[rigtorp]

The standard says that a thread must eventually terminate, do an atomic operation or do IO. So the while(lock.exchange(true)); loop is different.

Also keep in mind that C++11 specifies std::mutex::lock() to have acquire semantics and unlock() to have release semantics on the lock object. In order for std::mutex to actually work the reordering of m1.unlock(); m2.lock(); to m2.lock(); m1.unlock(); must be disallowed. But since m1 and m2 are separate objects m1.unlock() has no happens before relationship with m2.lock(). This seems to be a problem in the C++11 memory model. The arguments I have heard from some WG21 people is that there is no problem since transforming a wellformed terminating program into a non-terminating program is not allowed. I can't find the wording in the C++ standard that asserts this. But oh well, it works right now on gcc/llvm/msvc.