Re: const shared_ptr<>

From: John Harris (john.harris_at_[hidden])
Date: 2003-04-30 11:33:11

--- In Boost-Users_at_[hidden], Hossein Haeri <powerprogman_at_y...> 
wrote:
> Dear all,
> 
> I'm a newbie in boost!
> 
> I want to know what the following means:
> 
> int a = 5;
> const boost::shared_ptr<int> sp(&a);
> 
> Does it mean that the following line is an error?
> 
> *sp = 7;
> 
> If so, what are the guarantees? Is there any
> documented
> guarantee? (I couldn't find anything in the 
> documentation. Excuseme if I'm careless!)

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.

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.

jh