Re: Can copy assignment operator be const?
Paavo Helde wrote:
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 ;-)
Sorry, but mutable members should never form part of the observable
state of an object. If the assignment operator is defined const - it
will not and should not work for non-const objects. And if the object
is const and trying to invoke the assignment is trying to change the
observable state. Which can not be very intuitive. Also, if one wants
to disable the copy assignment then they always have the option of
putting the declaration in the private/protected section.
So, I think what you are trying to achieve is flawed in its
implementation. To propagate const-ness with smart ptrs and objects
pointed to, you may want to consider taking a look at Scott Meyers,
MEC++ Item 28 - Smart pointers (last section - smart pointers and
const). Unless, I mis-understood you. :)
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