Re: Can copy assignment operator be const?

Keith H Duggar wrote:


In retrospect, we could have followed the STL iterator design and
define two separate classes, e.g. Ptr and Const_Ptr. However, the
current design has spread over large codebase by now and I would not
want to change this, unless somebody convinces me that I'm doing
something illegal.

In essence, the top-level const of the smartpointer is recycled to
indicate the constness of the pointed object. After all, the top-level
const is not very much used for value types like pointers (there was a
recent thread about this as well; I think the consensus was that one
has to keep the functions short enough anyway so there is not much need
to use top-level const for local data and function arguments). In order
to bypass language constness checks on top-level const, assignment ops
have to be declared const. So the smartpointer we have something like
the following:

class Base; // class hierarchy of dynamically created objects.

struct PVoid {
    PVoid(): p_(0) {}
    PVoid(const Base* p);
    PVoid(const PVoid& b);
    ~PVoid();
    const PVoid& operator=(const Base* p) const;
    const PVoid& operator=(const PVoid& b) const;
    const Base* operator->() const { return p_; }
    Base* operator->() { return p_; }
    const Base& operator*() const { return *p_;}
    Base& operator*() { return *p_;}
private:
    mutable Base* p_;
};

In the end, I have something like following (simplified, in real code
only derived classes are used both for pointers and entity objects).

PVoid a = new Base();
a->f(); // calls non-const f
const PVoid b;
b = a; // would not work without const assignment.
b->f(); // calls const f

This corresponds to:

Base* a = new Base();
a->f(); // calls non-const f
const Base* b;
b = a;
a->f(); // calls const f

So I prefer to think that the smartpointer class is designed this way
in order to keep semantics *unchanged* for this similar syntax. Yes I
know this is not really the case ;-)

Regards
Paavo

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