[rbmm]

"other threads can share-acquire the same lock without waiting." can but not always and mandatory. you really allow to system enter another shared requestor. but only allow,not demand. system not let shared requestor enter to lock, if before it the exclusive request will be. after first exclusive request - all next shared request will block, util this exclusive request not acquire and then release lock. can be and some another reason, in the end - this is only optimization - you only *allow* multiple shared enter at once. so i be not say that this is implementation bug

[quotemstr]

So we have to treat shared acquires as exclusive ones just in case an exclusive acquire comes along? What if I know that my program doesn't do that?

[rbmm]

yes. after any thread request exclusive access to lock, the sequential shared acquire request will block. this iswell known fact, i hope. this is example when shared request will be blocked, despite shared owner(s) now inside lock

also look my 2 comments on reddit: https://www.reddit.com/r/cpp/comments/1b55686/comment/ktfhjs... and https://www.reddit.com/r/cpp/comments/1b55686/comment/ktfggu...

if say true - i here in very strange position. i am by self under debugger research exacly what happens in concrete case and create repro code. i think that implementation of the RtlReleaseSRWLockExclusive is not the best and may be really containing "bug" as i in more details describe in comment on reddit. but from another side, if you want pure formal c++ rules - can you explain - in what exactly was bug in concre case ? what rule/guarantee is violated ? why demo code decide that ALL shared waiters can at once acquire the lock. formal documentation not state this. this intuitive must be true, because no exclusive request more. but.. i really dont know. and i also for fun create own implementation of SRW lock ( if someone interested to look -https://github.com/rbmm/PushLock ) which free from this roblem - i always ( hope) do only single atomic change to lock state during api call. and finally - sorry for my english and too long answer