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fix: combine date and time to convert loc datetime
Normally DST begins at A.M. 3 or 4. If we re-use conversion operator of local_date and local_time independently, the conversion fails if it is the day when DST begins or ends. Since local_date considers the time is 00:00 A.M. and local_time does not consider DST because it does not have any date information. We need to consider both date and time information at the same time to convert it correctly.
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@ -398,10 +398,31 @@ struct local_datetime
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{
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using internal_duration =
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typename std::chrono::system_clock::time_point::duration;
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// Normally DST begins at A.M. 3 or 4. If we re-use conversion operator
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// of local_date and local_time independently, the conversion fails if
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// it is the day when DST begins or ends. Since local_date considers the
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// time is 00:00 A.M. and local_time does not consider DST because it
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// does not have any date information. We need to consider both date and
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// time information at the same time to convert it correctly.
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std::tm t;
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t.tm_sec = static_cast<int>(this->time.second);
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t.tm_min = static_cast<int>(this->time.minute);
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t.tm_hour = static_cast<int>(this->time.hour);
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t.tm_mday = static_cast<int>(this->date.day);
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t.tm_mon = static_cast<int>(this->date.month);
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t.tm_year = static_cast<int>(this->date.year) - 1900;
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t.tm_wday = 0; // the value will be ignored
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t.tm_yday = 0; // the value will be ignored
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t.tm_isdst = -1;
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// std::mktime returns date as local time zone. no conversion needed
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auto dt = std::chrono::system_clock::time_point(this->date);
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auto dt = std::chrono::system_clock::from_time_t(std::mktime(&t));
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dt += std::chrono::duration_cast<internal_duration>(
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std::chrono::nanoseconds(this->time));
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std::chrono::milliseconds(this->time.millisecond) +
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std::chrono::microseconds(this->time.microsecond) +
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std::chrono::nanoseconds (this->time.nanosecond));
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return dt;
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}
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