why () is () and [] is [] work in other way?

[Ian Kelly]

On Thu, Apr 26, 2012 at 3:51 PM, Chris Angelico <rosuav at gmail.com> wrote:
> On Fri, Apr 27, 2012 at 7:39 AM, Ian Kelly <ian.g.kelly at gmail.com> wrote:
>> I'm not sure precisely what you mean by "temporary object", so I am
>> taking it to mean an object that is referenced only by the VM stack
>> (or something equivalent for other implementations).
>>
>> In that case: no, you can't.  Take "f() is g()", where the code
>> objects of f and g are supplied at runtime.  Are the objects returned
>> by either of those expressions "temporary"?  Without being able to do
>> static analysis of the code of f and g, there is no way to know.
>
> The expression itself will have references to all its operands (at
> least conceptually).

Yes, the references on the VM stack.

> If their refcounts are precisely 1, then the
> objects are temporaries and will be disposable as soon as the
> expression's fully evaluated.

You can't check ref counts at parse time.