nghttp2/src/timegm.c

76 lines
2.3 KiB
C

/*
* Spdylay - SPDY Library
*
* Copyright (c) 2013 Tatsuhiro Tsujikawa
*
* Permission is hereby granted, free of charge, to any person obtaining
* a copy of this software and associated documentation files (the
* "Software"), to deal in the Software without restriction, including
* without limitation the rights to use, copy, modify, merge, publish,
* distribute, sublicense, and/or sell copies of the Software, and to
* permit persons to whom the Software is furnished to do so, subject to
* the following conditions:
*
* The above copyright notice and this permission notice shall be
* included in all copies or substantial portions of the Software.
*
* THE SOFTWARE IS PROVIDED "AS IS", WITHOUT WARRANTY OF ANY KIND,
* EXPRESS OR IMPLIED, INCLUDING BUT NOT LIMITED TO THE WARRANTIES OF
* MERCHANTABILITY, FITNESS FOR A PARTICULAR PURPOSE AND
* NONINFRINGEMENT. IN NO EVENT SHALL THE AUTHORS OR COPYRIGHT HOLDERS BE
* LIABLE FOR ANY CLAIM, DAMAGES OR OTHER LIABILITY, WHETHER IN AN ACTION
* OF CONTRACT, TORT OR OTHERWISE, ARISING FROM, OUT OF OR IN CONNECTION
* WITH THE SOFTWARE OR THE USE OR OTHER DEALINGS IN THE SOFTWARE.
*/
#include "timegm.h"
#ifndef HAVE_TIMEGM
#include <stdint.h>
/* Counter the number of leap year in the range [0, y). The |y| is the
year, including century (e.g., 2012) */
static int count_leap_year(int y)
{
y -= 1;
return y/4-y/100+y/400;
}
/* Returns nonzero if the |y| is the leap year. The |y| is the year,
including century (e.g., 2012) */
static int is_leap_year(int y)
{
return y%4 == 0 && (y%100 != 0 || y%400 == 0);
}
/* The number of days before ith month begins */
static int daysum[] = {
0, 31, 59, 90, 120, 151, 181, 212, 243, 273, 304, 334
};
/* Based on the algorithm of Python 2.7 calendar.timegm. */
time_t timegm(struct tm *tm)
{
int days;
int num_leap_year;
int64_t t;
if(tm->tm_mon > 11) {
return -1;
}
num_leap_year = count_leap_year(tm->tm_year + 1900) - count_leap_year(1970);
days = (tm->tm_year - 70) * 365 +
num_leap_year + daysum[tm->tm_mon] + tm->tm_mday-1;
if(tm->tm_mon >= 2 && is_leap_year(tm->tm_year + 1900)) {
++days;
}
t = ((int64_t)days * 24 + tm->tm_hour) * 3600 + tm->tm_min * 60 + tm->tm_sec;
if(sizeof(time_t) == 4) {
if(t < INT32_MIN || t > INT32_MAX) {
return -1;
}
}
return t;
}
#endif /* !HAVE_TIMEGM */