Difference between rdtscp, rdtsc : memory and cpuid / rdtsc?

Question

Assume we're trying to use the tsc for performance monitoring and we we want to prevent instruction reordering.

These are our options:

1: rdtscp is a serializing call. It prevents reordering around the call to rdtscp.

__asm__ __volatile__("rdtscp; "         // serializing read of tsc
                     "shl $32,%%rdx; "  // shift higher 32 bits stored in rdx up
                     "or %%rdx,%%rax"   // and or onto rax
                     : "=a"(tsc)        // output to tsc variable
                     :
                     : "%rcx", "%rdx"); // rcx and rdx are clobbered

However, rdtscp is only available on newer CPUs. So in this case we have to use rdtsc. But rdtsc is non-serializing, so using it alone will not prevent the CPU from reordering it.

So we can use either of these two options to prevent reordering:

2: This is a call to cpuid and then rdtsc. cpuid is a serializing call.

volatile int dont_remove __attribute__((unused)); // volatile to stop optimizing
unsigned tmp;
__cpuid(0, tmp, tmp, tmp, tmp);                   // cpuid is a serialising call
dont_remove = tmp;                                // prevent optimizing out cpuid

__asm__ __volatile__("rdtsc; "          // read of tsc
                     "shl $32,%%rdx; "  // shift higher 32 bits stored in rdx up
                     "or %%rdx,%%rax"   // and or onto rax
                     : "=a"(tsc)        // output to tsc
                     :
                     : "%rcx", "%rdx"); // rcx and rdx are clobbered

3: This is a call to rdtsc with memory in the clobber list, which prevents reordering

__asm__ __volatile__("rdtsc; "          // read of tsc
                     "shl $32,%%rdx; "  // shift higher 32 bits stored in rdx up
                     "or %%rdx,%%rax"   // and or onto rax
                     : "=a"(tsc)        // output to tsc
                     :
                     : "%rcx", "%rdx", "memory"); // rcx and rdx are clobbered
                                                  // memory to prevent reordering

My understanding for the 3rd option is as follows:

Making the call __volatile__ prevents the optimizer from removing the asm or moving it across any instructions that could need the results (or change the inputs) of the asm. However it could still move it with respect to unrelated operations. So __volatile__ is not enough.

Tell the compiler memory is being clobbered: : "memory"). The "memory" clobber means that GCC cannot make any assumptions about memory contents remaining the same across the asm, and thus will not reorder around it.

So my questions are:

1: Is my understanding of __volatile__ and "memory" correct?
2: Do the second two calls do the same thing?
3: Using "memory" looks much simpler than using another serializing instruction. Why would anyone use the 3rd option over the 2nd option?

You seem to confuse reordering of instructions generated by the compiler, which you can avoid by using `volatile` and `memory` and reordering of instructions executed by the processor (aka _out of order execution_), which you avoid by using `cpuid`. — Gunther Piez, Sep 28 '12 at 00:17
@hirschhornsalz but won't having `memory` in the clobber list prevent the processor reordering the instructions? Doesn't `memory` act like a memory fence? — Steve Lorimer, Sep 28 '12 at 00:41
or perhaps the `memory` in the clobber list is only emitted to gcc, and the resulting machine code doesn't expose this to the processor? — Steve Lorimer, Sep 28 '12 at 03:24
No, memory fences are a different thing, and the compiler will not insert those if you use a "memory" clobber. These are about reordering loads/stores by the processors and are used in conjunction with instructions with weak memory ordering in respect to multithreaded environments, like `movntdq`. Most of the time you do _not_ need a memory fence on Intel/AMD processors, as these processors have strong memory ordering by default. And yes, `memory` only affects the order in which instructions are emitted by the compiler, it does not make the compiler emit additional instructions. — Gunther Piez, Sep 28 '12 at 07:03
`rdtscp` doesn't prevent reordering, it only ensures all previous instructions have finished executing: `The RDTSCP instruction waits until all previous instructions have been executed before reading the counter. However, subsequent instructions may begin execution before the read operation is performed.`, I suggest you read this whitepaper from intel if you are considering using this for benchmarking etc: http://download.intel.com/embedded/software/IA/324264.pdf (it actually shows that you need both `rdtsc` + `cpuid` **and** `rdtscp` + `cpuid` for correct measurements) — Necrolis, Sep 28 '12 at 07:25

janneb · Accepted Answer · 2019-11-25T12:15:40.810

50

As mentioned in a comment, there's a difference between a compiler barrier and a processor barrier. volatile and memory in the asm statement act as a compiler barrier, but the processor is still free to reorder instructions.

