Reputation: 1460
I'm trying to get the time in a different timezone (PST) using C++.
#define PST (-8);
char* Time::getSecondSystemTime() {
time_t rawtime;
struct tm * timeinfo;
char buffer[80];
time(&rawtime);
timeinfo = gmtime(&rawtime);
timeinfo->tm_hour = timeinfo->tm_hour + PST;
strftime(buffer, 80, "%I:%M %p", timeinfo);
std::string temp = std::string(buffer); // to get rid of extra stuff
std::string extraInfo = " Pacific Time ( US & Canada )";
temp.append(extraInfo);
return (char*) (temp.c_str());
}
The problem here is that when the GMT time is less than 8 hours (for example, right now, the time there is 3AM in the morning), subtracting 8 hours from it does not work!
What is the proper way to get time at a different time zone in Unix?
Upvotes: 1
Views: 6563
Reputation: 21
Here is a cleaner way of doing this (this example gets GMT time including DST bias):
struct STimeZoneFromRegistry
{
long Bias;
long StandardBias;
long DaylightBias;
SYSTEMTIME StandardDate;
SYSTEMTIME DaylightDate;
};
static SYSTEMTIME GmtNow()
{
FILETIME UTC;
GetSystemTimeAsFileTime(&UTC);
SYSTEMTIME GMT;
TIME_ZONE_INFORMATION tz = {0};
STimeZoneFromRegistry binary_data;
DWORD size = sizeof(binary_data);
HKEY hk = NULL;
TCHAR zone_key[] = _T("SOFTWARE\\Microsoft\\Windows NT\\CurrentVersion\\Time Zones\\GMT Standard Time");
if ((RegOpenKeyEx(HKEY_LOCAL_MACHINE, zone_key, 0, KEY_QUERY_VALUE, &hk) == ERROR_SUCCESS) &&
(RegQueryValueEx(hk, "TZI", NULL, NULL, (BYTE *) &binary_data, &size) == ERROR_SUCCESS))
{
tz.Bias = binary_data.Bias;
tz.DaylightBias = binary_data.DaylightBias;
tz.DaylightDate = binary_data.DaylightDate;
tz.StandardBias = binary_data.StandardBias;
tz.StandardDate = binary_data.StandardDate;
}
SystemTimeToTzSpecificLocalTime(&tz, &UTC, &GMT);
return GMT;
}
Upvotes: 2
Reputation: 753515
I have the following C code stashed away to deal with the problem. Efficiency isn't the first word that springs to mind (two calls to setenv()
, two calls to tzset()
), but the standard C library doesn't make it easy to do better:
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <time.h>
#include <unistd.h>
static void time_convert(time_t t0, char const *tz_value)
{
char old_tz[64];
strcpy(old_tz, getenv("TZ"));
setenv("TZ", tz_value, 1);
tzset();
char new_tz[64];
strcpy(new_tz, getenv("TZ"));
char buffer[64];
struct tm *lt = localtime(&t0);
strftime(buffer, sizeof(buffer), "%Y-%m-%d %H:%M:%S", lt);
setenv("TZ", old_tz, 1);
tzset();
printf("%ld = %s (TZ=%s)\n", (long)t0, buffer, new_tz);
}
int main(void)
{
time_t t0 = time(0);
char *tz = getenv("TZ");
time_convert(t0, tz);
time_convert(t0, "UTC0");
time_convert(t0, "IST-5:30");
time_convert(t0, "EST5");
time_convert(t0, "EST5EDT");
time_convert(t0, "PST8");
time_convert(t0, "PST8PDT");
}
In your original code, you have to worry about normalizing the time structure after changing the hour offset. You can do that with the mktime()
function. Here's a program based on the function in the question, which is pure C and avoids the problems of returning a pointer to a local variable (and the #define
ending with a semi-colon):
#include <assert.h>
#include <stdio.h>
#include <string.h>
#include <stdlib.h>
#include <time.h>
#define PST (-8)
extern int getSecondSystemTime(char *buffer, size_t buflen);
int getSecondSystemTime(char *buffer, size_t buflen)
{
time_t rawtime = time(0);;
struct tm *timeinfo;
char t_buff[32];
timeinfo = gmtime(&rawtime);
timeinfo->tm_hour = timeinfo->tm_hour + PST;
time_t pst_time = mktime(timeinfo);
assert(pst_time != (time_t)-1);
int len = strftime(t_buff, sizeof(t_buff), "%Y-%m-%d %H:%M:%S", timeinfo);
assert(len != 0);
int rv = snprintf(buffer, buflen, "%ld = %s (%s)", (long)rawtime, t_buff,
"Pacific Time (US & Canada)");
assert(rv > 0);
return rv;
}
int main(void)
{
char buffer[128];
getSecondSystemTime(buffer, sizeof(buffer));
printf("%s\n", buffer);
return(0);
}
Clearly, a better interface would pass the UTC time value and the time zone offset (in hours and minutes) as arguments. Despite the fact that my computer runs in US/Pacific (or America/Los_Angeles) time zone by default, I tested with TZ set to various values (including US/Eastern, IST-05:30) and got the correct values out; I'm reasonably convinced based on past experience that the calculation is correct.
I have another program that attempts to dissect whether the -1
returned from mktime()
is because of an error or because the converted time corresponds to (time_t)-1
:
/* Attempt to determine whether time is really 1969-12-31 23:59:59 +00:00 */
static int unix_epoch_minus_one(const struct tm *lt)
{
printf("tm_sec = %d\n", lt->tm_sec);
if (lt->tm_sec != 59)
return(0);
printf("tm_min = %d\n", lt->tm_min);
/* Accounts for time zones such as Newfoundland (-04:30), India (+05:30) and Nepal (+05:45) */
if (lt->tm_min % 15 != 14)
return(0);
/* Years minus 1900 */
printf("tm_year = %d\n", lt->tm_year);
if (lt->tm_year != 69 && lt->tm_year != 70)
return(0);
printf("tm_mday = %d\n", lt->tm_mday);
if (lt->tm_mday != 31 && lt->tm_mday != 1)
return(0);
/* Months 0..11 */
printf("tm_mon = %d\n", lt->tm_mon);
if (lt->tm_mon != 11 && lt->tm_mon != 0)
return(0);
/* Pretend it is valid after all - though there is a small chance we are incorrect */
return 1;
}
Upvotes: 4
Reputation: 16379
Since you said "UNIX", this uses TZ, but, TZ=[what goes here]
You need to find out [what goes here] on your system. It might be "America/LosAngeles" or one of several other strings for PST.
If your system is POSIX: TZ=PST8PST is guaranteed to work. But it may not be optimal.
Primitive non-production code assumes TZ is not currently in use. This is in C, not C++ since your tag was C:
setenv("TZ", "PST8PST", 1); // set TZ
tzset(); // recognize TZ
time_t lt=time(NULL); //epoch seconds
struct tm *p=localtime(<); // get local time struct tm
char tmp[80]={0x0};
strftime(tmp, 80, "%c", p); // format time use format string, %c
printf("time and date PST: %s\n", tmp); // display time and date
// you may or may not want to remove the TZ variable at this point.
Upvotes: 7