每字节的负周期? rdtsc C.

时间:2013-07-31 12:19:52

标签: c performance benchmarking cpu-usage

我写了一些代码来测量每个字节的cpu周期。我得到否定cpb但不知道为什么......它向我显示cpb = -0.855553 cycles/byte

我的伪代码

#include <stdio.h>
#include <string.h>
#include <stdlib.h>

uint64_t rdtsc(){
    unsigned int lo,hi;
    __asm__ __volatile__ ("rdtsc" : "=a" (lo), "=d" (hi));
    return ((uint64_t)hi << 32) | lo;
}

int main()
{
    long double inputsSize = 1024;
    long double counter = 1;

    long double cpuCycleStart = rdtsc();

        while(counter < 3s)
            function(args);

    long double cpuCycleEnd = rdtsc();

        long double cpb = ((cpuCycleEnd - cpuCycleStart) / (counter *  inputsSize));

    printf("%Lf cycles/byte\n", cpb);

    return 0;
}

编辑,改进的代码,结果是相同的(负面):

#include <stdio.h>
#include <string.h>
#include <stdlib.h>

unsigned long rdtsc( void )
    {
        unsigned long lo, hi;
        asm( "rdtsc" : "=a" (lo), "=d" (hi) );
        return( lo );
    }

int main()
{
    long double counter;
    long double inputsSize = 1024;
    char *buff = createInput(inputsSize);

    long double cpuCycleStart = rdtsc();
        countDownTime(3.0);
    for(counter=1; !secondsElapsed; counter++)
            function(args);
    long cpuCycleEnd = rdtsc();

        long double cpb = ((cpuCycleEnd - cpuCycleStart) / (counter *  inputsSize));

    printf("%Lf cycles/byte\n", cpb);

    return 0;
}

真的很奇怪。写了测试代码:

printf("\n%lu cpuCycleEnd \n%lu cpuCycleStart \n", cpuCycleEnd, cpuCycleStart);
    printf("\n%lu counter\n%lu inputsSize \n\n", counter, inputsSize);

        long double cpb = (((long double)cpuCycleEnd - (long double)cpuCycleStart) / ((long double)counter *  (long double)inputsSize));

    printf("%Lf cycles/byte\n", cpb);

显示:

30534991 cpuCycleEnd 
1139165971 cpuCycleStart 

1273029 counter
1024 inputsSize 

-0.850450 cycles/byte

任何想法?

1 个答案:

答案 0 :(得分:4)

您正在为unsigned long为32位的目标进行编译。

您应#include <stdint.h>并使用uint64_t代替unsigned long。此外,您可能希望编译unsigned long为64位的目标。

(注意:打印uint64_t#include <inttypes.h>并使用printf("%" PRIu64 "\n", value);。)