How to release pointer from boost::shared_ptr?
Don't. Boost's FAQ entry:
Q. Why doesn't shared_ptr provide a release() function?
A. shared_ptr cannot give away ownership unless it's unique() because the other copy will still destroy the object.
Consider:
shared_ptr<int> a(new int); shared_ptr<int> b(a); // a.use_count() == b.use_count() == 2 int * p = a.release(); // Who owns p now? b will still call delete on it in its destructor.
Furthermore, the pointer returned by release() would be difficult to deallocate reliably, as the source shared_ptr could have been created with a custom deleter.
So, this would be safe in case it's the only shared_ptr instance pointing to your object (when unique() returns true) and the object doesn't require a special deleter. I'd still question your design, if you used such a .release() function.
You could use fake deleter. Then pointers will not be deleted actually.
struct NullDeleter {template<typename T> void operator()(T*) {} };
// pp of type some_t defined somewhere
boost::shared_ptr<some_t> x(pp, NullDeleter() );
Kids, don't do this at home:
// set smarty to point to nothing
// returns old(smarty.get())
// caller is responsible for the returned pointer (careful)
template <typename T>
T* release (shared_ptr<T>& smarty) {
// sanity check:
assert (smarty.unique());
// only one owner (please don't play games with weak_ptr in another thread)
// would want to check the total count (shared+weak) here
// save the pointer:
T *raw = &*smarty;
// at this point smarty owns raw, can't return it
try {
// an exception here would be quite unpleasant
// now smash smarty:
new (&smarty) shared_ptr<T> ();
// REALLY: don't do it!
// the behaviour is not defined!
// in practice: at least a memory leak!
} catch (...) {
// there is no shared_ptr<T> in smarty zombie now
// can't fix it at this point:
// the only fix would be to retry, and it would probably throw again
// sorry, can't do anything
abort ();
}
// smarty is a fresh shared_ptr<T> that doesn't own raw
// at this point, nobody owns raw, can return it
return raw;
}
Now, is there a way to check if total count of owners for the ref count is > 1?