) {}

var obj1 = new MyClass(2,3,4)

var tmp = function() {}
tmp.prototype = MyClass.prototype;
var obj2 = new tmp;
obj2.constructor = MyClass;
MyClass.call(obj2, 2,3,4);

- edit1 -

Does your answers mean that the above code is wrong or is not equivalent to an object construction ?

there is no ES4 you want ES3. ES4 is abandoned.

Raynos
– Raynos

2011-04-06 17:10:11 +00:00
Commented Apr 6, 2011 at 17:10 — Raynos
– Raynos, Commented Apr 6, 2011 at 17:10

KooiInc · Accepted Answer · 2011-04-07 04:57:17Z

1

Using the following syntax, you should be able to apply/call a constructor directly without too much fuzz. Or instantiate it without using the new operator for that matter.

function MyClass(a,b,c){
  if (!(this instanceof MyClass)){
    return new MyClass(a,b,c)
  }
  // to demonstrate the arguments are usable using call/apply
  this.arg1 = a;
  this.arg2 = b;
  this.arg3 = c;
}

var obj2 = MyClass.call(null,1,2,3); // or MyClass.apply(null,[1,2,3])
alert([obj2.arg1,obj2.arg2,obj2.arg3].join(',')); //=> 1,2,3

edited Apr 7, 2011 at 4:57

answered Apr 6, 2011 at 16:50

KooiInc

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Comments

Alnitak · Accepted Answer · 2011-04-06 16:57:02Z

1

If you're using ES5 you can do this properly with Object.create():

function MyClass(a) {
    console.log('in constructor with param: ' + a);
    this.foo = a;
}

var a = new MyClass('a');

var b = Object.create(MyClass.prototype);
b.constructor.apply(b, ['b']);

console.log('a.foo = ', a.foo, ', b.foo = ', b.foo);
console.log(b instanceof MyClass);

output:

in constructor with param: a
in constructor with param: b
a.foo =  a , b.foo =  b
true

See http://jsfiddle.net/Qjfyt/1/, which shows the constructor only being called once, as expected.

edited Apr 6, 2011 at 16:57

answered Apr 6, 2011 at 16:48

Alnitak

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1 Comment

Franck Freiburger Over a year ago

Right, Object.create is actually the solution for ES5. I have just modified the question to make it clear. Thanks.

Raynos · Accepted Answer · 2011-04-06 17:24:13Z

1

What about this :

var o = new MyClass;
MyClass.apply(o, args);

Your creating an object which inherits from the prototype properly through the new keyword then your calling the constructor again on that object with your parameters.

All this requires is that your constructor is well formed, doesn't crash on null parameters and doesn't have side effects when called more then once on the same object. Of course all these conditions are true because you write proper units, right?

To answer your actual question if you add

tmp.prototype.constructor = MyClass

then yes obj1 and obj2 are "the same".

edited Apr 6, 2011 at 17:24

answered Apr 6, 2011 at 16:44

Raynos

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2 Comments

Franck Freiburger Over a year ago

Thanks for the suggestion, but calling the constructor twice could be a bad idea.

Raynos Over a year ago

@Soubok yes. Your constructor has to be well written, cheap and must not have side effects

afshin · Accepted Answer · 2011-04-06 17:14:57Z

0

Well, you can use apply and get an object that is almost exactly like one instantiated with new like this:

function MyClass(a, b, c) {this.foo = a;}
var a, b = {}, args = [2, 3, 4]; // notice that b is an Object!
a = new MyClass(2,3,4);
MyClass.prototype.constructor.apply(b, [2, 3, 4]);
console.log('a.foo', a.foo, 'b.foo', b.foo);

As you'll see, those are equivalent, but if you check further:

console.log('b.constructor === a.constructor', b.constructor === a.constructor);

They don't have the same constructor. This is because b was originally an instance of "Object" since it was set to an object literal, and it was used as the context of the apply to the MyClass prototype's constructor. They will have the same attributes/methods, so maybe that's good enough for your needs?

EDIT: See @Raynos' comment below. This technique means you lose out on everything defined using the MyClass.prototype.

edited Apr 6, 2011 at 17:14

answered Apr 6, 2011 at 16:22

afshin

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1 Comment

Raynos Over a year ago

your b object has none of the methods/properties defined on the prototype

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