Re: Rewriting clone() to return a boost::shared_ptr ?

helix wrote:


Given the code you posted I think everything should work as you expect.
On the other hand, as others pointed out it is customary to exploit
covariant return types when implementing cloning mechanisms, so that
clone() returns objects with the same type as the static type of the
object they were called from. Given a hyerarchy such as

struct A {
// ...
};

struct B : A {
// ...
};

You expect the following to work:

A a;
A * pa = a.clone();
B b;
B * pb = b.clone();

For this to be possible the clone member functions have to be declared
as follows:

struct A {
  virtual A * clone() const;
};

struct B : A {
  virtual B * clone() const;
};

C++ guarantees that B::clone() const is an override of A::clone()
const, as long as the return types are either pointers or references to
classes whose base/derived relationship goes in the same direction as
that of the classes of which these functions are members.

This doesn't work for smart pointers, but the following appears to me
as a reasonable workaround. I made the original clone member functions
private and defined a clone function template that wraps the pointers
to the cloned objects in shared_ptr's:

#include <boost/shared_ptr.hpp>

template <typename T> boost::shared_ptr<T> clone(const T & t) {
    return boost::shared_ptr<T>(t.clone());
}

class A {
    public:
        virtual ~A() {}

    private:
        friend boost::shared_ptr<A> clone(const A & t);
        virtual A * clone() const { return new A(*this); }
};

class B : public A {
    public:
        virtual ~B() {}

    private:
        friend boost::shared_ptr<B> clone(const B & t);
        virtual B * clone() const { return new B(*this); }
};

int main() {
    A a;
    boost::shared_ptr<A> spa(clone(a));
    B b;
    boost::shared_ptr<B> spb(clone(b));
    spa = clone(b);
    A * pa = &b;
    spa = clone(*pa);
// spb = clone(*pa); Error!
    spb = boost::dynamic_pointer_cast<B>(clone(*pa));
}

Cheers,
Nicola Musatti

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