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