Processor barrier are special instructions that must be explicitly given, e.g. rdtscp, cpuid, memory fence instructions (mfence, lfence, ...) etc.

As an aside, while using cpuid as a barrier before rdtsc is common, it can also be very bad from a performance perspective, since virtual machine platforms often trap and emulate the cpuid instruction in order to impose a common set of CPU features across multiple machines in a cluster (to ensure that live migration works). Thus it's better to use one of the memory fence instructions.

The Linux kernel uses mfence;rdtsc on AMD platforms and lfence;rdtsc on Intel. If you don't want to bother with distinguishing between these, mfence;rdtsc works on both although it's slightly slower as mfence is a stronger barrier than lfence.

Edit 2019-11-25: As of Linux kernel 5.4, lfence is used to serialize rdtsc on both Intel and AMD. See this commit "x86: Remove X86_FEATURE_MFENCE_RDTSC": https://git.kernel.org/pub/scm/linux/kernel/git/torvalds/linux.git/commit/?id=be261ffce6f13229dad50f59c5e491f933d3167f

edited Nov 25 '19 at 12:15

answered Sep 28 '12 at 06:41

janneb

36,249
2
81
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6

The `cpuid; rdtsc` is not about memory fences, it's about serializing the instruction stream. Usually it is used for benchmarking purposes to make sure no "old" instructions remain in the reorder buffer/reservation station. The execution time of `cpuid` (which is quite long, I remember >200 cycles) is then to be subtracted. If the result is more "exact" this way is not quite clear to me, I experimented with and without and the differences seems less the the natural error of measurement, even in single user mode with nothing else running at all. – Gunther Piez Sep 28 '12 at 07:15
I am not sure, but I possibly the fence instruction used this way in the kernel are not useful at all ^^ – Gunther Piez Sep 28 '12 at 07:20
6

@hirschhornsalz: According to the git commit logs, AMD and Intel confirmed that the m/lfence will serialize rdtsc on currently available CPU's. I suppose Andi Kleen can provide more details on what exactly was said, if you're interested and ask him. – janneb Sep 29 '12 at 20:02
1

@hirschhornsalz: ... IIRC the argument basically goes that while the fence instructions only serialize wrt. instructions that read/write memory, in practice there's no point in reordering non-mem instructions wrt rdtsc and thus it's not done. Although per the architecture manual it's in principle allowed. – janneb Sep 29 '12 at 20:06
That's exactly what I think, in practice (=non-benchmarking code) there is no point in avoiding the reordering of instructions. I would even go one step further and argue that there isn't even a point in avoiding the reordering of memory instructions, since rdtsc is only used as a non memory depended timer source here and so drop the fences. But I should really ask Andy :-) – Gunther Piez Oct 01 '12 at 08:06
Is the memory clobber part of the asm still necessary? I notice the code in Intel's white paper makes no mention of it: http://www.intel.com/content/dam/www/public/us/en/documents/white-papers/ia-32-ia-64-benchmark-code-execution-paper.pdf – Joseph Garvin Mar 24 '17 at 17:53
Are you *sure* that `mfence; rdtsc` on Intel really serializes the instruction stream? `lfence` is now officially / more-clearly documented as serializing, (so it can be used to mitigate Spectre mis-speculation of bounds-check branches). But I'm not sure `mfence` serializes the instruction stream on Intel. (Maybe it does, but it's not clearly documented). Fun fact: on Core2, `mfence` has better throughput than `lfence` (when that's *all* the machine is running, no other instructions mixed in. source: Agner Fog's tests). – Peter Cordes Jan 29 '18 at 14:37
2

