有没有一种方法可以使用跨平台方法在 C 语言中获得毫秒精度、准确(至少在几毫秒内)的时间?
在 POSIX 系统上我可以使用 sys/time.h,但这不是跨平台的。
stdlib time() 函数只提供秒级精度
最佳答案
我还没有找到一个跨平台的解决方案来测量 C 中的时间,就其本身而言。然而,我所做的是对 Unix 和 Windows 使用几乎相同的函数。我创建了这个 gist因为我总是每次都要重新查找。简而言之:
Unix
#include <time.h>
long diff_micro(struct timespec *start, struct timespec *end)
{
/* us */
return ((end->tv_sec * (1000000)) + (end->tv_nsec / 1000)) -
((start->tv_sec * 1000000) + (start->tv_nsec / 1000));
}
long diff_milli(struct timespec *start, struct timespec *end)
{
/* ms */
return ((end->tv_sec * 1000) + (end->tv_nsec / 1000000)) -
((start->tv_sec * 1000) + (start->tv_nsec / 1000000));
}
int main(int argc, char **argv)
{
struct timespec start, end;
clock_gettime(CLOCK_MONOTONIC, &start);
// Activity to be timed
sleep(1000);
clock_gettime(CLOCK_MONOTONIC, &end);
printf("%ld us\n", diff_micro(&start, &end));
printf("%ld ms\n", diff_milli(&start, &end));
return 0;
}
Win32
#include <Windows.h>
long diff_micro(LARGE_INTEGER *start, LARGE_INTEGER *end)
{
LARGE_INTEGER Frequency, elapsed;
QueryPerformanceFrequency(&Frequency);
elapsed.QuadPart = end->QuadPart - start->QuadPart;
elapsed.QuadPart *= 1000000;
elapsed.QuadPart /= Frequency.QuadPart;
return elapsed.QuadPart;
}
long diff_milli(LARGE_INTEGER *start, LARGE_INTEGER *end)
{
LARGE_INTEGER Frequency, elapsed;
QueryPerformanceFrequency(&Frequency);
elapsed.QuadPart = end->QuadPart - start->QuadPart;
elapsed.QuadPart *= 1000;
elapsed.QuadPart /= Frequency.QuadPart;
return elapsed.QuadPart;
}
int main(int argc, char **argv)
{
LARGE_INTEGER StartingTime, EndingTime;
QueryPerformanceCounter(&StartingTime);
// Activity to be timed
Sleep(1000);
QueryPerformanceCounter(&EndingTime);
printf("%ld us\n", diff_micro(&StartingTime, &EndingTime));
printf("%ld ms\n", diff_milli(&StartingTime, &EndingTime));
return 0;
}
关于c - C - 跨平台中函数的毫秒精度计时,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/17639213/