问题
I was told when writing Microsoft specific C++ code that writing Sleep(1) is much better than Sleep(0)
for spinlocking, due to the fact that Sleep(0)
will use more of the CPU time, moreover, it only yields if there is another equal-priority thread waiting to run.
However, with the C++11 thread library, there isn't much documentation (at least that I've been able to find) about the effects of std::this_thread::yield()
vs. std::this_thread::sleep_for( std::chrono::milliseconds(1) )
; the second is certainly more verbose, but are they both equally efficient for a spinlock, or does it suffer from potentially the same gotchas that affected Sleep(0)
vs. Sleep(1)
?
An example loop where either std::this_thread::yield()
or std::this_thread::sleep_for( std::chrono::milliseconds(1) )
would be acceptable:
void SpinLock( const bool& bSomeCondition )
{
// Wait for some condition to be satisfied
while( !bSomeCondition )
{
/*Either std::this_thread::yield() or
std::this_thread::sleep_for( std::chrono::milliseconds(1) )
is acceptable here.*/
}
// Do something!
}
回答1:
The Standard is somewhat fuzzy here, as a concrete implementation will largely be influenced by the scheduling capabilities of the underlying operating system.
That being said, you can safely assume a few things on any modern OS:
yield
will give up the current timeslice and re-insert the thread into the scheduling queue. The amount of time that expires until the thread is executed again is usually entirely dependent upon the scheduler. Note that the Standard speaks of yield as an opportunity for rescheduling. So an implementation is completely free to return from a yield immediately if it desires. A yield will never mark a thread as inactive, so a thread spinning on a yield will always produce a 100% load on one core. If no other threads are ready, you are likely to lose at most the remainder of the current timeslice before you get scheduled again.sleep_*
will block the thread for at least the requested amount of time. An implementation may turn asleep_for(0)
into ayield
. Thesleep_for(1)
on the other hand will send your thread into suspension. Instead of going back to the scheduling queue, the thread goes to a different queue of sleeping threads first. Only after the requested amount of time has passed will the scheduler consider re-inserting the thread into the scheduling queue. The load produced by a small sleep will still be very high. If the requested sleep time is smaller than a system timeslice, you can expect that the thread will only skip one timeslice (that is, one yield to release the active timeslice and then skipping the one afterwards), which will still lead to a cpu load close or even equal to 100% on one core.
A few words about which is better for spin-locking. Spin-locking is a tool of choice when expecting little to no contention on the lock. If in the vast majority of cases you expect the lock to be available, spin-locks are a cheap and valuable solution. However, as soon as you do have contention, spin-locks will cost you. If you are worrying about whether yield or sleep is the better solution here spin-locks are the wrong tool for the job. You should use a mutex instead.
For a spin-lock, the case that you actually have to wait for the lock should be considered exceptional. Therefore it is perfectly fine to just yield here - it expresses the intent clearly and wasting CPU time should never be a concern in the first place.
回答2:
I just did a test with Visual Studio 2013 on Windows 7, 2.8GHz Intel i7, default release mode optimizations.
sleep_for(nonzero) appears sleep for a minimium of around one millisecond and takes no CPU resources in a loop like:
for (int k = 0; k < 1000; ++k)
std::this_thread::sleep_for(std::chrono::nanoseconds(1));
This loop of 1,000 sleeps takes about 1 second if you use 1 nanosecond, 1 microsecond, or 1 millisecond. On the other hand, yield() takes about 0.25 microseconds each but will spin the CPU to 100% for the thread:
for (int k = 0; k < 4,000,000; ++k) (commas added for clarity)
std::this_thread::yield();
std::this_thread::sleep_for((std::chrono::nanoseconds(0)) seems to be about the the same as yield() (test not shown here).
In comparison, locking an atomic_flag for a spinlock takes about 5 nanoseconds. This loop is 1 second:
std::atomic_flag f = ATOMIC_FLAG_INIT;
for (int k = 0; k < 200,000,000; ++k)
f.test_and_set();
Also, a mutex takes about 50 nanoseconds, 1 second for this loop:
for (int k = 0; k < 20,000,000; ++k)
std::lock_guard<std::mutex> lock(g_mutex);
Based on this, I probably wouldn't hesitate to put a yield in the spinlock, but I would almost certainly wouldn't use sleep_for. If you think your locks will be spinning a lot and are worried about cpu consumption, I would switch to std::mutex if that's practical in your application. Hopefully, the days of really bad performance on std::mutex in Windows are behind us.
回答3:
if you are interested in cpu load while using yield - it's very bad, except one case-(only your application is running, and you are aware that it will basically eat all your resources)
here is more explanation:
- running yield in loop will ensure that cpu will release execution of thread, still, if system try to come back to thread it will just repeat yield operation. This can make thread use full 100% load of cpu core.
- running
sleep()
orsleep_for()
is also a mistake, this will block thread execution but you will have something like wait time on cpu. Don't be mistaken, this IS working cpu but on lowest possible priority. While somehow working for simple usage examples ( fully loaded cpu on sleep() is half that bad as fully loaded working processor ), if you want to ensure application responsibility, you would like something like third example: combining! :
std::chrono::milliseconds duration(1); while (true) { if(!mutex.try_lock()) { std::this_thread::yield(); std::this_thread::sleep+for(duration); continue; } return; }
something like this will ensure, cpu will yield as fast as this operation will be executed, and also sleep_for() will ensure that cpu will wait some time before even trying to execute next iteration. This time can be of course dynamicaly (or staticaly) adjusted to suits your needs
cheers :)
回答4:
What you want is probably a condition variable. A condition variable with a conditional wake up function is typically implemented like what you are writing, with the sleep or yield inside the loop a wait on the condition.
Your code would look like:
std::unique_lock<std::mutex> lck(mtx)
while(!bSomeCondition) {
cv.wait(lck);
}
Or
std::unique_lock<std::mutex> lck(mtx)
cv.wait(lck, [bSomeCondition](){ return !bSomeCondition; })
All you need to do is notify the condition variable on another thread when the data is ready. However, you cannot avoid a lock there if you want to use condition variable.
来源:https://stackoverflow.com/questions/17325888/c11-thread-waiting-behaviour-stdthis-threadyield-vs-stdthis-thread