Weak binding using c++17 - c++

Question

I am working on a processing framework where callbacks are registered to events and to ensure that no callback is invoked on an object, which has been deleted, I would like to use weak capture rather than capture by reference. It was no problem to make this work using C++14 and shared_from_this(), but how is this correctly achieved using C++17 and weak_from_this().
The example below prints nothing when C++17 is used. I am using g++ 6.3.0-18
#define CXX17 // When this is defined, nothing is printed
#ifdef CXX17
# include <experimental/memory>
# include <experimental/functional>
template <typename T>
using enable_shared_from_this = std::experimental::enable_shared_from_this<T>;
#else
# include <memory>
# include <functional>
template <typename T>
using enable_shared_from_this = std::enable_shared_from_this<T>;
#endif
#include <thread>
#include <mutex>
#include <condition_variable>
#include <atomic>
#include <iostream>
struct A : enable_shared_from_this<A> {
int a;
A() : a(7) {}
auto getptr() {
#ifdef CXX17
return this->weak_from_this();
#else
auto sptr = shared_from_this();
auto wptr = std::weak_ptr<decltype(sptr)::element_type>(sptr);
sptr.reset(); // Drop strong referencing
return wptr;
#endif
}
};
std::condition_variable condition;
std::mutex mutex;
std::atomic<bool> start0{false};
std::atomic<bool> start1{false};
std::shared_ptr<A> g_a;
static void thread_func0() {
auto w_a = g_a->getptr();
std::unique_lock<std::mutex> lock {mutex};
condition.wait(lock, [&]() {
return start0.load();
});
std::this_thread::sleep_for(std::chrono::microseconds(10));
if (auto t = w_a.lock()) {
std::cout << t->a << std::endl;
}
}
static void thread_func1() {
std::unique_lock<std::mutex> lock {mutex};
condition.wait(lock, [&]() {
return start1.load();
});
std::this_thread::sleep_for(std::chrono::microseconds(10000));
g_a = nullptr;
}
int main() {
g_a = std::make_shared<A>();
std::thread thread0(thread_func0);
std::thread thread1(thread_func1);
start0 = true;
start1 = true;
condition.notify_all();
thread0.join();
thread1.join();
return 0;
}

Answer 1

Here's a way more reduced example:
#include <experimental/memory>
#include <iostream>
template <typename T>
using enable_shared_from_this = std::experimental::enable_shared_from_this<T>;
struct A : enable_shared_from_this<A> {
int a;
A() : a(7) {}
};
int main() {
auto sp = std::make_shared<A>();
auto wp = sp->weak_from_this();
if (auto s = wp.lock()) {
std::cout << s->a << std::endl;
}
}
This prints nothing. Why? The reason is ultimately the reason why it's std::enable_shared_from_this and not some other type that you yourself can provide: the shared_ptr class needs to opt-in to this functionality. The new functionality is experimental, so std::shared_ptr was not opting in - so the underlying weak_ptr was never initialized. It just doesn't happen, so wp is always an "empty" weak_ptr here.
On the other hand, std::experimental::shared_ptr does opt-in to this functionality. You need to use the shared_ptr corresponding to your enable_shared_from_this - which is std::experimental::shared_ptr.
There's no std::experimental::make_shared (or at least, as far as I could find), but the opt-in mechanism isn't based on that anyway - it's just based on any shared_ptr construction. So if you change:
auto sp = std::make_shared<A>();
to:
auto sp = std::experimental::shared_ptr<A>(new A);
Then the opt-in mechanism matches the shared_ptr type and does the right thing, you get a valid weak_ptr (a std::experimental::weak_ptr), lock() gives you shared ownership of the underlying A, and the program prints 7.