Conversion from boost::shared_ptr to std::shared_ptr?

Question

I got a library that internally uses Boost's version of shared_ptr and exposes only those. For my application, I'd like to use std::shared_ptr whenever possible though. Sadly, there is no direct conversion between the two types, as the ref counting stuff is implementation dependent.

Is there any way to have both a boost::shared_ptr and a std::shared_ptr share the same ref-count-object? Or at least steal the ref-count from the Boost version and only let the stdlib version take care of it?

I do not see reasons to mix memory management from different library. There might be internal objects link to the boost:shared_ptr and if you take over the ref-count, you will need to make sure the update is reflected on your std::shared_ptr — YeenFei
– YeenFei, Commented Jun 13, 2011 at 5:56

Fozi · Accepted Answer · 2018-11-15 14:49:18Z

Based on janm's response at first I did this:

template<class T> std::shared_ptr<T> to_std(const boost::shared_ptr<T> &p) {
    return std::shared_ptr<T>(p.get(), [p](...) mutable { p.reset(); });
}

template<class T> boost::shared_ptr<T> to_boost(const std::shared_ptr<T> &p) {
    return boost::shared_ptr<T>(p.get(), [p](...) mutable { p.reset(); });
}

But then I realized I could do this instead:

namespace {
    template<class SharedPointer> struct Holder {
        SharedPointer p;

        Holder(const SharedPointer &p) : p(p) {}
        Holder(const Holder &other) : p(other.p) {}
        Holder(Holder &&other) : p(std::move(other.p)) {}

        void operator () (...) { p.reset(); }
    };
}

template<class T> std::shared_ptr<T> to_std_ptr(const boost::shared_ptr<T> &p) {
    typedef Holder<std::shared_ptr<T>> H;
    if(H *h = boost::get_deleter<H>(p)) {
        return h->p;
    } else {
        return std::shared_ptr<T>(p.get(), Holder<boost::shared_ptr<T>>(p));
    }
}

template<class T> boost::shared_ptr<T> to_boost_ptr(const std::shared_ptr<T> &p){
    typedef Holder<boost::shared_ptr<T>> H;
    if(H * h = std::get_deleter<H>(p)) {
        return h->p;
    } else {
        return boost::shared_ptr<T>(p.get(), Holder<std::shared_ptr<T>>(p));
    }
}

This solution leaves no reason for not using it without restrictions since you get the original pointer back if you convert back to the original type.

janm · Accepted Answer · 2022-03-24 11:59:22Z

Update to an "off the top of my head" answer, almost eleven years later:

As pointed out in this answer, an implementation is allowed to extend the lifetime the deleter beyond the lifetime of the shared_ptr. For example, until after all weak_ptr instances are also destroyed. This would cause a problem where the presence of weak_ptr instances prevent the destruction of the underlying object, obviously a problem.

To avoid this, either use the approach in the answer by @Fozi with an explicit call to reset(), or the aliasing constructor approach in the linked answer.

Original Answer:

You can carry the boost::shared_ptr "inside" the std::shared_ptr by using the destructor to carry the reference around:

template<typename T>
void do_release(typename boost::shared_ptr<T> const&, T*)
{
}

template<typename T>
typename std::shared_ptr<T> to_std(typename boost::shared_ptr<T> const& p)
{
    return
        std::shared_ptr<T>(
                p.get(),
                boost::bind(&do_release<T>, p, _1));

}

The only real reason to do this is if you have a bunch of code that expects std::shared_ptr<T>.

To be honest, at first I didn't exactly understand what you meant with "carry the reference arond", but this is indeed a very neat was, thanks. :)
@Xeo TIL that there is a problem with unspecified lifetime, especially in the presence of weak_ptrs: stackoverflow.com/a/71575543/85371

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