Say a thread in one core is spinning on a variable which will be updated by a thread running on another core. My question is what is the overhead at cache level. Will the waiti开发者_JAVA百科ng thread cache the variable and therefore does not cause any traffic on the bus until the writing thread writes to that variable?
How can this overhead be reduced. Does x86 pause
instruction help?
I believe all modern x86 CPUs use the MESI protocol. So the spinning "reader" thread will likely have a cached copy of the data in either "exclusive" or "shared" mode, generating no memory bus traffic while you spin.
It is only when the other core writes to the location that it will have to perform cross-core communication.
[update]
A "spinlock" like this is only a good idea if you will not be spinning for very long. If it may be a while before the variable gets updated, use a mutex + condition variable instead, which will put your thread to sleep so that it adds no overhead while it waits.
(Incidentally, I suspect a lot of people -- including me -- are wondering "what are you actually trying to do?")
If you spin lock for short intervals you are usually fine. However there is a timer interrupt on Linux (and I assume similar on other OSes) so if you spin lock for 10 ms or close to it you will see a cache disturbance.
I have heard its possible to modify the Linux kernel to prevent all interrupts on specific cores and this disturbance goes away, but I don't know what is involved in doing this.
In the case of two threads the overhead may be ignored, anyway it could be a good idea make a simple benchmark. For instance, if you implement spinlocks, how much time the thread spends into the spin. This effect on the cache it's called cache line bouncing.
I tested this extensively in this post. The overhead in general is incurred by the bus-locking component of the spinlock, usually the instruction "xchg reg,mem" or some variant of it. Since that particular overhead cannot be avoided you have the options of economizing on the frequency with which you invoke the spinlock and performing the absolute minimum amount of work necessary - once the lock is in place - before releasing it.
精彩评论