Can multiple threads join the same boost::thread? - c++

Question

pthreads has undefined behavior if multiple threads try to join the same thread:
If multiple threads simultaneously try to join with the same thread,
the results are undefined.
Is the same true for boost::threads? The documentation does not appears to specify this.
If it is undefined, then what would be a clean way for multiple threads to wait on one thread completing?

Answer 1

If it is undefined, then what would be a clean way for multiple threads to wait on one thread completing?
The clean way would be for that one thread to inform the others that it is complete. A packaged_task contains a future which can be waited on, which can help us here.
Here's one way of doing that. I have used std::thread and std::packaged_task, but you could use the boost equivalents just as well.
#include <thread>
#include <mutex>
#include <future>
#include <vector>
#include <iostream>
void emit(const char* msg) {
static std::mutex m;
std::lock_guard<std::mutex> l(m);
std::cout << msg << std::endl;
std::cout.flush();
}
int main()
{
using namespace std;
auto one_task = std::packaged_task<void()>([]{
emit("waiting...");
std::this_thread::sleep_for(std::chrono::microseconds(500));
emit("wait over!");
});
// note: convert future to a shared_future so we can pass it
// to two subordinate threads simultaneously
auto one_done = std::shared_future<void>(one_task.get_future());
auto one = std::thread(std::move(one_task));
std::vector<std::thread> many;
many.emplace_back([one_done] {
one_done.wait();
// do my thing here
emit("starting thread 1");
});
many.emplace_back([one_done] {
one_done.wait();
// do my thing here
emit("starting thread 2");
});
one.join();
for (auto& t : many) {
t.join();
}
cout << "Hello, World" << endl;
return 0;
}
expected output:
waiting...
wait over!
starting thread 2
starting thread 1
Hello, World

Answer 2

I ended up using a boost::condition_variable... roughly:
class thread_wrapper {
boost::mutex mutex;
boost::condition_variable thread_done_condition;
bool thread_done = false;
void the_func() {
// ...
// end of the thread
{
boost:unique_lock<boost::mutex> lock(mutex);
thread_done = true;
}
thread_done_condition.notify_all();
}
void wait_until_done() {
boost::unique_lock<boost::mutex> lock(mutex);
thread_done_condition.wait(lock, [this]{ return thread_done; });
}
}
Then multiple callers can safely call wait_until_done().

Answer 3

It strikes me now that something like the following would also have worked:
class thread_wrapper {
public:
thread_wrapper() : thread([this]() { this->the_func(); }) { }
void wait_until_done() {
boost::unique_lock<boost::mutex> lock(join_mutex);
thread.join();
}
private:
void the_func() {
// ...
}
boost::mutex join_mutex;
boost::thread thread;
}