Learning Javascript bind method explanation

Question

I'm using Ruby at daily basis, but Javascript is nowadays everywhere, and I need to learn this language too.

I began with "Learning JavaScript Design Patterns", now I read the 6th edition of "JavaScript: The Definitive Guide".

I'm reading some blogs too.

I found a code snippet, which I totally not understood:

if (!Function.prototype.bind) {
  Function.prototype.bind = function(obj) {
    var slice = [].slice,
        args  = slice.call(arguments, 1),
        self  = this,
        nop   = function () {},
        bound = function () {
          return self.apply(this instanceof nop ? this : (obj || {}), args.concat(slice.call(arguments)));   
        };
    nop.prototype   = self.prototype;
    bound.prototype = new nop();
    return bound;
  };
}

I know it checks if a function with the name bind is defined, and when not, define it.

But why it is checked on the prototype of the Function object ?

Why it is not a simple check of:

if(typeof bind != 'function')

Because that's the purpose of that script, to polyfill (add missing functionality) on browsers that dont' support `Function.prototype.bind` — elclanrs, Nov 10 '13 at 21:34
Bind can be used to set the value of `this` no matter what the invoking object is. You can use closures for this as well. The value of `this` is explained here: http://stackoverflow.com/a/16063711/1641941 — HMR, Nov 11 '13 at 01:52

score 2 · Accepted Answer · answered Nov 10 '13 at 21:34

2

Because bind is a class method on Objects of type Function, not a global function provided by Javascript.

answered Nov 10 '13 at 21:34

musically_ut

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Thanks, that was the kind of explanation I was seeking. – astropanic Nov 10 '13 at 21:42
Glad to have been of help. :) – musically_ut Nov 10 '13 at 21:52

Learning Javascript bind method explanation

1 Answers1