Re: [boost] Patch for smart pointers assignation of plain pointers

From: me22 (me22.ca_at_[hidden])
Date: 2006-01-31 11:06:51

On 31/01/06, Pierre THIERRY <nowhere.man_at_[hidden]> wrote:
> +    shared_ptr & operator=(T * r)
> +    {
> +        this_type(r).swap(*this);
> +        return *this;
> +    }
> +

That results in implicitly creating a shared_ptr from a raw pointer,
which as I understand it is explicitly disallowed because it means
that simple code such as the following will fail:
int *p = new int;
shared_ptr<int> sp;
sp = p;
sp = p;

Since it would be the same as:
int *p = new int;
shared_ptr<int> sp;
sp.reset(p);
sp.reset(); // which has the same effect as delete p, in this case
sp.reset(p); // sp is now holding a pointer to unallocated memory

My understanding is that the operator=(T*) is purposefully not there
to make this more explicit.  It should help in migrating to shared_ptr
as well, since if sp was previously an int* that was manually deleted
at the end of the scope, the code would have been fine.

- Scott McMurray