Hello, Well, the subject says it all ;) This question comes because boost::signal's documentation says it's not thread safe, but I have trouble finding exactly why it isn't. Let's look the following example: Each thread is connected to the same signal, then the signal is emmited, and each threads receives the message.... Why would this not be thread safe ? Of course in this code the issue would be that I don't mutex the "finished" variable but that's another story. #include <iostream> #include <boost/bind.hpp> #include <boost/thread.hpp> #include <boost/signals.hpp> struct thread1 { void operator()() { finished = false; while(!finished); } void slot() { finished = true; } bool finished; }; struct thread2 { void operator()() { finished = false; while(!finished); } void slot() { finished = true; } bool finished; }; int main() { boost::signal<void ()> sig; thread1 t1; thread2 t2; sig.connect(boost::bind(&thread1::slot, &t1)); sig.connect(boost::bind(&thread2::slot, &t2)); boost::thread trd1(boost::ref(t1)); boost::thread trd2(boost::ref(t2)); usleep(10); sig(); trd1.join(); trd2.join(); return 0; } Thank you for insight. Philippe