Re: [Boost-users] Does lock-less queue wait on empty queue ?

Subject: Re: [Boost-users] Does lock-less queue wait on empty queue ?
From: Gottlob Frege (gottlobfrege_at_[hidden])
Date: 2015-05-20 17:15:47

Using condition variables does require mutexes, but you only need to
grab the mutex when someone is actually waiting.  So you can have a
separate (atomic) flag 'waiting' and then only grab the mutex when it
is true.

But that is an extra atomic op on every push (checking the flag).  Of
course, it could probably be a relaxed atomic (I think), which is
almost free on many platforms.

And it gets tricky ensuring that you don't have a case where someone
is waiting but every pusher missed seeing the flag.

If the waiting is inside the queue implementation, you can possibly
hide the flag inside the head pointer, getting the extra atomic op for
free.

I touched on some of this in my Lock-free talk at BoostCon/C++Now last week.

Tony

On Mon, May 18, 2015 at 9:28 PM, Gavin Lambert <gavinl_at_[hidden]> wrote:
> On 19/05/2015 01:02, Sebastian Messerschmidt wrote:
>>
>> Am 18.05.2015 um 13:51 schrieb Ran Shalit:
>>>
>>> Hello,
>>>
>>> I'm newbie of boost library.
>>>
>>> I would like to ask if the lockless queue waits on empty queue ? (I
>>> actually need it to wait).
>>
>> You can easily wait yourself, and everyone would want to wait
>> differently, so there is no point in putting into the queue's
>> implementation.
>> For instance you could give up your scheduler time-slice, or you might
>> want to spin-lock
>>
>> while(!queue.pop(my_data))
>> {
>>        boost::this_thread::yield(); //give up time slice
>> }
>
>
> You can also do a "real" wait until something has been pushed (which can be
> important as yield() does not let lower-priority threads run), but currently
> doing that requires using either Win32 events or using condition variables
> -- but the latter requires adding a mutex on both ends, which may defeat the
> point of using a lock-free queue in the first place.
>
>
>
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