从UNIX时间戳获取同月第一天的时间戳，而无需进行任何到本地时间的转换

Question

Suppose that I have UNIX timestamp like this

1252301772

and I need to convert it to another timestamp representing the first day of the same month (without any conversions to the local time or something like this). Example:

1252301772 -> 07 Sep 2009 05:36:12 GMT
x -> 01 Sep 2009 00:00:00 GMT

Answer 1

#include <time.h>

time_t first_of_the_month(time_t t)
{
  struct tm tmt;
  if (!gmtime_r(&t, &tmt))
    return -1;
  tmt.tm_sec = 0;
  tmt.tm_min = 0;
  tmt.tm_hour = 0;
  tmt.tm_mday = 1;
  tmt.tm_isdst = -1;
  return timegm(&tmt);
}

This uses the non-standard timegm function (thanks to Jeremy Friesner for pointing that out). If that function isn't available on your platform you can use mktime but that depends on the local timezone, so you would need to adjust the program's timezone (by setting TZ in the environment and calling tzset) before and after calling mktime , which is not thread-safe because it alters global program state.