Question

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?

Answer 1

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(&lt); // 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.