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给定输入日期和时间（以字符串格式），我试图使用like中time给出的函数获取它的纪元时间。将纪元时间转换回日期和时间会导致日期和时间比原始时间少一小时。我经历了一些讨论，说在夏令时的情况下可能会调整一小时。这是代码： ctimemktimetime_t

//sample strptime program.

#include <iostream>
#include <ctime>
#include <string>
using namespace std;

long parseTime(string time) {

  cout << "Time entered = " << time << endl;

  long timeSinceEpoch;

  struct tm t;

  if(time.find("/") != string::npos) {
    //format of date is mm/dd/yyyy. followed by clock in hh:mm (24 hour clock).
    if(strptime(time.c_str(), "%m/%e/%Y %H:%M", &t) == NULL) {
      cout << "Error. Check string for formatting." << endl;
    }
  } else if(time.find("-") != string::npos) {
    //format of date is yyyy-mm-dd hh:mm:ss (hh in 24 hour clock format).
    cout << "I am here." << endl;
    if(strptime(time.c_str(), "%Y-%m-%e %H:%M:%S", &t) == NULL) {
      cout << "Error. Check string for formatting of new date." << endl;
    }
  }

  cout << "Details of the time structure:" << endl;
  cout << "Years since 1900 = " << t.tm_year << endl;
  cout << "Months since January = " << t.tm_mon << endl;
  cout << "Day of the month = " << t.tm_mday << endl;
  cout << "Hour = " << t.tm_hour << " Minute = " << t.tm_min << " second = " << t.tm_sec << endl;

  timeSinceEpoch = mktime(&t);
  time_t temp = mktime(&t);
  cout << "Time since epoch = " << timeSinceEpoch << endl;

  cout << "Reconverting to the time structure:" << endl;
  struct tm* t2 = localtime(&temp);
  cout << "Details of the time structure:" << endl;
  cout << "Years since 1900 = " << t2->tm_year << endl;
  cout << "Months since January = " << t2->tm_mon << endl;
  cout << "Day of the month = " << t2->tm_mday << endl;
  cout << "Hour = " << t2->tm_hour << " Minute = " << t2->tm_min << " second = " << t2->tm_sec << endl;

  return timeSinceEpoch;
}

int main(int argc, char *argv[]) {

  string date, t;
  cout << "Enter date: " << endl;
  cin >> date;
  cout << "Enter time: " << endl;
  cin >> t;

  struct tm time;
  string overall = date + " " + t;

  long result = parseTime(overall);
  cout << "Time in date + time = " << overall << " and since epoch = " << result << endl;

return 0;
}

麻烦的输入是：日期：2013-03-11 时间：04:41:53

我的问题：
1. 检查tm_idst标志返回非零，表示 DST 有效。但是，我怎么知道正在谈论什么时区？
2. 上面给出的时间戳可能与我所在的时区不同。有没有办法指定时区以便tm_idst正确设置标志？
3. 不确定时间戳记录在哪个时区时，如何处理夏令时？

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1 回答 1

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PlainC++的时区数据非常稀疏，并且在要格式化的时间中没有时区规范，有几个时期您会得到不一致的结果 - 例如，时钟回溯后的重复时间。这就是为什么总是建议记录所有时间戳的原因UTC- 即永远不要将时区应用于记录的时间戳，将其记录在 GMT 中，然后将该值作为显示变量来回切换，您可以完全控制它。

Linux/BSD 有一些额外的字段可用于确定时区和与 UTC 的偏移量 - 例如在 linux 上它是__tm_gmtoff字段，而在 BSD(/Mac OS X) 中它被称为tm_gmtoff.

__tm_zone在 linux 和BSD(/Mac OS X) 上还有一个标记时区的附加字段tm_zone，但该字段仅在您获取本地时间时才会填充。

我稍微改变了你的例子，得到了以下输出：

Time entered = 2013-04-05 15:00
I am here.
Error. Check string for formatting of new date.
Details of the time structure:
Years since 1900 = 113
Months since January = 3
Day of the month = 5
Hour = 15 Minute = 0 second = 0
gmtoff = 0
Time since epoch = 1365174000
Reconverting to the time structure:
Details of the time structure:
Years since 1900 = 113
Months since January = 3
Day of the month = 5
Hour = 16 Minute = 0 second = 0
gmtoff = 3600
Zone = IST
Time in date + time = 2013-04-05 15:00 and since epoch = 1365174000

但是，如果您在 Windows 中使用此结构，您将不得不使用另一种机制，因为它没有这两个额外的字段。

于 2013-03-27T19:36:17.103 回答