It's probably important to use `lfence` on Intel and `mfence` on AMD; any argument about "stronger barrier" is totally inapplicable because we're talking about the instruction stream and additional micro-architectural effects, *not* the well-documented memory-ordering effects. For example, LFENCE isn't fully serializing on AMD: it has 4-per-clock throughput Bulldozer-family / Ryzen! Maybe it does serialize `rdtsc` but not itself or some other instructions? Or more likely it's very cheap on AMD because their memory-ordering implementation works differently. – Peter Cordes Jan 29 '18 at 14:40
@JosephGarvin: A "memory clobber" is an explicit notice to a *compiler* that a piece of code may be dependent upon memory ordering in ways the compiler should not expect to understand. Some compilers are prone to assume that memory order only matters in situations where they can see explicit reasons why it might; others assume it may matter in cases where they can't prove it doesn't. Such considerations are orthogonal to anything a processor might do with memory ordering. – supercat Mar 23 '21 at 22:19
@supercat: I understand that, but confusingly the kernel code linked does *not* use the `memory` constraint. Maybe because GCC understands it is implied by `lfence`? – Joseph Garvin Aug 09 '21 at 21:08
@JosephGarvin: If the kernel code uses lfence and mfence without memory clobbers, it likely does so because the authors thought it obvious that any quality compiler should recognize them as including an implied memory clobber; whether a gratuitously clever compiler would regard them likewise, or instead exploit the fact that more "optimizations" would be possible without a memory clobber, is a separate issue. – supercat Aug 09 '21 at 21:35

score 5 · Answer 2 · answered Jan 08 '13 at 11:44

5

you can use it like shown below:

asm volatile (
"CPUID\n\t"/*serialize*/
"RDTSC\n\t"/*read the clock*/
"mov %%edx, %0\n\t"
"mov %%eax, %1\n\t": "=r" (cycles_high), "=r"
(cycles_low):: "%rax", "%rbx", "%rcx", "%rdx");
/*
Call the function to benchmark
*/
asm volatile (
"RDTSCP\n\t"/*read the clock*/
"mov %%edx, %0\n\t"
"mov %%eax, %1\n\t"
"CPUID\n\t": "=r" (cycles_high1), "=r"
(cycles_low1):: "%rax", "%rbx", "%rcx", "%rdx");

In the code above, the first CPUID call implements a barrier to avoid out-of-order execution of the instructions above and below the RDTSC instruction. With this method we avoid to call a CPUID instruction in between the reads of the real-time registers

The first RDTSC then reads the timestamp register and the value is stored in memory. Then the code that we want to measure is executed. The RDTSCP instruction reads the timestamp register for the second time and guarantees that the execution of all the code we wanted to measure is completed. The two “mov” instructions coming afterwards store the edx and eax registers values into memory. Finally a CPUID call guarantees that a barrier is implemented again so that it is impossible that any instruction coming afterwards is executed before CPUID itself.

answered Jan 08 '13 at 11:44

Pranjal Verma

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1
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19

Hi, it appears that you copied this answer from Gabriele Paolinis white paper "How to Benchmark Code Execution Times on Intel® IA-32 and IA-64 Instruction Set Architectures" (you missed a line break though). You're using someone else's work without giving the author credit. Why not add an attribution? – Jonatan Lindén May 12 '16 at 06:13
Yes, indeed, it is coped. I'm also wondering if the two movs in reading the start time is necessary: http://stackoverflow.com/questions/38994549/is-intels-timestamp-reading-asm-code-example-using-two-more-registers-than-are – Edd Barrett Aug 17 '16 at 11:13
Is there a specific reason to have two variables high and low? – ExOfDe Oct 14 '16 at 14:45
1

Yes, @ExOfDe, there is a reason. The `RDTSC[P]` instruction returns a 64-bit value, but it returns it in two 32-bit halves: the upper half in the `EDX` register and the lower half in the `EAX` register (as is the common convention for returning 64-bit values on 32-bit x86 systems). You can, of course, combine those two 32-bit halves into a single 64-bit value if you want, but that requires either (A) a 64-bit processor (and the `RDTSC[P]` instruction was introduced to the ISA *long* before 64-bit integers were natively supported), or (B) compiler/library support for 64-bit ints. – Cody Gray - on strike Mar 29 '17 at 08:22
4

If you're going to use your own inline asm instead of a builtin/intrinsic, at least write efficient inline asm that uses constraints to tell the compiler which registers to look at, instead of using `mov` instructions. – Peter Cordes Jan 29 '18 at 14:45

Difference between rdtscp, rdtsc : memory and cpuid / rdtsc?

2 Answers2

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