[Javascript] Unexpected Result

[Terry Riegel]

> If you want to do the equivalent of passing around
> references, then you need to pass around a reference type from which
> you read a property, and store new values in a property of that
> object.

This is where you lost me. How would that look in my example?

Thanks,

Terry










On Aug 3, 2010, at 5:14 PM, liorean wrote:

> On 3 August 2010 22:40, Terry Riegel <riegel at clearimageonline.com> wrote:
>> If I have something like this from the console...
>> 
>> # QQ=$.a.LASTLOSER[2]
>> 5
>> 
>> # QQ=2
>> 2
>> 
>> # QQ
>> 2
>> 
>> # $.a.LASTLOSER[2]
>> 5
>> 
>> I would have expected the last line to be 2. Anyone have advise on why this is and how I can avoid this misconception in the future?
> 
> Simple: Javascript *always* passes things by value, not by reference.
> The value may be a reference-type (objects, not primitives), but it's
> the value that is passed around, not the reference.
> 
> So, what's happening is that you're extracting the value from
> $.a.LASTLOSER[2] and store that value in QQ. Then you change the value
> of QQ, read the value of QQ and read the value of $.a.LASTLOSER[2]. At
> no time did you change the value of $.a.LASTLOSER[2], so naturally it
> stays the same. If you want to do the equivalent of passing around
> references, then you need to pass around a reference type from which
> you read a property, and store new values in a property of that
> object.
> -- 
> David "liorean" Andersson
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