如何以毫秒为单位获得以C为单位的时间? (视窗)

时间:2013-06-22 12:52:17

标签: c time elapsed

我在网上搜索过,但我只是找到了一种方法,但这样就会以秒为单位而不是毫秒。

我的代码是:

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

int main(void)
{
    int solucion;

    time_t start, stop;
    clock_t ticks;
    long count;

    time(&start);
    solucion = divisores_it(92000000, 2);
    time(&stop);

    printf("Finnished in %f seconds. \n", difftime(stop, start));
    return 0;
}

6 个答案:

答案 0 :(得分:24)

跨平台的方式是使用ftime。

Windows特定链接:http://msdn.microsoft.com/en-us/library/aa297926(v=vs.60).aspx

以下示例。

#include <stdio.h>
#include <sys\timeb.h> 

int main()     
{ 
    struct timeb start, end;
    int diff;
    int i = 0;
    ftime(&start);

    while(i++ < 999) {
        /* do something which takes some time */
        printf(".");    
    }

    ftime(&end);
    diff = (int) (1000.0 * (end.time - start.time)
        + (end.millitm - start.millitm));

    printf("\nOperation took %u milliseconds\n", diff);
    return 0;
}

我运行上面的代码并使用VS2008跟踪它并看到它实际上调用了Windows GetSystemTimeAsFileTime函数。

无论如何,ftime会给你毫秒精度。

答案 1 :(得分:9)

以下解决方案对我来说似乎没问题。你觉得怎么样?

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

long timediff(clock_t t1, clock_t t2) {
    long elapsed;
    elapsed = ((double)t2 - t1) / CLOCKS_PER_SEC * 1000;
    return elapsed;
}

int main(void) {
    clock_t t1, t2;
    int i;
    float x = 2.7182;
    long elapsed;

    t1 = clock();
    for (i=0; i < 1000000; i++) {
           x = x * 3.1415; 
    }
    t2 = clock();

    elapsed = timediff(t1, t2);
    printf("elapsed: %ld ms\n", elapsed);


    return 0;
}

参考:http://www.acm.uiuc.edu/webmonkeys/book/c_guide/2.15.html#clock

答案 2 :(得分:2)

对于Windows,GetSystemTime()是您想要的。对于POSIX,gettimeofday()

答案 3 :(得分:1)

GetSystemTime()使用结构SYSTEMTIME,它提供毫秒级的分辨率。

More on this here

答案 4 :(得分:1)

此代码段有效。这是基于Angus Comber的回答:

#include <sys/timeb.h>

uint64_t system_current_time_millis()
{
#if defined(_WIN32) || defined(_WIN64)
    struct _timeb timebuffer;
    _ftime(&timebuffer);
    return (uint64_t)(((timebuffer.time * 1000) + timebuffer.millitm));
#else
    struct timeb timebuffer;
    ftime(&timebuffer);
    return (uint64_t)(((timebuffer.time * 1000) + timebuffer.millitm));
#endif
}

答案 5 :(得分:0)

DWORD start = GetTickCount();
executeSmth();
printf("Elapsed: %i ms", GetTickCount() - start);

P.S。此方法有一些限制。参见GetTickCount