Question: A coworker had an issue recently that boiled down to what we believe was the following sequence of events in a C++ application with two threads:

* Thread A holds a mutex.
* While thread A is holding the mutex, thread B attempts to lock it. Since it is held, thread B is suspended.
* Thread A finishes the work that it was holding the mutex for, thus releasing the mutex.
* Very shortly thereafter, thread A needs to touch a resource that is protected by the mutex, so it locks it again.
* It appears that thread A is given the mutex again; thread B is still waiting, even though it "asked" for the lock first.

Does this sequence of events fit with the semantics of, say, C++11's `std::mutex` and/or pthreads? I can honestly say I've never thought about this aspect of mutexes before.

Are there any fairness guarantees to prevent starvation of other threads for too long, or any way to get such guarantees?

Answer: 
Known problem. C++ mutexes are thin layer on top of OS-provided mutexes, and OS-provided mutexes are often not fair. They do not care for FIFO.

The other side of the same coin is that threads are usually not pre-empted until they run out of their time slice. As a result, thread A in this scenario was likely to continue to be executed, and got the mutex right away because of that.
You can use a *fair mutex* to solve your task, i.e. a mutex that will guarantee the FIFO order of your operations. Unfortunately, C++ standard library doesn't have a fair mutex.

Thankfully, there are open-source implementations, for example [yamc](https://github.com/yohhoy/yamc/blob/master/include/fair_mutex.hpp) (a header-only library).