C `clock()` 函数只返回一个零

问题描述

C clock() 函数只返回零.我尝试使用不同的类型，但没有任何改进......这是一种以高精度测量时间的好方法吗?

#include #include int main(){clock_t 开始，结束；双 cpu_time_used;字符 [32];开始 = 时钟()；printf("
睡眠 3 秒...

");睡眠(3)；结束 = 时钟()；cpu_time_used = ((double)(end - start))/((double)CLOCKS_PER_SEC);printf("start = %.20f
end = %.20f
", start, end);printf("delta = %.20f
", ((double) (end - start)));printf("cpu_time_used = %.15f
", cpu_time_used);printf("CLOCKS_PER_SEC = %i

", CLOCKS_PER_SEC);返回0；}

<块引用>

睡眠3秒...开始 = 0.00000000000000000000结束 = 0.00000000000000000000增量 = 0.00000000000000000000cpu_time_used = 0.000000000000000CLOCKS_PER_SEC = 1000000

平台:Intel 32 位，RedHat Linux，gcc 3.4.6

clock() 报告 CPU 使用时间.sleep() 不使用任何 CPU 时间.所以你的结果可能完全正确，只是不是你想要的.

The C clock() function just returns me a zero. I tried using different types, with no improvement... Is this a good way to measure time with good precision?

#include <time.h>
#include <stdio.h>

int main()
{
    clock_t start, end;
    double cpu_time_used;

    char s[32];

    start = clock();

    printf("
Sleeping 3 seconds...

");
    sleep(3);

    end = clock();

    cpu_time_used = ((double)(end - start)) / ((double)CLOCKS_PER_SEC);

    printf("start = %.20f
end   = %.20f
", start, end);
    printf("delta = %.20f
", ((double) (end - start)));
    printf("cpu_time_used  = %.15f
", cpu_time_used);
    printf("CLOCKS_PER_SEC = %i

", CLOCKS_PER_SEC);

    return 0;
}

Sleeping 3 seconds...

start = 0.00000000000000000000
end   = 0.00000000000000000000
delta = 0.00000000000000000000
cpu_time_used  = 0.000000000000000
CLOCKS_PER_SEC = 1000000

Platform: Intel 32 bit, RedHat Linux, gcc 3.4.6

clock() reports CPU time used. sleep() doesn't use any CPU time. So your result is probably exactly correct, just not what you want.

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