1

I have a basic understanding of instanceof in JavaScript, testing if the left hand side object "is of" the right hand side object type. The following 2 examples help me understand that...

var demo1 = function() {};
demo1.prototype = {
foo: "hello"
};

var demo2 = function() {
var pub = {
    bar:"world"
};
return this.pub;
};

var obj1 = new demo1();
var obj2 = new demo2();

console.log(obj1 instanceof demo1);  //returns true
console.log(obj2 instanceof demo2);  //returns true

But on this 3rd example, I get false and I don't understand why....

var o = {}; // new Object;
o.toString(); // [object Object]
console.log(o instanceof toString); //returns false

Thanks for any help in understanding whats going on. Also...is is possible to make the 3rd example true?

Thanks again

klewis
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  • Your `demo2` function is screwed up. There is no `this.pub`, so it will return `undefined`, and when called with the `new` operator this will result in an empty object (beeing an `instanceof demo2` at least) – Bergi Oct 15 '12 at 14:08
  • It returns true for me, I'm using an example based off of T.J. Crawford's blog here - http://blog.niftysnippets.org/2010/03/anonymouses-anonymous.html – klewis Oct 15 '12 at 14:12
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    Yes, `obj2 instanceof demo2` is `true` for me, too. Still, your constructor is useless and could be replaced by an empty function. I can't see how yours came from any example in that very good article. – Bergi Oct 15 '12 at 14:18
  • Bergi I didn't post example 2 as a best practice for patterns, I know that. It's based on a response from a blog comment on that article. That's all. I plan on using standard patterns to achieve my goals. Thanks for the heavy critique on staying away from the style mentioned above. And yes its a great article. – klewis Oct 15 '12 at 14:23

3 Answers3

4

toString does not cause a type change of o. It just returns a string representation of the object, without altering it. So, o is still a simple object and no instanceof String.

var o = {}; // new Object object
var ostring = o.toString(); // "[object Object]"
typeof o; // object
typeof ostring; // string - this is a primitive value, not an object; so that
ostring instanceof String; // is false
var stringobj = new String(ostring); // new String object
typeof stringobj; // object
stringobj instanceof String; // true!
o instanceof Object; // also true!

See also MDN reference for the String constructor.

Bergi
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1

With an object literal like

var o = {}; // new Object;

you create an object whose prototype is Object. Testing with instanceof will not yield any useful information. The only comparison that will yield true is 

o instanceof Object
Ingo Kegel
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1

o is an object; toString is a function. They are different types in JavaScript.

alert(typeof(o));  //"object"
alert(typeof(toString)); //"function"

JavaScript makes a distinction between objects and functions. Therefore, that's why you get false returned in your example.

David Hoerster
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  • obj2 is a object and demo2() is a function. I just tested it within my web console, and yet I get a "return true" with them. So there has to be more than what you are saying... – klewis Oct 15 '12 at 14:16
  • Check out this SO question - http://stackoverflow.com/questions/2449254/what-is-the-instanceof-operator-in-javascript . In the accepted answer, a variable on the LHS returns `true` if the object inherit's from the RHS's prototype. That's what you're doing in your `obj2 = new demo2()` statement. So `obj2` inherits from `demo2` prototype. Therefore, `obj2` is an instance of `demo2`. `o` and `toString` aren't related by a common prototype, so the `instanceof` check returns `false`. I hope that helps. – David Hoerster Oct 15 '12 at 14:23