Re: [Boost-Users] Re: const shared_ptr<>

From: Hossein Haeri (powerprogman_at_[hidden])
Date: 2003-05-01 06:32:05

Dear John,

> I'd say you could call any of the 'const' member
> functions (e.g. 
> operator*(), operator->(), get(),...,etc), but the
> returned type is a 
> non-const pointer.
> 
> Since operator*() is a const member function, it is
> legal on a const 
> shared_ptr<T>.  The '=', then, applies to the
> non-const shared ptr 
> returned, not to the shared_ptr<T> object, itself.

I can't understand! Can you please describe your idea
more? Do you mean that calling a non-const member 
function for shared_ptr<T> will result in acting the
desired operatio on a temporal mutable copy of the 
original const version?

> Note also, that you're supposed to initialized
> shared_ptr<T> with a 
> *dynamically* allocated T (use new), because it will
> call 'delete' on 
> the ptr when the last reference disappears.  So you
> are wrong to 
> initialize it with the address of a stack variable.

Thank you very much because of recalling that to me.
But a subtle point here is that the asserion you made
is not true for all types. If T is not a scaler type,
it has some destructor, so I can call that destructor
explicitely, before the compiler does it, and before
the end of shared_ptr<T>. This way, there would be no
problem had I initiialised the share_ptr<T> with an
object of type T.

Wishes,
--Hossein

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