如何从gettimeofday创建NTP时间戳?

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我需要使用gettimeofday计算一个ntp时间戳。

#include <unistd.h>
#include <sys/time.h>

const unsigned long EPOCH = 2208988800UL; // delta between epoch time and ntp time
const double NTP_SCALE_FRAC = 4294967295.0; // maximum value of the ntp fractional part

int main()
{
  struct timeval tv;
  uint64_t ntp_time;
  uint64_t tv_ntp;
  double tv_usecs;

  gettimeofday(&tv, NULL);
  tv_ntp = tv.tv_sec + EPOCH;

  // convert tv_usec to a fraction of a second
  // next, we multiply this fraction times the NTP_SCALE_FRAC, which represents
  // the maximum value of the fraction until it rolls over to one. Thus,
  // .05 seconds is represented in NTP as (.05 * NTP_SCALE_FRAC)
  tv_usecs = (tv.tv_usec * 1e-6) * NTP_SCALE_FRAC;

  // next we take the tv_ntp seconds value and shift it 32 bits to the left. This puts the 
  // seconds in the proper location for NTP time stamps. I recognize this method has an 
  // overflow hazard if used after around the year 2106
  // Next we do a bitwise OR with the tv_usecs cast as a uin32_t, dropping the fractional
  // part
  ntp_time = ((tv_ntp << 32) | (uint32_t)tv_usecs);
}
提问于
用户回答回答于

这里不需要使用uint64_t- unsigned long long,保证至少64位宽。

你也不需要去往double,因为NTP_SCALE_FRAC * 1000000很容易适应unsigned long long

EPOCH应该是unsigned long long,不是unsigned long,所以添加tv.tv_sec

全部:

const unsigned long long EPOCH = 2208988800ULL;
const unsigned long long NTP_SCALE_FRAC = 4294967296ULL;

unsigned long long tv_to_ntp(struct timeval tv)
{
    unsigned long long tv_ntp, tv_usecs;

    tv_ntp = tv.tv_sec + EPOCH;
    tv_usecs = (NTP_SCALE_FRAC * tv.tv_usec) / 1000000UL;

    return (tv_ntp << 32) | tv_usecs;
}
用户回答回答于
extern uint64_t tvtontp64(struct timeval *tv) {
    uint64_t ntpts;

    ntpts = (((uint64_t)tv->tv_sec + 2208988800u) << 32) + ((uint32_t)tv->tv_usec * 4294.967296);

    return (ntpts);
}

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