Why does mktime give me an hour less?

Question

I would like to see if at 00:00:00 on January 1, 1970 it actually corresponds to 0 seconds, and I wrote the following:

#include <stdio.h>
#include <time.h>

int main(void) {
    int year = 1970;
    
    struct tm t = {0};
    
    t.tm_mday = 1; // January
    t.tm_year = year - 1900;
    t.tm_hour = 0;
    t.tm_isdst = -1;
    
    printf("%ld\n", mktime(&t));
    
    return 0;
}

it gives me a value of -3600. Where am I wrong?

PS: tested with GCC v.10.1. I tried with another compiler under another architecture and it gives me back the correct value.

Shouldn't it result in seconds from that date? Regardless of the time zone? — Roberto Rocco, Jul 12 '20 at 23:08
The number-of-seconds time is defined to be UTC, but `mktime()` is defined to work with *local* time; that's the difference — Steve Friedl, Jul 12 '20 at 23:26

score 2 · Answer 1 · answered Jul 12 '20 at 23:07

2

The mktime function takes a time in local time. Apparently, 00:00:00 at your local time was one hour before the epoch. Launch the program with TZ set to UTC.

answered Jul 12 '20 at 23:07

David Schwartz

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score 2 · Accepted Answer · answered Jul 12 '20 at 23:08

The time info you provide to mktime() is in local time, so the timezone matters even if summer time / daylight savings time does not.

You can fool your program by telling it you're in UTC:

$ gcc mytime.c -o mytime
$ ./mytime
28800          <-- Pacific time in the US
$ TZ=GMT0 ./mytime
0

chux - Reinstate Monica · Answer 3 · 2020-07-12T23:34:24.957

I would like to see if at 00:00:00 on January 1, 1970 it actually corresponds to 0 seconds, and I wrote the following:

00:00:00 on January 1, 1970 GMT, UTC corresponds to 0 seconds.

00:00:00 on January 1, 1970 Italia time corresponds to -3600 seconds.

Set timezone to UTC and then call mktime(). Unfortunately C does not have a portable way to do this, so the suggested code is only illustrative.

setenv("TZ", "UTC", 1);
tzset();
....
mktime(&t)

time_t does not necessarily match long. Recommend casting to a wide type.

// printf("%ld\n", mktime(&t));
printf("%lld\n", (long long) mktime(&t));

t.tm_mday = 1; // January misleads. .tm_mday is the day of the month, not January.

.tm_mon is the months since January so the initialization to 0 matches January.

Concerns about DST apply here only if the local time was using DST in January.

score 1 · Answer 4 · answered Jul 13 '20 at 02:06

As other answers indicate, mktime works in your local time zone. However, many operating systems offer a related function timegm that works in UTC. This slight modification of your program prints 0, as expected, on my computer:

#include <stdio.h>
#include <time.h>

int main(void)
{
    int year = 1970;
    
    struct tm t = {0};
    
    t.tm_mday = 1; // January
    t.tm_year = year - 1900;
    t.tm_hour = 0;
    t.tm_isdst = -1;
    
    printf("%ld\n", timegm(&t));
    
    return 0;
}

Regrettably, this function is not standardized. You may have to define a special "feature selection macro" to get your time.h to declare it.

Why does mktime give me an hour less?

4 Answers4