Graham Reitz wrote:
Is there a method to release all threads waiting on a mutex?
Yes.
Or, more generally, a portable way to release a bunch of threads at the same time?
Try condition::notify_all (see attached code for a sample). Regards, Tobias #include <string> #include <iostream> #include <boost/bind.hpp> #include <boost/thread.hpp> #include <boost/noncopyable.hpp> class runner : boost::noncopyable { boost::condition & go; boost::mutex & mutex; boost::thread thread; public: explicit runner(boost::condition & go_signal, boost::mutex & m) : go(go_signal), mutex(m) , thread( boost::bind(& runner::run, this) ) { } ~runner() { thread.join(); } private: void run() { std::cout << std::hex; { boost::mutex::scoped_lock lock(mutex); std::cout << "Runner @ 0x" << (std::ptrdiff_t) this << " ready to run." << std::endl; go.wait(lock); } for (int i = 0; i < 1000; ++i) { // locking here so the output doesn't get scrambled // and so that non thread-safe iostream variants won't crash boost::mutex::scoped_lock lock(mutex); std::cout << "Runner @ 0x" << (std::ptrdiff_t) this << " running." << std::endl; } } }; int main() { boost::condition go_signal; boost::mutex m; runner x(go_signal,m); runner y(go_signal,m); runner z(go_signal,m); std::string s; std::getline(std::cin, s); go_signal.notify_all(); return 0; }