内存屏障/栅栏的开销

Question

I'm currently writing C++ code and use a lot of memory barriers / fences in my code. I know, that a MB tolds the compiler and the hardware to not reorder write/reads around it. But i don't know how complex this operation is for the processor at runtime.

My Question is: What is the runtime-overhead of such a barrier? I didn't found any useful answer with google... Is the overhead negligible? Or leads heavy usage of MBs to serious performance problems?

Best regards.

Answer 1

Compared to arithmetic and "normal" instructions I understand these to be very costly, but do not have numbers to back up that statement. I like jalf's answer by describing effects of the instructions, and would like to add a bit.

There are in general a few different types of barriers, so understanding the differences could be helpful. A barrier like the one that jalf mentioned is required for example in a mutex implementation before clearing the lock word (lwsync on ppc, or st4.rel on ia64 for example). All reads and writes must be complete, and only instructions later in the pipeline that have no memory access and no dependencies on in progress memory operations can be executed.

Another type of barrier is the sort that you'd use in a mutex implementation when acquiring a lock (examples, isync on ppc, or instr.acq on ia64). This has an effect on future instructions, so if a non-dependent load has been prefetched it must be discarded. Example:

if ( pSharedMem->atomic.bit_is_set() ) // use a bit to flag that somethingElse is "ready"
{
   foo( pSharedMem->somethingElse ) ;
}

Without an acquire barrier (borrowing ia64 lingo), your program may have unexpected results if somethingElse made it into a register before the check of the flagging bit check is complete.

There is a third type of barrier, generally less used, and is required to enforce store load ordering. Examples of instructions for such an ordering enforcing instruction are, sync on ppc (heavyweight sync), MF on ia64, membar #storeload on sparc (required even for TSO).

Using ia64 like pseudocode to illustrate, suppose one had

st4.rel
ld4.acq

without an mf in between one has no guarentee that the load follows the store. You know that loads and stores preceding the st4.rel are done before that store or the "subsequent" load, but that load or other future loads (and perhaps stores if non-dependent?) could sneak in, completing earlier since nothing prevents that otherwise.

Because mutex implementations very likely only use acquire and release barriers in thier implementations, I'd expect that an observable effect of this is that memory access following lock release may actually sometimes occur while "still in the critical section".

Answer 2

Try thinking about what the instruction does. It doesn't make the CPU do anything complicated in terms of logic, but it forces it to wait until all reads and writes have been committed to main memory. So the cost really depends on the cost of accessing main memory (and the number of outstanding reads/writes).

Accessing main memory is generally pretty expensive (10-200 clock cycles), but in a sense, that work would have to be done without the barrier as well, it could just be hidden by executing some other instructions simultaneously so you didn't feel the cost so much.

It also limits the CPU's (and compilers) ability to reschedule instructions, so there may be an indirect cost as well in that nearby instructions can't be interleaved which might otherwise yield a more efficient execution schedule.