I remember encountering this concept before, but can't find it in Google now.
If I have an object of type A, which directly embeds an object of type B:
class A {
B b;
};
How can I have a smart pointer to B
, e. g. boost::shared_ptr<B>
, but use reference count of A
? Assume an instance of A
itself is heap-allocated I can safely g开发者_开发问答et its shared count using, say, enable_shared_from_this
.
D'oh!
Found it right in shared_ptr
documentation. It's called aliasing (see section III of shared_ptr improvements for C++0x).
I just needed to use a different constructor (or a corresponding reset
function overload):
template<class Y> shared_ptr( shared_ptr<Y> const & r, T * p );
Which works like this (you need to construct shared_ptr to parent first):
#include <boost/shared_ptr.hpp>
#include <iostream>
struct A {
A() : i_(13) {}
int i_;
};
struct B {
A a_;
~B() { std::cout << "B deleted" << std::endl; }
};
int
main() {
boost::shared_ptr<A> a;
{
boost::shared_ptr<B> b(new B);
a = boost::shared_ptr<A>(b, &b->a_);
std::cout << "ref count = " << a.use_count() << std::endl;
}
std::cout << "ref count = " << a.use_count() << std::endl;
std::cout << a->i_ << std::endl;
}
I have not tested this, but you should be able to use a custom deallocator object to keep a shared_ptr to the parent around as long as the child is still needed. Something along these lines:
template<typename Parent, typename Child>
class Guard {
private:
boost::shared_ptr<Parent> *parent;
public:
explicit Guard(const boost::shared_ptr<Parent> a_parent) {
// Save one shared_ptr to parent (in this guard object and all it's copies)
// This keeps the parent alive.
parent = new boost::shared_ptr<Parent>(a_parent);
}
void operator()(Child *child) {
// The smart pointer says to "delete" the child, so delete the shared_ptr
// to parent. As far as we are concerned, the parent can die now.
delete parent;
}
};
// ...
boost::shared_ptr<A> par;
boost::shared_ptr<B> ch(&par->b, Guard<A, B>(par));
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