Get date and time in UTC format in c++ [duplicate] - c++

This question already has answers here:
Convert time_t from localtime zone to UTC
(4 answers)
Closed 2 years ago.
How can I get a datetime in UTC time in this format ('2002-05-30T09:30:10Z') in C++?
This is the function I have made.
void settime(){
// Current date/time based on current system
time_t now = time(0);
// Convert now to tm struct for local timezone
tm* localtm = localtime(&now);
cout << "The local date and time is: " << asctime(localtm) << endl;
// Convert now to tm struct for UTC
tm* gmtm = gmtime(&now);
if (gmtm != NULL) {
cout << "The UTC date and time is: " << asctime(gmtm) << endl;
}
else {
cerr << "Failed to get the UTC date and time" << endl;
}
}
This function print in this format 'The UTC date and time is: Mon Oct 12 18:56:59 2020' right now.

Using this free, open-source, header-only preview of C++20 <chrono>:
#include "date/date.h"
#include <chrono>
#include <iostream>
int
main()
{
namespace cr = std::chrono;
std::cout << date::format("%FT%TZ", cr::floor<cr::seconds>(cr::system_clock::now())) << '\n';
}
This will easily port to C++20 by:
Drop #include "date/date.h"
Change date::format to std::format
Change "%FT%TZ" to "{:%FT%TZ}"

Related

Getting Current Date inside a C++ process running 24*7

I have a backend process running 24*7 mostly built using C++ and I need to validate if an input date (in format YYYYMMDD) belongs in a set of next 5 business days. The input date is not a clear indicator of the current date so I am using the following function to get the current date and then calculating the next 5 business days from it.
const std::string& CurrentDateStr() {
static const std::string sDate = []() {
time_t currTime = time(NULL);
struct tm timeinfo;
localtime_r(&currTime, &timeinfo);
char buffer[16]="";
strftime(buffer, sizeof(buffer), "%Y%m%d", &timeinfo);
return std::string(buffer);
} ();
return sDate;
}
This function returns me the correct current date if the process was started today but if the process continues running till tomorrow then it will return me yesterday's date as current date due to which calculation of next 5 business days from current date goes for a toss.
Is this expected ? Is there some workaround for it or is there a better way to implement the requirement using standard C++
Your issue is the static variable. You should read up on that, because you're going to encounter it a lot. This is what the comments were trying to get you to do. You can fix your issue by just removing it:
const std::string& CurrentDateStr() {
time_t currTime = time(NULL);
struct tm timeinfo;
localtime_r(&currTime, &timeinfo);
char buffer[16]="";
strftime(buffer, sizeof(buffer), "%Y%m%d", &timeinfo);
return std::string(buffer);
}
For a more modern solution, as suggested in the comments as well, read up on chrono. Especially system_clock::now().
one way to do it using chrono:
#include <iostream>
#include <ctime>
#include <chrono>
#include <thread>
int main()
{
while (true)
{
theTime currentTime = time(nullptr);
tm* date = gmtime(&currentTime);
// Print the date and time
std::cout << "Current date and time: " << date->theDay << "/" << date->theMon + 1 << "/" << date->theYear + 1900;
std::cout << " " << date->theHour << ":" << date->theMmin << ":" << date->theSec << std::endl;
// Wait for 1 minute
std::this_thread::sleep_for(std::chrono::minutes(1));
}
}
OR Use the sleep method.
#include <iostream>
#include <ctime>
#include <unistd.h>
int main()
{
while (true)
{
time_t currentTime = time(nullptr);
tm* date = gmtime(&currentTime);
std::cout << "Current date and time: " << date->tm_mday << "/" << date->tm_mon + 1 << "/" << date->tm_year + 1900;
std::cout << " " << date->tm_hour << ":" << date->tm_min << std::endl;
// Wait for 1 minute (60 seconds)
sleep(60);
}
}

From xs:dateTime to std::chrono::timepoint [duplicate]

