Get time_t / timeval corresponding to ( local_timezone ) midnight on a given date

Question

Given a time_t or struct timeval, how do I get the timeval or time_t of midnight EST/EDT ( local timezone ) on that day ? As in assuming local timezone is EST/EDT, given a time_t corresponding to say 2010-11-30 08:00:00 EST/EDT, the expected answer is a time_t that corresponds to 2010-11-30 00:00:00 EST/EDT

Attempt 1 ( Incorrect: since it does not handle DST, and assumes EST/EDT is always 5 hours behind UTC ):

time_t RewindToMidnight ( const time_t temp_time_t_ )
{
  return ( (5*3600) + ((( temp_time_t_ - 5*3600 )/86400 ) * 86400) );
}

Attempt 2 ( Incorrect: since it returns a time_t that corresponds to when it was midnight UTC and not EST/EDT, local timezone ):

time_t RewindToMidnight ( const time_t temp_time_t_ )
{
   boost::posix_time::ptime temp_ptime_ = boost::posix_time::from_time_t ( temp_time_t_ );
   boost::gregorian::date temp_date_ = temp_ptime_.date();
   boost::posix_time::ptime temp_ptime_midnight_ ( temp_date_,
                                                   boost::posix_time::time_duration ( 0, 0, 0 ) );
   return to_time_t ( temp_ptime_midnight_ );
}

time_t to_time_t ( const boost::posix_time::ptime & temp_ptime_ )
{
   boost::posix_time::ptime temp_epoch_ptime_(boost::gregorian::date(1970,1,1));
   boost::posix_time::time_duration::sec_type temp_sec_type_ = ( temp_ptime_ - temp_epoch_ptime_ ).total_seconds();
   return time_t ( temp_sec_type_ );
}

I feel there should be a solution involving (i) struct tm, mktime or (ii) boost::local_date_time perhaps ?

Answer 1

Current solution :

time_t RewindToMidnight ( const time_t & temp_time_t_ )
{
    int temp_yyyymmdd_ = iso_date_from_time_t ( temp_time_t_ );
    return time_t_from_iso_date ( temp_yyyymmdd_ );
}

int iso_date_from_time_t ( const time_t & in_time_t_ ) 
{
     tm temp_this_tm_;

     { // the following to set local dst fields of struct tm ?
         time_t tvsec_ = time(NULL);
         localtime_r ( & tvsec_, & temp_this_tm_ ) ;
     }
     localtime_r ( & in_time_t, & temp_this_tm_ ) ;

     return ( ( ( ( 1900 + temp_this_tm_.tm_year ) * 100 + ( 1 + temp_this_tm_.tm_mon ) ) * 100 ) + temp_this_tm_.tm_mday ) ;
}

time_t time_t_from_iso_date ( const int & temp_yyyymmdd_ ) 
{ 

     boost::gregorian::date d1 ( (int)( temp_yyyymmdd_/10000), 
                                 (int)( ( temp_yyyymmdd_/100) % 100), 
                                 (int)( temp_yyyymmdd_ % 100) );

     std::tm this_tm_ = to_tm ( d1 );
     return ( mktime ( & this_tm_ ) ) ;
}

Please advise.

Answer 2

As time_t is time in seconds since the Epoch (00:00:00 UTC, January 1, 1970), you just need to get rid of the seconds of the day. There are 86400 seconds in a day (leap seconds are normally ignored) so the result should be a multiple of 86400. Hence:

time_t now = time();
time_t midnight = now / 86400 * 86400

Answer 3

time_t local_midnight(time_t x) {
  struct tm t;
  localtime_r(&x, &t);
  t.tm_sec = t.tm_min = t.tm_hour = 0;
  return mktime(&t);
}

I used localtime_r since it must be available as you also used it in your answer.

Example:

int main() {
  time_t now = time(0);
  cout << "local: " << asctime(localtime(&now));
  cout << "UTC:   " << asctime(gmtime(&now));
  time_t midnight = local_midnight(now);
  cout << "\n       " << asctime(localtime(&midnight));
  return 0;
}