If you assign to members or add members to the prototype, you will still have
an existing f.constructor.
In the code below, you'll now also have
Foo.prototype.function_name.prototype.constructor as well.
On Jun 4, 2012, at 9:08 PM, dhruvbird wrote:
> ah! Thanks! I didn't realize that. So the correct way to do it would
> be to say:
>
> Foo.prototype.function_name = function foo() { /* blah */ };
>
> eh?
>
> Regards,
> -Dhruv.
>
> On Jun 3, 5:59 am, Michael Schwartz <[email protected]> wrote:
>> constructor is part of the prototype. You wiped it out by assigning the
>> prototype to {}, which has a constructor of object.
>>
>> On Jun 3, 2012, at 12:00 AM, dhruvbird wrote:
>>
>>
>>
>>
>>
>>
>>
>>> Hello,
>>
>>> I see different output for these 2 snippets of code. The only
>>> difference between them is that a prototype is defined on one of them,
>>> but not on the other.
>>
>>> // Snippet-1:
>>> function Foo() { }
>>> var f = new Foo;
>>> console.log(f.constructor.name); // Prints "Foo"
>>
>>> // Snippet-2:
>>> function Foo() { }
>>> Foo.prototype = { };
>>> var f = new Foo;
>>> console.log(f.constructor.name); // Prints "Object"
>>
>>> Any idea why the behaviour changes by just adding a prototype?
>>
>>> Regards,
>>> -Dhruv.
>>
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>>> v8-users mailing list
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>>> http://groups.google.com/group/v8-users
>
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