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2012-04-13T13:45:47.870
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On your comments: <pre><code>> //Object.create equals an anonymous function that accepts one parameter, 'o'. </code></pre> Better to say that a function is assigned to the <code>create</code> property of <code>Object</code>. All functions can be considered anonymous, just that some are assigned to named properties or variables, others aren't. <pre><code>> //Create a new function called 'F' which is just an empty object. </code></pre> Declare a function. Yes, it's an object too. <pre><code>> F.prototype = o; > //the prototype of the 'F' function should point to the > //parameter of the anonymous function. </code></pre> A reference to <code>o</code> is assigned to <code>F.prototype</code> is a bit shorter to write. <pre><code>> //create a new constructor function based off of the 'F' function. </code></pre> No, should be "Return an instance of F". So the returned object has an internal <code>[[Prototype]]</code> that references the object passed to the function. The messy part is that a useless F function had to be created to perform the trick, and the returned object's constructor will not have a useful value as it references the empty <code>F</code>. Not that the constructor property is very reliable or particularly useful normally anyway. <blockquote> This way, the prototype of the 'stooge' object can't be overwritten when we augment stuff to 'another_stooge'. No need to reset the 'stooge' prototype using 'constructor'. </blockquote> That's a strange statement. *another_stooge* has <code>stooge</code> as it's private <code>[[Prototype]]</code>, it does not inherit from <code>stooge.prototype</code> but from <code>stooge.[[Prototype]]</code>. If you want <code>another_stooge</code> to inherit from <code>stooge.prototype</code>, use <code>Object.create(stooge.prototype)</code> or <code>Object.create(new stooge())</code>, the former is probably more suitable. I hope that all makes sense.
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1. POUnderstanding Crockford's Object.create shim
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1. USRobG
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1. COHi RobG. Thanks for responding! I have to reread to full understand the syntax but I certainly know more than I did an hour ago. Thanks again!
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