Javascript: Why am I able to access Function.prototype.call without mentioning the prototype?

Question

Here is a snippet to explain my question:

+function(str) {
  return str.replace(/^[a-z]|\s[a-z]/g,
            Function.call.bind(String.prototype.toUpperCase));
}('foo bar baz.');  //Returns Foo Bar Baz.

Function.call works, but String.toUpperCase does not. I have to instead write, String.prototype.toUpperCase.

http://blog.slaks.net/2015-06-18/code-snippets-and-in-the-darkness-bind-them/ — SLaks, Mar 09 '16 at 15:42
Pointy's answer below is perfectly right. Note that this is still a roundabout way of accessing `call`, so please don't do this in actual code =p — Kyll, Mar 09 '16 at 15:42
@Kyll Yes, this was just an experiment. I cannot accept the answer yet! :) — s4san, Mar 09 '16 at 15:44

score 8 · Accepted Answer · answered Mar 09 '16 at 15:40

The Function() constructor is itself a function. Therefore, it inherits from the same prototype object as any other function.

Instances of String() inherit from the prototype, but the String() constructor is not an instance of String(). It too is a function.

Javascript: Why am I able to access Function.prototype.call without mentioning the prototype?

1 Answers1