两个语句执行之间的时间差不一致

Question

Could you please tell me why the value of timediff printed by the following program is often 4 microseconds (in the range 90 to 1000 times for different runs), but sometimes 70 or more microseconds for a few cases (in the range of 2 to 10 times for different runs):

#include <iostream>
using namespace std;
#include<sys/time.h>
#define MAXQ 1000000
#define THRDS 3
double GetMicroSecond()
{
    timeval tv;
    gettimeofday (&tv, NULL);
    return (double) (((double)tv.tv_sec * 1000000) + (double)tv.tv_usec);
}

int main()
{
        double timew, timer, timediff;
        bool flagarray[MAXQ];
        int x=0, y=0;
        for(int i=0; i<MAXQ; ++i)
            flagarray[i] = false;
        while(y <MAXQ)
       {
            x++;
            if(x%1000 == 0)
            {
                    timew = GetMicroSecond();
                    flagarray[y++]=true;
                    timer = GetMicroSecond();
                    timediff = timer - timew;
                    if(timediff > THRDS) cout << timer-timew << endl;
            }
       }
}

Compiled using: g++ testlatency.cpp -o testlatency

Note: In my system there are 12 cores. The performance is checked with only this program running in the system.

Answer 1

Generally, there are many threads sharing a small number of cores. Unless you take steps to ensure that your thread has uninterrupted use of a core, you can't guarantee that the OS won't decide to preempt your thread between the two calls GetMicroSecond() calls, and let some other thread use the core for a bit.

Even if your code runs uninterrupted, the line you're trying to time:

flagarray[y++]=true;

likely takes much less time to execute than the measurement code itself.

Answer 2

There are many things happening inside of modern OS at the same time as Your program executes. Some of them may may "steal" CPU from Your program as it is stated in NPE's answer. A few more examples of what can influence timing:

• interrups from devices (timer, HDD, network interfaces a few to mention);
• access to RAM (caching)

None of these are easily predictable.

You can expect consistency if You run Your code on some microcontroller, or maybe using real time OS .

Answer 3

timew = GetMicroSecond();
flagarray[y++]=true;
timer = GetMicroSecond();

The statement flagarray[y++]=true; will take much less than a microsecond to execute on a modern computer if flagarray[y++ ] happens to be in the level 1 cache. The statement will take longer to execute if that location is in level 2 cache but not in level 1 cache, much longer if it is in level 3 cache but not in level 1 or level 2 cache, and much, much longer yet if it isn't in any of the caches.

Another thing that can make timer-timew exceed three milliseconds is when your program yields to the OS. Cache misses can result in a yield. So can system calls. The function gettimeofday is a system call. As a general rule, you should expect any system call to yield.

Note: In my system there are 12 cores. The performance is checked with only this program running in the system.

This is not true. There are always many other programs, and many, many other threads running on your 12 core computer. These include the operating system itself (which comprises many threads in and of itself), plus lots and lots of little daemons. Whenever your program yields, the OS can decide to temporarily suspend your program so that one of the myriad other threads that are temporarily suspended but are asking for use of the CPU.

One of those daemons is the Network Time Protocol daemon (ntpd). This does all kinds of funky little things to your system clock to keep it close to in sync with atomic clocks. With a tiny little instruction such as flagarray[y++]=true being the only thing between successive calls to gettimeofday , you might even see time occasionally go backwards.

---

When testing for timing, its a good idea to do the timing at a coarse level. Don't time an individual statement that doesn't involve any function calls. It's much better to time a loop than it is to time than it is to time individual executions of the loop body. Even then, you should expect some variability in timing because of cache misses and because the OS temporarily suspends execution of your program.

Modern Unix-based systems have better timers (eg, clock_gettime ) than gettimeofday that are not subject to changes made by the Network Time Protocol daemon. You should use one of these rather than gettimeofday .

Answer 4

There are a lot of variables that might explain different time values seen. I would focus more on

• Cache miss/fill
• Scheduler Events

• Interrupts

  bool flagarray[MAXQ];

Since you defined MAXQ to 1000000, let's assume that flagarray takes up 1MB of space.

You can compute how many cache-misses can occur, based on your L1/L2 D-cache sizes. Then you can correlate with how many iterations it takes to fill all of L1 and start missing and same with L2. OS may deschedule your process and reschedule it - but, that I am hoping is less likely due to the number of cores you have. Same is the case with interrupts. An idle system is never completely idle. You may choose to affine your process to a core number, say N by doing

taskset 0x<MASK> ./exe and control its execution.

If you are really curious, I would suggest that you use "perf" tool available on most Linux distros.

You may do

perf stat -e L1-dcache-loadmisses

or

perf stat -e LLC-load-misses

Once you have these numbers and the number of iterations you start building a picture of the activity that causes the noticed lag. You may also monitor OS scheduler events using "perf stat".