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Raman Tennetieaae52c2021-03-11 16:17:50 -08001//===-- Implementation of mktime function ---------------------------------===//
2//
3// Part of the LLVM Project, under the Apache License v2.0 with LLVM Exceptions.
4// See https://llvm.org/LICENSE.txt for license information.
5// SPDX-License-Identifier: Apache-2.0 WITH LLVM-exception
6//
7//===----------------------------------------------------------------------===//
8
9#include "src/time/time_utils.h"
10#include "src/__support/common.h"
11
12#include <limits.h>
13
14namespace __llvm_libc {
15namespace time_utils {
16
17using __llvm_libc::time_utils::TimeConstants;
18
19static int64_t computeRemainingYears(int64_t daysPerYears,
20 int64_t quotientYears,
21 int64_t *remainingDays) {
22 int64_t years = *remainingDays / daysPerYears;
23 if (years == quotientYears)
24 years--;
25 *remainingDays -= years * daysPerYears;
26 return years;
27}
28
29// First, divide "total_seconds" by the number of seconds in a day to get the
30// number of days since Jan 1 1970. The remainder will be used to calculate the
31// number of Hours, Minutes and Seconds.
32//
33// Then, adjust that number of days by a constant to be the number of days
34// since Mar 1 2000. Year 2000 is a multiple of 400, the leap year cycle. This
35// makes it easier to count how many leap years have passed using division.
36//
37// While calculating numbers of years in the days, the following algorithm
38// subdivides the days into the number of 400 years, the number of 100 years and
39// the number of 4 years. These numbers of cycle years are used in calculating
40// leap day. This is similar to the algorithm used in getNumOfLeapYearsBefore()
41// and isLeapYear(). Then compute the total number of years in days from these
42// subdivided units.
43//
44// Compute the number of months from the remaining days. Finally, adjust years
45// to be 1900 and months to be from January.
46int64_t UpdateFromSeconds(int64_t total_seconds, struct tm *tm) {
47 // Days in month starting from March in the year 2000.
48 static const char daysInMonth[] = {31 /* Mar */, 30, 31, 30, 31, 31,
49 30, 31, 30, 31, 31, 29};
50
51 if (sizeof(time_t) == 4) {
52 if (total_seconds < 0x80000000)
53 return time_utils::OutOfRange();
54 if (total_seconds > 0x7FFFFFFF)
55 return time_utils::OutOfRange();
56 } else {
57 if (total_seconds <
58 INT_MIN * static_cast<int64_t>(
59 TimeConstants::NumberOfSecondsInLeapYear) ||
60 total_seconds > INT_MAX * static_cast<int64_t>(
61 TimeConstants::NumberOfSecondsInLeapYear))
62 return time_utils::OutOfRange();
63 }
64
65 int64_t seconds = total_seconds - TimeConstants::SecondsUntil2000MarchFirst;
66 int64_t days = seconds / TimeConstants::SecondsPerDay;
67 int64_t remainingSeconds = seconds % TimeConstants::SecondsPerDay;
68 if (remainingSeconds < 0) {
69 remainingSeconds += TimeConstants::SecondsPerDay;
70 days--;
71 }
72
73 int64_t wday = (TimeConstants::WeekDayOf2000MarchFirst + days) %
74 TimeConstants::DaysPerWeek;
75 if (wday < 0)
76 wday += TimeConstants::DaysPerWeek;
77
78 // Compute the number of 400 year cycles.
79 int64_t numOfFourHundredYearCycles = days / TimeConstants::DaysPer400Years;
80 int64_t remainingDays = days % TimeConstants::DaysPer400Years;
81 if (remainingDays < 0) {
82 remainingDays += TimeConstants::DaysPer400Years;
83 numOfFourHundredYearCycles--;
84 }
85
86 // The reminder number of years after computing number of
87 // "four hundred year cycles" will be 4 hundred year cycles or less in 400
88 // years.
89 int64_t numOfHundredYearCycles =
90 computeRemainingYears(TimeConstants::DaysPer100Years, 4, &remainingDays);
91
92 // The reminder number of years after computing number of
93 // "hundred year cycles" will be 25 four year cycles or less in 100 years.
94 int64_t numOfFourYearCycles =
95 computeRemainingYears(TimeConstants::DaysPer4Years, 25, &remainingDays);
96
97 // The reminder number of years after computing number of "four year cycles"
98 // will be 4 one year cycles or less in 4 years.
99 int64_t remainingYears = computeRemainingYears(
100 TimeConstants::DaysPerNonLeapYear, 4, &remainingDays);
101
102 // Calculate number of years from year 2000.
103 int64_t years = remainingYears + 4 * numOfFourYearCycles +
104 100 * numOfHundredYearCycles +
105 400LL * numOfFourHundredYearCycles;
106
107 int leapDay =
108 !remainingYears && (numOfFourYearCycles || !numOfHundredYearCycles);
109
110 int64_t yday = remainingDays + 31 + 28 + leapDay;
111 if (yday >= TimeConstants::DaysPerNonLeapYear + leapDay)
112 yday -= TimeConstants::DaysPerNonLeapYear + leapDay;
113
114 int64_t months = 0;
115 while (daysInMonth[months] <= remainingDays) {
116 remainingDays -= daysInMonth[months];
117 months++;
118 }
119
120 if (months >= TimeConstants::MonthsPerYear - 2) {
121 months -= TimeConstants::MonthsPerYear;
122 years++;
123 }
124
125 if (years > INT_MAX || years < INT_MIN)
126 return time_utils::OutOfRange();
127
128 // All the data (years, month and remaining days) was calculated from
129 // March, 2000. Thus adjust the data to be from January, 1900.
130 tm->tm_year = years + 2000 - TimeConstants::TimeYearBase;
131 tm->tm_mon = months + 2;
132 tm->tm_mday = remainingDays + 1;
133 tm->tm_wday = wday;
134 tm->tm_yday = yday;
135
136 tm->tm_hour = remainingSeconds / TimeConstants::SecondsPerHour;
137 tm->tm_min = remainingSeconds / TimeConstants::SecondsPerMin %
138 TimeConstants::SecondsPerMin;
139 tm->tm_sec = remainingSeconds % TimeConstants::SecondsPerMin;
140 // TODO(rtenneti): Need to handle timezone and update of tm_isdst.
141 tm->tm_isdst = 0;
142
143 return 0;
144}
145
146} // namespace time_utils
147} // namespace __llvm_libc