How to release pointer from boost::shared_ptr?

Question

Can boost::shared_ptr release the stored pointer without deleting it?

I can see no release function exists in the documentation, also in the FAQ is explained why it

Answer 1

If your pointers are indeed unique do use std::unique_ptr or boost::scoped_ptr if the former is not available for your compiler. Otherwise consider combining the use of boost::shared_ptr with boost::weak_ptr. Check out the Boost documentation for details.

Answer 2

I am using Poco::HTTPRequestHandlerFactory which expects to return a raw HTTPRequestHandler*, the Poco framework deletes the handler once the request finishes.

Also using DI Sauce project to create the controllers, however the Injector returns shared_ptr which I cannot return directly, and returning handler.get() is no good either since the as soon as this function returns the shared_ptr goes out of scope and deletes then handler before its executed, so here is a reasonable (I think) reason to have a .release() method. I ended up creating a HTTPRequestHandlerWrapper class as follows :-

class HTTPRequestHandlerWrapper : public HTTPRequestHandler {
private:
    sauce::shared_ptr<HTTPRequestHandler> _handler;

public:
    HTTPRequestHandlerWrapper(sauce::shared_ptr<HTTPRequestHandler> handler) {
        _handler = handler;
    }

    virtual void handleRequest(HTTPServerRequest& request, HTTPServerResponse& response) {
        return _handler->handleRequest(request, response);
    }
};

and then the factory would

HTTPRequestHandler* HttpHandlerFactory::createRequestHandler(const HTTPServerRequest& request) {
    URI uri = URI(request.getURI());
    auto path = uri.getPath();
    auto method = request.getMethod();

    sauce::shared_ptr<HTTPRequestHandler> handler = _injector->get<HTTPRequestHandler>(method + ":" + path);

    return new HTTPRequestHandlerWrapper(handler);
}

which satisfied both Sauce and Poco and works nicely.

Answer 3

You need to use a deleter that you can request not to delete the underlying pointer.

See this answer (which has been marked as a duplicate of this question) for more information.

Answer 4

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?

Answer 5

You can delete the shared pointer, which seems much the same to me. If pointers are always unique, then std::auto_ptr<> is a good choice. Bear in mind that unique pointers can't be used in STL containers, since operations on them do a lot of copying and temporary duplication.

Answer 6

Here's a hack that might work. I wouldn't recommend it unless you're in a real bind.

template<typename T>
T * release_shared(std::shared_ptr<T> & shared)
{
    static std::vector<std::shared_ptr<T> > graveyard;
    graveyard.push_back(shared);
    shared.reset();
    return graveyard.back().get();
}