RE: [boost] [smart_ptr] const-correctness as function argument

From: Rozental, Gennadiy (gennadiy.rozental_at_[hidden])
Date: 2002-10-02 14:15:09

> The correct solution, of course, is the following:
> 
> void g(shared_ptr<foo const> p)
> {
>     p->bar();
>     p->baz(3); // Error!
> }
> 
> Now, once again, we are prevented from calling foo::baz(), because
> operator->() returns foo const*.  However, there are a few undesirable
> features of this setup.  First, we can't declare the proper 
> type from our
> typedef foo_ptr.  So we have to refer to the original type, 
> which binds
> us to shared_ptr explicitly (prevents us from swapping it out with a
> different pointer).  

I do not believe that's a problem you can always do

typedef ..ptr<T> T_pointer;
typedef ..ptr<T const> T_const_pointer;

and define 
void g( T_const_pointer )

ptr<T const> should be constructible from ptr<T> and everything will work as
expected.

Now about the class member. Let see how raw pointers behave:
              In const member function     | In non-const member function | 
____________________________________________________
T*           could not change the pointer | could change the pointer 
              could change the pointee      | could change the pointee
____________________________________________________

ptr<T> behave the same

____________________________________________________
T* const  could not change the pointer | could not change the pointer
              could change the pointee      | could change the pointee
____________________________________________________

ptr<T> const behave the same
____________________________________________________
T const*  could not change the pointer | could change the pointer
              could not change the pointee| could not change the pointee
____________________________________________________

ptr<T const> behave the same
____________________________________________________
T const* const
             could not change the pointer  | could not change the pointer
             could not change the pointee | could not change the pointee
____________________________________________________

ptr<T const> const behave the same

So I do not see any reason to blame ptr<T>.

Gennadiy.