潜望镜时间戳

本文介绍了潜望镜时间戳的处理方法，对大家解决问题具有一定的参考价值，需要的朋友们下面随着小编来一起学习吧！ 问题描述

当我执行以下请求：POST signer.periscope.tv/sign HTTP在/1.1应用程式潜望镜它生成3时间戳。

When I perform the following request: POST signer.periscope.tv/sign HTTP/1.1 in Periscope app it generates 3 timestamps.

下面是那些时间戳的3个例子，但我想不出他们的意思。

Below are 3 examples of those timestamps, but I can't figure out what they mean.

我知道，3号永远是Linux的时间戳，但什么是1安培; 2，他们如何与Linux的时间戳？

I know that number 3 is always the Linux timestamp, but what are 1 & 2 and how are they related to the Linux timestamp?

什么是tpForBroadcasterFrame和时间戳是如何关联的？

What is "tpForBroadcasterFrame" and how is related to timestamp?

• 1 ntpForBroadcasterFrame：15704244410975025152
• 2 ntpForLiveFrame：15704244125303865344

• 3时间戳：1447440545

• 1 ntpForBroadcasterFrame: 15704244410975025152
• 2 ntpForLiveFrame : 15704244125303865344

• 3 timestamp : 1447440545

＃

3时间戳：1447440553

3 timestamp : 1447440553

＃

推荐答案

前两个看起来是网络时间协议的时间戳。第三个是更正确地称为POSIX时间戳。

The first two look to be Network Time Protocol time stamps. The third one is more correctly called a POSIX timestamp.

在一个无符号的64位整数，再presents自1900年以来的秒32位，小数秒32位NTP存储数据，所以......

NTP stores data in an unsigned 64 bit integer that represents 32 bits of seconds since 1900 and 32 bits of fractional seconds, so...

屏蔽关的15704244410975025152前32位给出了3656429334秒自1900年以来其他32位都没有映射到POSIX时间戳作为它的最小分辨率为1秒。

Masking off the first 32 bits of 15704244410975025152 gives 3656429334 seconds since 1900. The other 32 bits have no mapping to a POSIX timestamp as its minimum resolution is 1 second.

减去2208988800，the 1900年到1970年之间的秒数，从3656429334给人1447440534秒因为POSIX的时代，或星期五，2015年11月13日18时48分54秒GMT

Subtracting 2208988800, the number of seconds between 1900 and 1970, from 3656429334 gives 1447440534 seconds since the Posix epoch, or Fri, 13 Nov 2015 18:48:54 GMT

快速劈code：

#include <iostream> constexpr uint64_t epochdelta = 2208988800L; // number of seconds between 1900 and 1970 int main() { uint64_t num= 15704244410975025152ULL; uint32_t seconds = (uint32_t)(num >> 32); std::cout << seconds << " seconds since 1900" << std::endl; std::cout << seconds - epochdelta << " seconds since 1970" << std::endl; return 0; }