[SOLVED] problem with time zone change in c++

cbaurand · 09-18-2012, 01:59 AM

Hi all,

I have written a little function which can get the time zone value from a time zone name.
Here is the code:

Code:

   std::string getTimeZone(const std::string& zoneName)
    {
        struct tm *t1, *t2;
        time_t tm1, tm2;

        // LOCAL TIME (Europe/Zurich)
        tm1 = time(NULL);
        string tz = "TZ=";
        
        // set the new TZ value (TZ=zoneName)
        putenv((char *)(tz+zoneName).c_str());  
        
        // check
        envTz = getenv("TZ");
        if (envTz != NULL)
                cout << "TZ2 = "<<envTz<<endl; 
        
        long dtime = 0;

        // GMT time
        t2 = gmtime(&tm1);
        tm2 = mktime(t2);
        t1 = localtime(&tm1);
        dtime = (long)(tm1 - tm2);
        short min = (short)(abs(dtime) %3600/60);

        ostringstream os;
        os <<boost::format("%|0+3d|")%(short)(dtime/3600)<<":"<< boost::format("%|02|")%min;
        
        return os.str();
        
    }

Here is the main which calls the funcion :

Code:

int main(int	argc,char *	argv[])
{
    cout << "America/Denver : "<<getTimeZone("America/Denver")<<endl;
    cout << "America/Denver : "<<getTimeZone("America/Denver")<<endl;
    cout << "Indian/Comoro : " <<getTimeZone("Indian/Comoro")<<endl;
    cout << "Asia/Iran : " << getTimeZone("Asia/Iran")<<endl;;
    cout << "Canada/Atlantic : "<<getTimeZone("Canada/Atlantic")<<endl;
    
    return 0;

}

Result :

Code:

TZ2 = America/Denver
America/Denver : +01:00     <---- Problem
TZ2 = America/Denver
America/Denver : -07:00     <---- Same is correct !!!??!!
TZ2 = Indian/Comoro
Indian/Comoro : +03:00
TZ2 = Asia/Iran
Asia/Iran : +00:00
TZ2 = Canada/Atlantic
Canada/Atlantic : -04:00

As you can see, the first result is bad !

Any idea ?

Snark1994 · 09-18-2012, 03:57 AM

Code:

   string getTimeZone(const string& zoneName){
...
        string tz = "TZ=";
        
        // set the new TZ value (TZ=zoneName)
        putenv((char *)(tz+zoneName).c_str());  
        
...
    }

I'm afraid you can't concatenate strings like that in C++. You want something like:

Code:

#include <sstream>
#include <string>

using namespace std;

string getTimeZone(const string& zoneName){
...
    stringstream ss;
    ss << "TZ=" << zoneName;
        
    putenv(ss.str().c_str());  
        
...
}

Hope this helps,

cbaurand · 09-18-2012, 04:16 AM

You can concatenate two std::string without any problem....
It's not the problem here.

pan64 · 09-18-2012, 05:36 AM

just a question: what will happen if you change the values: only the first answer is not ok, or only this pecific TZ is not ok?

dwhitney67 · 09-18-2012, 05:49 AM

Quote:

Originally Posted by cbaurand

You can concatenate two std::string without any problem....
It's not the problem here.

Yes, you are correct... concatenating strings is possible using the + operator. However, your second statement is incorrect (the problem is with the usage of putenv()).

The putenv() function accepts a pointer to char (a string), which becomes part of the environment. In other words, it takes ownership of the pointer. You are passing pointer to char of a temporary object.

From the man-page:

Code:

DESCRIPTION
       The  putenv()  function adds or changes the value of environment variables.  The argument string is
       of the form name=value.  If name does not already exist in the environment, then string is added to
       the  environment.   If  name  does  exist,  then the value of name in the environment is changed to
       value.  The string pointed to by string becomes part of the environment,  so  altering  the  string
       changes the environment.

I've attempted to fix your code to the best of my ability, however valgrind complains whether I allocate on the heap using "new" or whether using the C function strdup(). Here's that code:

Code:

std::string getTimeZone(const std::string& zoneName)
{
    // set the new TZ value (TZ=zoneName)
    std::string tz = "TZ=" + zoneName;

/*
    char* env_tz = new char[tz.size() + 1];
    memcpy(env_tz, tz.c_str(), tz.size());
    env_tz[tz.size()] = '\0';
*/
    char* env_tz = strdup(tz.c_str());
    putenv(env_tz);

    // LOCAL TIME (Europe/Zurich)
    time_t tm1 = time(NULL);

    // GMT time
    struct tm* t2 = gmtime(&tm1);
    time_t tm2 = mktime(t2);
    localtime(&tm1);

    std::ostringstream os;
    long dtime = (long)(tm1 - tm2);
    short min = (short)(abs(dtime) % 3600/60);
    os << boost::format("%|0+3d|") % (short)(dtime / 3600) << ":" << boost::format("%|02|") % min;

    return os.str();
}

Snark1994 · 09-18-2012, 06:04 AM

Tell you what's irritating: when you spend your morning trying to work out what the heck is wrong with the code (sorry for my wrong pointer earlier) only to find dwhitney67's done it first *eyeroll* At least I now think I know a lot more about C's timezone functions...

(I jest, of course)

cbaurand · 09-18-2012, 06:50 AM

Thanks dwhitney67,

It seems to work "correctly" now. More one thing is bad now: the result for America/Denver should be -06:00. The function localtime() doesn't consider the Daylight Saving Time.
I need now investigate about that.

dwhitney67 · 09-18-2012, 07:16 AM

Quote:

Originally Posted by cbaurand

Thanks dwhitney67,

It seems to work "correctly" now. More one thing is bad now: the result for America/Denver should be -06:00. The function localtime() doesn't consider the Daylight Saving Time.
I need now investigate about that.

What value does the following report on your system? On my system, it reports that DST is 0 (which is correct). Maybe it is coincidence that it works. I was always under the impression that Linux developers gave up on trying to support DST readings because it changes often (by governments).

Code:

#include <ctime>
#include <cstring>
#include <cstdlib>
#include <string>
#include <iostream>

int main()
{
    std::string timezone = "TZ=America/New York";

    char* env_tz = strdup(timezone.c_str());

    putenv(env_tz);

    time_t theTime = time(0);

    struct tm* theTimeTM = localtime(&theTime);

    std::cout << "DST = " << theTimeTM->tm_isdst << std::endl;
}

cbaurand · 09-18-2012, 07:47 AM

I got DST = 0.
After some investigations, i found that the problem comes from mktime. To avoid mismatch between winter and summer hour, just set the tm_isdst to -1.
It works fine now.

Thanks for all.

cbaurand · 09-18-2012, 07:54 AM

Just for dwhitney67:
You've obtained DST=0 because "New York" is not recognised. Try "New_York". ;-)

Snark1994 · 09-19-2012, 04:37 AM

If you consider the problem solved, then please mark the thread 'SOLVED' (in thread tools, or a link at the top of the page) and mark any of dwhitney's posts that you found helpful as helpful

Thanks,