On Aug 21, 2007, at 6:02 PM, Yuval Ronen wrote:
Howard Hinnant wrote:
On Aug 21, 2007, at 3:22 PM, Yuval Ronen wrote:
Howard Hinnant wrote:
On Aug 21, 2007, at 8:46 AM, Yuval Ronen wrote:
1. I couldn't understand what defer_lock is good for, even after reading Q.9 of the FAQ. I believe the use-case shown in Q.9 should actually use accept_ownership instead. Can you elaborate please? See if this is any clearer:
http://home.twcny.rr.com/hinnant/cpp_extensions/concurrency_rationale.html#u... I'm afraid not...
This example has 3 lines, the first 2 create unique_lock with defer_lock, and the 3rd calls std::lock. Those unique_lock don't lock, because std::lock to lock. OK. But who unlocks? The unique_locks don't own the mutexes, and therefore don't unlock them. But someone needs to unlock, and it sounds logical that the unique_locks would... Had we used accept_ownership, the unique_locks would have owned the mutexes, and unlock them. That's the difference between defer_lock and accept_ownership, the ownership, isn't it?
Ok, the lightbulb went off in my head and I think I understand your question now. Thanks for not giving up on me.
I've tried again here:
http://home.twcny.rr.com/hinnant/cpp_extensions/concurrency_rationale.html#u...
and see the next question (#10) as well. If that doesn't do it, see:
http://home.twcny.rr.com/hinnant/cpp_extensions/mutex_base
and search for "defer_lock_type" and "accept_ownership_type" for the unique_lock implementation of these constructors. Neither constructor does anything to the mutex, and simply sets the owns() flag to false or true respectively. The unique_lock destructor will unlock the mutex iff it owns()'s the mutex.
Not yet... :)
One of the added sentences is "After std::lock locks l1 and l2, these locks now own their respective mutexes". How does that happen? I looked at the implementation code, but could see nothing that changes owns() from false to true after construction. Have I missed it?
I think so, but it is rather subtle. Re-quoting std::lock (the two- lock version) here: template <class _L1, class _L2> void lock(_L1& __l1, _L2& __l2) { while (true) { { unique_lock<_L1> __u1(__l1); if (__l2.try_lock()) { __u1.release(); break; } } std::this_thread::yield(); { unique_lock<_L2> __u2(__l2); if (__l1.try_lock()) { __u2.release(); break; } } std::this_thread::yield(); } } And I'm going to answer your second question below before continuing:
Another related question is why std::lock works with locks and not mutexes? Can't see the benefit in that.
You're absolutely right. This is my fault due to lack of proper documentation at the moment. All that is required of _L1 and _L2 in the above algorithm is that they support: void lock(); bool try_lock(); void unlock(); It will work for _L1 being a lock or mutex (and same for _L2). Ok, now back to how std::lock works: This line: unique_lock<_L1> __u1(__l1); implicitly calls __.l1.lock() inside of the unique_lock constructor. If __l1 is a mutex, the deed is done. If __l1 is a lock, hopefully that will forward to the referenced mutex's lock() function in the proper manner. And in the process, that should set the lock's owns() data to true as well. template <class _Mutex> void unique_lock<_Mutex>::lock() { if (__m == 0 || __owns) throw lock_error(); __m->lock(); __owns = true; // owns() set here!!! } Similarly for: if (__l2.try_lock()) And if all that fails, then the second half of the algorithm just does the same thing in the reverse order. The reason for the local unique_lock is because I'm anticipating that the try_lock() could throw an exception. If it does, the local unique_lock will unlock its referenced lock/mutex as the exception propagates out. If an exception isn't thrown, and the try_lock succeeds, then the unique_lock just releases the the lock/mutex (much like auto_ptr). Both locks remain locked and the algorithm returns normally. This algorithm (the std::lock) is actually my cornerstone of "why I like uncoupled - non-nested - lock types". The local unique_lock is applied to any mutex, or any lock, even another unique_lock, transparently. It just does its job without caring what type of mutex or lock it needs to temporarily hold on to. unique_lock is very, very analogous to unique_ptr (formerly auto_ptr). One holds lock/mutex ownership state, the other holds heap allocated memory. Both are resource holders. Both have virtually the same responsibilities and design. Their names are purposefully similar.
Either way, I believe this design wouldn't meet the use case which I didn't effectively communicate in #14:
Given a read-write mutex and its associated condition variable:
my::shared_mutex rw_mut; std::condition<my::shared_mutex> cv(rw_mut);
client code wants to wait on that cv in two different ways:
1. With rw_mut read-locked. 2. With rw_mut write-locked.
If we initialized the condition variable with cv(rw_mut.exclusive()), then cv.wait() would wait with rw_mut write-locked, but we wouldn't be able to wait on cv with rw_mut read-locked.
If we initialized the condition variable with cv(rw_mut.shared()), then cv.wait() would wait with rw_mut read-locked, but we wouldn't be able to wait on cv with rw_mut write-locked.
This use case desires *both* types of waits on the *same* mutex/cv pair. The last sentence starts with "This use case", but I see no use case. Do we really have such a use case? I haven't seen one yet. But even if we had, then maybe the solution is the same solution to the requirement you phrased as "The freedom to dynamically associate mutexes with condition variables" or "The ability to wait on general mutex / lock types" (what's the difference between those two sentences anyway?) in your response to Peter. Add a 'set_mutex(mutex_type &)', or maybe even 'set_mutex(mutex_type *)' to std::condition. I think it will solve this rare case.
Ok, perhaps I'll clean the following use case up, and include it in the faq. First I'll try it out here. :-)
I've got a "communication node" class. It serves as a node in a network. It gets data from somewhere and temporarily stores it while it uses several threads to forward (or relay) the information to other nodes in the network. For simplicity I'm using vector<int> for the data, and only two relay threads. The example is incomplete (not even compiled), just illustrative right now:
class communication_node { std::vector<int>* from_; std::vector<int> data_; std::vector<int> to_[2];
typedef std::tr2::shared_mutex Mutex; Mutex mut_; std::condition<Mutex> cv_; bool get_data_; bool fresh_data_; bool data_relayed_[2]; public: void supplier() { while (true) { std::unique_lock<Mutex> write_lock(mut_); while (!get_data_ || !data_relayed_[0] || ! data_relayed_[1]) cv_.wait(write_lock); std::copy(from_->begin(), from_->end(), data_.begin()); get_data_ = false; fresh_data_ = true; data_relayed_[0] = false; data_relayed_[1] = false; cv_.notify_all(); } } void relayer(int id) { while (true) { std::tr2::shared_lock<Mutex> read_lock(mut_); while (data_relayed_[id]) cv_.wait(read_lock); std::copy(data_.begin(), data_.end(), to[id].begin()); data_relayed_[id] = true; cv_.notify_all(); } } };
The relayer, which supposed to be a reader, actually is a writer, to data_relayed_[id], so I believe a read_lock is not enough...
I should have made to_: std::vector<int>* to_[2]; I.e. a vector to a pointer to data outside the node. And now that I think more about it, the relayer thread is the "missing supplier" thread not demonstrated above, but for the next node downstream. It only needs read access to this node, but it will need write access to a downstream node. The example definitely needs more work.
You keep on improving. I hope we all are ;-)
:-) -Howard