unsigned int lo = 0;
unsigned int hi = 0;
__asm__ __volatile__ (
\"mfence;rdtsc\" : \"=a\"(lo), \"=d\"(hi) : : \"memory\"
);
mfence
mfence is there to force serialization in CPU before rdtsc.
Usually you will find cpuid there (which is also serializing instruction).
Quote from Intel manuals about using rdtsc will make it clearer
Starting with the Intel Pentium processor, most Intel CPUs support out-of-order execution of the code. The purpose is to optimize the penalties due to the different instruction latencies. Unfortunately this feature does not guarantee that the temporal sequence of the single compiled C instructions will respect the sequence of the instruction themselves as written in the source C file. When we call the RDTSC instruction, we pretend that that instruction will be executed exactly at the beginning and at the end of code being measured (i.e., we don’t want to measure compiled code executed outside of the RDTSC calls or executed in between the calls themselves). The solution is to call a serializing instruction before calling the RDTSC one. A serializing instruction is an instruction that forces the CPU to complete every preceding instruction of the C code before continuing the program execution. By doing so we guarantee that only the code that is under measurement will be executed in between the RDTSC calls and that no part of that code will be executed outside the calls.
TL;DR version - without serializing instruction before rdtsc you have no idea when that instruction started to execute making measurements possibly incorrect.
HINT - use rdtscp when possible.
Based on my test, cpu reorder is not found.
Still no guarantee that it may happen - that's why original code had "memory"
to indicate possible memory clobber preventing compiler from reordering it.
What you need to perform a sensible measurement with rdtsc
is a serializing instruction.
As it is well known, a lot of people use cpuid
before rdtsc
.
rdtsc
needs to be serialized from above and below (read: all instructions before it must be retired and it must be retired before the test code starts).
Unfortunately the second condition is often neglected because cpuid
is a very bad choice for this task (it clobbers the output of rdtsc
).
When looking for alternatives people think that instructions that have a "fence" in their names will do, but this is also untrue. Straight from Intel:
MFENCE does not serialize the instruction stream.
An instruction that is almost serializing and will do in any measurement where previous stores don't need to complete is lfence
.
Simply put, lfence
makes sure that no new instructions start before any prior instruction completes locally. See this answer of mine for a more detailed explanation on locality.
It also doesn't drain the Store Buffer like mfence
does and doesn't clobbers the registers like cpuid
does.
So lfence / rdtsc / lfence
is a better crafted sequence of instructions than mfence / rdtsc
, where mfence
is pretty much useless unless you explicitly want the previous stores to be completed before the test begins/ends (but not before rdstc
is executed!).
If your test to detect reordering is assert(t2 > t1)
then I believe you will test nothing.
Leaving out the return
and the call that may or may not prevent the CPU from seeing the second rdtsc
in time for a reorder, it is unlikely (though possible!) that the CPU will reorder two rdtsc
even if one is right after the other.
Imagine we have a rdtsc2
that is exactly like rdtsc
but writes ecx:ebx
1.
Executing
rdtsc
rdtsc2
is highly likely that ecx:ebx > edx:eax
because the CPU has no reason to execute rdtsc2
before rdtsc
.
Reordering doesn't mean random ordering, it means look for other instruction if the current one cannot be executed.
But rdtsc
has no dependency on any previous instruction, so it's unlikely to be delayed when encountered by the OoO core.
However peculiar internal micro-architectural details may invalidate my thesis, hence the likely word in my previous statement.
1 We don't need this altered instruction: register renaming will do it, but in case you are not familiar with it, this will help.