Question

#include <stdlib.h>
#include <stdio.h>
#include <time.h>
#include <unistd.h>
#include <time.h>
typedef unsigned int uint32;
uint32 a;

int main()
{
            struct timespec start;

            if((clock_gettime( CLOCK_REALTIME, &start)) == -1 )
            {
              perror("clock gettime\n");

            }

/* I am calculating the Clock granularity here the granularity is basically how long that timer interrupt will last while it's processing the background task.*/
            //micro seconds output
        a = (1000000 * start.tv_sec + start.tv_nsec / 1000);

         printf( "%u\n", a);
            return EXIT_SUCCESS;
}

我创建了一个计时器来获取任何地点的时间戳，所以上面是一个自由运行的计时器来获取时间戳。我试图以微秒为单位得到输出并得到值：2847675807这是对的？我应该以微秒为单位获得该值。我认为，获得一些更大的价值观。有人请帮帮我。

typedef unsigned int uint64;
typedef unsigned int uint32;
uint32 a;

uint64 timestamp()
{
            struct timespec start;

            if((clock_gettime( CLOCK_REALTIME, &start)) == -1 )
            {
              perror("clock gettime\n");

            }

/* I am calculating the Clock granularity here the granularity is basically how long that timer interrupt
 * will last while it's processing the background task.*/
            //micro seconds output
        a = (uint32)(1e6 * start.tv_sec + start.tv_nsec * 1e-3);

         printf( "%u\n", a);
            return EXIT_SUCCESS;
}


int main()
{
timestamp();
return 1;
}

我像上面那样进行了修改但是后面也有相同的结果，比如更大的数字

Answer 1

也许你可以试试这个。它不是实时的，它不如gettimeofday准确。

#include <time.h>
#include <stdio.h>
int main(int ac, char **av)
 {
  clock_t       start, end;
  double        duration = 0;

  start = clock();
  for (int i = 0; i < 100000000; i++)
    {}
  end = clock();
  duration = (double)(end - start) / CLOCKS_PER_SEC;
  printf("%f ms\n", duration);
  return 0;
 }

如何以微秒为单位得到当前时间的结果？

1 个答案: