1 //===-- Implementation of mktime function ---------------------------------===//
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
7 //===----------------------------------------------------------------------===//
9 #include "src/time/time_utils.h"
10 #include "src/__support/CPP/limits.h" // INT_MIN, INT_MAX
11 #include "src/__support/common.h"
12 #include "src/__support/macros/config.h"
14 namespace LIBC_NAMESPACE_DECL
{
15 namespace time_utils
{
17 using LIBC_NAMESPACE::time_utils::TimeConstants
;
19 static int64_t computeRemainingYears(int64_t daysPerYears
,
20 int64_t quotientYears
,
21 int64_t *remainingDays
) {
22 int64_t years
= *remainingDays
/ daysPerYears
;
23 if (years
== quotientYears
)
25 *remainingDays
-= years
* daysPerYears
;
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.
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.
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
44 // Compute the number of months from the remaining days. Finally, adjust years
45 // to be 1900 and months to be from January.
46 int64_t update_from_seconds(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};
51 constexpr time_t time_min
=
54 : INT_MIN
* static_cast<int64_t>(
55 TimeConstants::NUMBER_OF_SECONDS_IN_LEAP_YEAR
);
56 constexpr time_t time_max
=
59 : INT_MAX
* static_cast<int64_t>(
60 TimeConstants::NUMBER_OF_SECONDS_IN_LEAP_YEAR
);
62 time_t ts
= static_cast<time_t>(total_seconds
);
63 if (ts
< time_min
|| ts
> time_max
)
64 return time_utils::out_of_range();
67 total_seconds
- TimeConstants::SECONDS_UNTIL2000_MARCH_FIRST
;
68 int64_t days
= seconds
/ TimeConstants::SECONDS_PER_DAY
;
69 int64_t remainingSeconds
= seconds
% TimeConstants::SECONDS_PER_DAY
;
70 if (remainingSeconds
< 0) {
71 remainingSeconds
+= TimeConstants::SECONDS_PER_DAY
;
75 int64_t wday
= (TimeConstants::WEEK_DAY_OF2000_MARCH_FIRST
+ days
) %
76 TimeConstants::DAYS_PER_WEEK
;
78 wday
+= TimeConstants::DAYS_PER_WEEK
;
80 // Compute the number of 400 year cycles.
81 int64_t numOfFourHundredYearCycles
= days
/ TimeConstants::DAYS_PER400_YEARS
;
82 int64_t remainingDays
= days
% TimeConstants::DAYS_PER400_YEARS
;
83 if (remainingDays
< 0) {
84 remainingDays
+= TimeConstants::DAYS_PER400_YEARS
;
85 numOfFourHundredYearCycles
--;
88 // The remaining number of years after computing the number of
89 // "four hundred year cycles" will be 4 hundred year cycles or less in 400
91 int64_t numOfHundredYearCycles
= computeRemainingYears(
92 TimeConstants::DAYS_PER100_YEARS
, 4, &remainingDays
);
94 // The remaining number of years after computing the number of
95 // "hundred year cycles" will be 25 four year cycles or less in 100 years.
96 int64_t numOfFourYearCycles
=
97 computeRemainingYears(TimeConstants::DAYS_PER4_YEARS
, 25, &remainingDays
);
99 // The remaining number of years after computing the number of
100 // "four year cycles" will be 4 one year cycles or less in 4 years.
101 int64_t remainingYears
= computeRemainingYears(
102 TimeConstants::DAYS_PER_NON_LEAP_YEAR
, 4, &remainingDays
);
104 // Calculate number of years from year 2000.
105 int64_t years
= remainingYears
+ 4 * numOfFourYearCycles
+
106 100 * numOfHundredYearCycles
+
107 400LL * numOfFourHundredYearCycles
;
110 !remainingYears
&& (numOfFourYearCycles
|| !numOfHundredYearCycles
);
112 // We add 31 and 28 for the number of days in January and February, since our
113 // starting point was March 1st.
114 int64_t yday
= remainingDays
+ 31 + 28 + leapDay
;
115 if (yday
>= TimeConstants::DAYS_PER_NON_LEAP_YEAR
+ leapDay
)
116 yday
-= TimeConstants::DAYS_PER_NON_LEAP_YEAR
+ leapDay
;
119 while (daysInMonth
[months
] <= remainingDays
) {
120 remainingDays
-= daysInMonth
[months
];
124 if (months
>= TimeConstants::MONTHS_PER_YEAR
- 2) {
125 months
-= TimeConstants::MONTHS_PER_YEAR
;
129 if (years
> INT_MAX
|| years
< INT_MIN
)
130 return time_utils::out_of_range();
132 // All the data (years, month and remaining days) was calculated from
133 // March, 2000. Thus adjust the data to be from January, 1900.
134 tm
->tm_year
= static_cast<int>(years
+ 2000 - TimeConstants::TIME_YEAR_BASE
);
135 tm
->tm_mon
= static_cast<int>(months
+ 2);
136 tm
->tm_mday
= static_cast<int>(remainingDays
+ 1);
137 tm
->tm_wday
= static_cast<int>(wday
);
138 tm
->tm_yday
= static_cast<int>(yday
);
141 static_cast<int>(remainingSeconds
/ TimeConstants::SECONDS_PER_HOUR
);
143 static_cast<int>(remainingSeconds
/ TimeConstants::SECONDS_PER_MIN
%
144 TimeConstants::SECONDS_PER_MIN
);
146 static_cast<int>(remainingSeconds
% TimeConstants::SECONDS_PER_MIN
);
147 // TODO(rtenneti): Need to handle timezone and update of tm_isdst.
153 } // namespace time_utils
154 } // namespace LIBC_NAMESPACE_DECL