On Aug 27, 2007, at 1:29 PM, David Abrahams wrote:
on Mon Aug 27 2007, Howard Hinnant <howard.hinnant-AT-gmail.com> wrote:
until further notice, no other object in the program has executed __l1.lock(), or __l1.try_lock() (or any of the other functions which a can put a mutex into a exclusive locked state) without having already executed __l1.unlock().
It sounds like you're saying that it's legit for some "other object in the program" to call __l1.unlock() while __u1 is uniquely holding __l1's lock state. Are you really saying that?
No. I'm saying __u1 owns the exclusively locked state of __l1. What that ownership means is __l1's business.
What was all that about executing __l1.unlock(), then? Should I just ignore it?
I was trying to be precise. Instead I was confusing. Please ignore it. unique_lock<Mutex> works just like boost::thread::detail::scoped_lock<Mutex>. -Howard