Question

When I use call() or apply(), I got a problem.

console.log(String.prototype.replace === String.replace);//false

I think String.replace should be equal with String.prototype.replace, because they are the same Object.

However,they are different from each other.

What happens when I run the code below:

var s = "a b c";
String.replace.call(s,'a','A');//return "a" 

Why doesn’t this code throw an error, but return a value?

Answer 1

I think there is a lot of mixed information here. First of all we have to clarify that the String constructor function does not have a replace method.

So, whatever String.replace is in Firefox, it is non-standard and therefore you should stay away from it. A quick test in Chrome shows that String.replace indeed does not exist there.

Unfortunately I cannot tell you where String.replace comes from in Firefox. The documentation does not mention it. But it seems that it is not an inherited property, as String.hasOwnProperty('replace') returns true.

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Now to some points in your question:

I think String.replace should be equal with String.prototype.replace, because they are the same Object.

Obviously they are not. If they were, it would return true. Also: String !== String.prototype.

The replace method which is used by the string instances is String.prototype.replace. So if you want to use call or apply, you them on this method.

Why doesn’t this code throw an error, but return a value?

In order to answer that, we would have to know what that method is doing. Maybe a look at the Firefox or Spidermonkey source provides some information.

If you are confused about how prototype inheritance works, have a look at the MDN JavaScript Guide – Details of the object model and Inheritance revisited.