找出当地时区与格林威治标准时间的时差分钟

Question

我的时区格式为 GMT+09:00 和 GMT+10:00。我正在使用 Linux (Open Suse) 和 C++17。我想在几分钟内找出当地时区与 GMT 的时差。

这是我的示例程序。

static constexpr time_t const NULL_TIME = -1;

// returns difference in mintues
long tz_offset(time_t when = NULL_TIME) {
  if (when == NULL_TIME)
    when = std::time(nullptr);
  auto const tm = *std::localtime(&when);
  std::ostringstream os;
  os << std::put_time(&tm, "%z");
  std::string s = os.str();
  // s is in ISO 8601 format: "±HHMM"
  int h = std::stoi(s.substr(0, 3), nullptr, 10);
  int m = std::stoi(s[0] + s.substr(3), nullptr, 10);
  return h * 60 + m;
}

int main() {
  for (auto &timezone :  {"GMT+08:00", "GMT+09:30", "GMT+10:00", "GMT+10:30", "GMT+11:00"}) {
    setenv("TZ", timezone, 1);
    tzset();
    std::cout << "----------- TZ Changed to " << timezone << "---------------\n";
    std::cout << "difference in mins =" << tz_offset() << std::endl;
    auto tz_env_var = getenv("TZ");
    if (tz_env_var != nullptr) {
      std::cout << "TZ=" << tz_env_var << "\n";
    }
  }

  return 0;
}

但我得到的是负数的差异，它们应该是正数。

----------- TZ Changed to GMT+08:00---------------
difference in mins =-480
TZ=GMT+08:00
----------- TZ Changed to GMT+09:30---------------
difference in mins =-570
TZ=GMT+09:30
----------- TZ Changed to GMT+10:00---------------
difference in mins =-600
TZ=GMT+10:00
----------- TZ Changed to GMT+10:30---------------
difference in mins =-630
TZ=GMT+10:30
----------- TZ Changed to GMT+11:00---------------
difference in mins =-660
TZ=GMT+11:00

为什么我让他们处于负面状态？

Answer 1

这些字符串被称为 POSIX 时区 ，并且是 defined here。这个定义说明了 UTC 偏移量：

If preceded by a '-', the timezone shall be east of the Prime Meridian; otherwise, it shall be west (which may be indicated by an optional preceding '+' ).

同时 std::put_time 是根据 C/Posix strftime 定义的，它说明了偏移量：

%z is replaced by the offset from UTC in the ISO 8601 format ‘‘−0430’’ (meaning 4 hours 30 minutes behind UTC, west of Greenwich), or by no characters if no time zone is determinable. [tm_isdst]

总而言之，posix 时区字符串有一个约定（负数表示本初子午线东），另一个约定适用于几乎所有其他人，包括其他人posix 的部分（负数表示本初子午线西）。

所以您的代码实际上得到了正确答案。

Fwiw，这是一个 C++17 free, open-source, header-only Posix time zone library，它可以在不改变全局环境变量 TZ 的情况下完成工作，并且使用更简单的语法：

#include "date/ptz.h"
#include <chrono>
#include <iostream>

int
main()
{
    using namespace std::chrono;
    using date::operator<<;

    for (auto &timezone : {"GMT+08:00", "GMT+09:30", "GMT+10:00", "GMT+10:30", "GMT+11:00"})
    {
        auto offset = Posix::time_zone{timezone}.get_info(system_clock::now()).offset;
        std::cout << "difference is " << duration_cast<minutes>(offset) << '\n';
        std::cout << "TZ is " << timezone << "\n\n";
    }
}

输出：

difference is -480min
TZ is GMT+08:00

difference is -570min
TZ is GMT+09:30

difference is -600min
TZ is GMT+10:00

difference is -630min
TZ is GMT+10:30

difference is -660min
TZ is GMT+11:00

找出当地时区与格林威治标准时间的时差分钟

find the local time zone’s difference from GMT in minutes

c++

linux

timezone-offset

chrono

c++17