io_service run within thread

Question

Why in this simple class if i use directly io.run() the function will be invoked otherwise if demand the run to other thread the print will not be invoked?

#include <iostream>
#include <boost/thread.hpp>
#include <boost/asio.hpp>

using namespace std;

class test
{
  public:
    test()
    {
      io.post(boost::bind(&test::print, this));
      //io.run();
      t = boost::thread(boost::bind(&boost::asio::io_service::run, &io));
    }

    void print()
    {
      cout << "test..." << endl;
    }

  private:
    boost::thread t;
    boost::asio::io_service io;
};

int main()
{
  test();
  return 0;
}

Answer 1

The thread object is being destroyed before allowing the io_service to completely run. The thread destructor documentation states:

[...] the programmer must ensure that the destructor is never executed while the thread is still joinable.

If BOOST_THREAD_PROVIDES_THREAD_DESTRUCTOR_CALLS_TERMINATE_IF_JOINABLE is defined, the program would abort as the thread destructor would call std::terminate().

---

If the io_service should run to completion, then consider joining the thread within Test's destructor. Here is a complete example that demonstrates synchronizing on the thread's completion:

#include <iostream>
#include <boost/asio.hpp>
#include <boost/thread.hpp>

class test
{
public:
  test()
  {
    io.post(boost::bind(&test::print, this));
    t = boost::thread(boost::bind(&boost::asio::io_service::run, &io));
  }

  ~test()
  {
    if (t.joinable())
      t.join();  
  }

  void print()
  {
    std::cout << "test..." << std::endl;
  }

private:
  boost::thread t;
  boost::asio::io_service io;
};

int main()
{
  test();
  return 0;
}

Output:

test...