将struct tm转换为time_t

Question

我有以下代码:

struct tm time;

strptime("27052010", "%d%m%Y", &time);

cout << "sec: " << time.tm_sec << "\n";
cout << "min: " << time.tm_min << "\n";
cout << "hour: " << time.tm_hour << "\n";
cout << "day: " << time.tm_mday << "\n";
cout << "month: " << (time.tm_mon + 1) << "\n";
cout << "year: " << time.tm_year << "\n";

time_t t = mktime(&time);

cout << "sec: " << time.tm_sec << "\n";
cout << "min: " << time.tm_min << "\n";
cout << "hour: " << time.tm_hour << "\n";
cout << "day: " << time.tm_mday << "\n";
cout << "month: " << (time.tm_mon + 1) << "\n";
cout << "year: " << time.tm_year << "\n";

cout << "time: " << t << "\n";

输出是:

sec: 1474116832
min: 32767
hour: 4238231
day: 27
month: 5
year: 110

sec: 52
min: 0
hour: 6
day: 2
month: 9
year: 640
time: 18008625652 (Fri, 02 Sep 2540 04:00:52 GMT)

我的问题是为什么mktime()更改值time和为什么转换time_t不等于我的输入日期.我希望输出是自1970年以来以秒为单位表示的日期(2010年5月27日= 1330905600).

提前致谢

Answer 1

mktime在转换为a之前规范化所有参数time_t.你有小时,分钟和秒的巨大价值,所以这些都被转换成适当的天数,将价值推向了未来.

您需要将tm之前调用的其他重要属性(包括小时/分钟/秒)归零mktime.正如注释中所述,只需将其初始化为零:( tm time = {0};标记为C++,因此struct不需要前导).另请注意,您可能希望设置tm_isdst为-1,以便它尝试确定夏令时值,而不是假定不是DST(如果初始化为零).