std::thread::join() hangs if called after main() exits when using VS2012 RC - c++

Question

The following example runs successfully (i.e. doesn't hang) if compiled using Clang 3.2 or GCC 4.7 on Ubuntu 12.04, but hangs if I compile using VS11 Beta or VS2012 RC.
#include <iostream>
#include <string>
#include <thread>
#include "boost/thread/thread.hpp"
void SleepFor(int ms) {
std::this_thread::sleep_for(std::chrono::milliseconds(ms));
}
template<typename T>
class ThreadTest {
public:
ThreadTest() : thread_([] { SleepFor(10); }) {}
~ThreadTest() {
std::cout << "About to join\t" << id() << '\n';
thread_.join();
std::cout << "Joined\t\t" << id() << '\n';
}
private:
std::string id() const { return typeid(decltype(thread_)).name(); }
T thread_;
};
int main() {
static ThreadTest<std::thread> std_test;
static ThreadTest<boost::thread> boost_test;
// SleepFor(100);
}
The issue appears to be that std::thread::join() never returns if it is invoked after main has exited. It is blocked at WaitForSingleObject in _Thrd_join defined in cthread.c.
Uncommenting SleepFor(100); at the end of main allows the program to exit properly, as does making std_test non-static. Using boost::thread also avoids the issue.
So I'd like to know if I'm invoking undefined behaviour here (seems unlikely to me), or if I should be filing a bug against VS2012?

Answer 1

Tracing through Fraser's sample code in his connect bug (https://connect.microsoft.com/VisualStudio/feedback/details/747145)
with VS2012 RTM seems to show a fairly straightforward case of deadlocking. This likely isn't specific to std::thread - likely _beginthreadex suffers the same fate.
What I see in the debugger is the following:
On the main thread, the main() function has completed, the process cleanup code has acquired a critical section called _EXIT_LOCK1, called the destructor of ThreadTest, and is waiting (indefinitely) on the second thread to exit (via the call to join()).
The second thread's anonymous function completed and is in the thread cleanup code waiting to acquire the _EXIT_LOCK1 critical section. Unfortunately, due to the timing of things (whereby the second thread's anonymous function's lifetime exceeds that of the main() function) the main thread already owns that critical section.
DEADLOCK.
Anything that extends the lifetime of main() such that the second thread can acquire _EXIT_LOCK1 before the main thread avoids the deadlock situation. That's why the uncommenting the sleep in main() results in a clean shutdown.
Alternatively if you remove the static keyword from the ThreadTest local variable, the destructor call is moved up to the end of the main() function (instead of in the process cleanup code) which then blocks until the second thread has exited - avoiding the deadlock situation.
Or you could add a function to ThreadTest that calls join() and call that function at the end of main() - again avoiding the deadlock situation.

Answer 2

I realize this is an old question regarding VS2012, but the bug is still present in VS2013. For those who are stuck on VS2013, perhaps due to Microsoft's refusal to provide upgrade pricing for VS2015, I offer the following analysis and workaround.
The problem is that the mutex (at_thread_exit_mutex) used by _Cnd_do_broadcast_at_thread_exit() is either not yet initialized, or has already been destroyed, depending on the exact circumstances. In the former case, _Cnd_do_broadcast_at_thread_exit() tries to initialize the mutex during shutdown, causing a deadlock. In the latter case, where the mutex has already been destroyed via the atexit stack, the program will crash on the way out.
The solution I found is to explicitly call _Cnd_do_broadcast_at_thread_exit() (which thankfully is declared publicly) early during program startup. This has the effect of creating the mutex before anyone else tries to access it, as well as ensuring that the mutex continues to exist until the last possible moment.
So, to fix the problem, insert the following code at the bottom of a source module, for instance somewhere below main().
#pragma warning(disable:4073) // initializers put in library initialization area
#pragma init_seg(lib)
#if _MSC_VER < 1900
struct VS2013_threading_fix
{
VS2013_threading_fix()
{
_Cnd_do_broadcast_at_thread_exit();
}
} threading_fix;
#endif

Answer 3

I believe your threads have already been terminated and their resources freed following the termination of your main function and before static destruction. This is the behavior of the VC runtimes dating back to at least VC6.
Do child threads exit when the parent thread terminates
boost thread and process cleanup on windows

Answer 4

My answer is too far late, but hope will help someone.
I was stucked by this bug, and i find a trick to solve this problem,it worked in my code.
int main()
{
ThreadTest trick_obj; //trick... You can put this line of code anywhere
static ThreadTest std_test;
return 1;
}

Answer 5

I have been battling this bug for a day, and found the following work-around, which turned out the be the least dirty trick:
Instead of returning, one can use the standard Windows API function call ExitThread() to terminate the thread. This method of course may mess up the internal state of the std::thread object and associated library, but since the program is going to terminate anyway, well, so be it.
#include <windows.h>
template<typename T>
class ThreadTest {
public:
ThreadTest() : thread_([] { SleepFor(10); ExitThread(NULL); }) {}
~ThreadTest() {
std::cout << "About to join\t" << id() << '\n';
thread_.join();
std::cout << "Joined\t\t" << id() << '\n';
}
private:
std::string id() const { return typeid(decltype(thread_)).name(); }
T thread_;
};
The join() call apparently works correctly. However, I chose to use a more safe method in our solution. One can get the thread HANDLE via std::thread::native_handle(). With this handle we can call the Windows API directly to join the thread:
WaitForSingleObject(thread_.native_handle(), INFINITE);
CloseHandle(thread_.native_handle());
Thereafter, the std::thread object must not be destroyed, as the destructor would try to join the thread a second time. So we just leave the std::thread object dangling at program exit.