Cpp thread object and shared_ptr issue

Question

When I read the document in cppreference here https://en.cppreference.com/w/cpp/memory/shared_ptr#Example

I am wondering what the possible value of the first lp.use_count() printed out is? I marked it with "<<<<<" in output content.

#include <iostream>
#include <memory>
#include <thread>
#include <chrono>
#include <mutex>
 
struct Base
{
    Base() { std::cout << "  Base::Base()\n"; }
    // Note: non-virtual destructor is OK here
    ~Base() { std::cout << "  Base::~Base()\n"; }
};
 
struct Derived: public Base
{
    Derived() { std::cout << "  Derived::Derived()\n"; }
    ~Derived() { std::cout << "  Derived::~Derived()\n"; }
};
 
void thr(std::shared_ptr<Base> p)
{
    std::this_thread::sleep_for(std::chrono::seconds(1));
    std::shared_ptr<Base> lp = p; // thread-safe, even though the
                                  // shared use_count is incremented
    {
        static std::mutex io_mutex;
        std::lock_guard<std::mutex> lk(io_mutex);
        std::cout << "local pointer in a thread:\n"
                  << "  lp.get() = " << lp.get()
                  << ", lp.use_count() = " << lp.use_count() << '\n';
    }
}
 
int main()
{
    std::shared_ptr<Base> p = std::make_shared<Derived>();
 
    std::cout << "Created a shared Derived (as a pointer to Base)\n"
              << "  p.get() = " << p.get()
              << ", p.use_count() = " << p.use_count() << '\n';
    std::thread t1(thr, p), t2(thr, p), t3(thr, p);
    p.reset(); // release ownership from main
    std::cout << "Shared ownership between 3 threads and released\n"
              << "ownership from main:\n"
              << "  p.get() = " << p.get()
              << ", p.use_count() = " << p.use_count() << '\n';
    t1.join(); t2.join(); t3.join();
    std::cout << "All threads completed, the last one deleted Derived\n";
}

Possible output:

Base::Base()
  Derived::Derived()
Created a shared Derived (as a pointer to Base)
  p.get() = 0x2299b30, p.use_count() = 1
Shared ownership between 3 threads and released
ownership from main:
  p.get() = 0, p.use_count() = 0
local pointer in a thread:
  lp.get() = 0x2299b30, lp.use_count() = 5   <<<<<<<< HERE <<<<<<
local pointer in a thread:
  lp.get() = 0x2299b30, lp.use_count() = 3
local pointer in a thread:
  lp.get() = 0x2299b30, lp.use_count() = 2
  Derived::~Derived()
  Base::~Base()
All threads completed, the last one deleted Derived

@user2452809 's answer is very appreciated, which pointed out an important feature of use_count().

Supposing use_count() would return an accurate count, what would be the answer?

Roman2452809 Roman2452809 · Accepted Answer · 2020-06-28T03:08:19

I wouldn't rely on that value anyway.

In multithreaded environment, the value returned by use_count is approximate (typical implementations use a memory_order_relaxed load)

Check the reference for more information: https://en.cppreference.com/w/cpp/memory/shared_ptr/use_count

Cpp thread object and shared_ptr issue

2 Answers