使用 boost::date_time 在当前时区获取当前时间的最简单方法?

2022-01-24 00:00:00 debian timezone c++ boost boost-date-time

如果我在命令行 (Debian/Lenny) 上执行 date +%H-%M-%S，我会得到一个用户友好的(不是 UTC，不是 DST-less，时间 a正常人在他们的手表上)打印时间.

If I do date +%H-%M-%S on the commandline (Debian/Lenny), I get a user-friendly (not UTC, not DST-less, the time a normal person has on their wristwatch) time printed.

使用 boost::date_time 获得相同内容的最简单方法是什么?

What's the simplest way to obtain the same thing with boost::date_time ?

如果我这样做:

std::ostringstream msg;

boost::local_time::local_date_time t = 
  boost::local_time::local_sec_clock::local_time(
    boost::local_time::time_zone_ptr()
  );

boost::local_time::local_time_facet* lf(
  new boost::local_time::local_time_facet("%H-%M-%S")
);

msg.imbue(std::locale(msg.getloc(),lf));
msg << t;

那么 msg.str() 比我想看的时间早了一个小时.我不确定这是否是因为它显示 UTC 或本地时区时间而没有 DST 更正(我在英国).

Then msg.str() is an hour earlier than the time I want to see. I'm not sure whether this is because it's showing UTC or local timezone time without a DST correction (I'm in the UK).

修改上述内容以产生 DST 校正的本地时区时间的最简单方法是什么?我有一个想法，它涉及 boost::date_time:: c_local_adjustor 但无法从示例中弄清楚.

What's the simplest way to modify the above to yield the DST corrected local timezone time ? I have an idea it involves boost::date_time:: c_local_adjustor but can't figure it out from the examples.

推荐答案

这就是我想要的:

  namespace pt = boost::posix_time;
  std::ostringstream msg;
  const pt::ptime now = pt::second_clock::local_time();
  pt::time_facet*const f = new pt::time_facet("%H-%M-%S");
  msg.imbue(std::locale(msg.getloc(),f));
  msg << now;

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