Consider a historic date string of format:
Thu Jan 9 12:35:34 2014
I want to parse such a string into some kind of C++ date representation, then calculate the amount of time that has passed since then.
From the resulting duration I need access to the numbers of seconds, minutes, hours and days.
Can this be done with the new C++11 std::chrono namespace? If not, how should I go about this today?
I'm using g++-4.8.1 though presumably an answer should just target the C++11 spec.
std::tm tm = {};
std::stringstream ss("Jan 9 2014 12:35:34");
ss >> std::get_time(&tm, "%b %d %Y %H:%M:%S");
auto tp = std::chrono::system_clock::from_time_t(std::mktime(&tm));
GCC prior to version 5 doesn't implement std::get_time. You should also be able to write:
std::tm tm = {};
strptime("Thu Jan 9 2014 12:35:34", "%a %b %d %Y %H:%M:%S", &tm);
auto tp = std::chrono::system_clock::from_time_t(std::mktime(&tm));
New answer for old question. Rationale for the new answer: The question was edited from its original form because tools at the time would not handle exactly what was being asked. And the resulting accepted answer gives a subtly different behavior than what the original question asked for.
I'm not trying to put down the accepted answer. It's a good answer. It's just that the C API is so confusing that it is inevitable that mistakes like this will happen.
The original question was to parse "Thu, 9 Jan 2014 12:35:34 +0000". So clearly the intent was to parse a timestamp representing a UTC time. But strptime (which isn't standard C or C++, but is POSIX) does not parse the trailing UTC offset indicating this is a UTC timestamp (it will format it with %z, but not parse it).
The question was then edited to ask about "Thu Jan 9 12:35:34 2014". But the question was not edited to clarify if this was a UTC timestamp, or a timestamp in the computer's current local timezone. The accepted answer implicitly assumes the timestamp represents the computer's current local timezone because of the use of std::mktime.
std::mktime not only transforms the field type tm to the serial type time_t, it also performs an offset adjustment from the computer's local time zone to UTC.
But what if we want to parse a UTC timestamp as the original (unedited) question asked?
That can be done today using this newer, free open-source library.
#include "date/date.h"
#include <iostream>
#include <sstream>
int
main()
{
using namespace std;
using namespace date;
istringstream in{"Thu, 9 Jan 2014 12:35:34 +0000"};
sys_seconds tp;
in >> parse("%a, %d %b %Y %T %z", tp);
}
This library can parse %z. And date::sys_seconds is just a typedef for:
std::chrono::time_point<std::chrono::system_clock, std::chrono::seconds>
The question also asks:
From the resulting duration I need access to the numbers of seconds, minutes, hours and days.
That part has remained unanswered. Here's how you do it with this library.
#include "date/date.h"
#include <chrono>
#include <iostream>
#include <sstream>
int
main()
{
using namespace std;
using namespace date;
istringstream in{"Thu, 9 Jan 2014 12:35:34 +0000"};
sys_seconds tp;
in >> parse("%a, %d %b %Y %T %z", tp);
auto tp_days = floor<days>(tp);
auto hms = hh_mm_ss<seconds>{tp - tp_days};
std::cout << "Number of days = " << tp_days.time_since_epoch() << '\n';
std::cout << "Number of hours = " << hms.hours() << '\n';
std::cout << "Number of minutes = " << hms.minutes() << '\n';
std::cout << "Number of seconds = " << hms.seconds() << '\n';
}
floor<days> truncates the seconds-precision time_point to a days-precision time_point. If you subtract the days-precision time_point from tp, you're left with a duration that represents the time since midnight (UTC).
The type hh_mm_ss<seconds> takes any duration convertible to seconds (in this case time since midnight) and creates a {hours, minutes, seconds} field type with getters for each field. If the duration has precision finer than seconds this field type will also have a getter for the subseconds. Prior to C++17, one has to specify that finer duration as the template parameter. In C++17 and later it can be deduced:
auto hms = hh_mm_ss{tp - tp_days};
Finally, one can just print out all of these durations. This example outputs:
Number of days = 16079d
Number of hours = 12h
Number of minutes = 35min
Number of seconds = 34s
So 2014-01-09 is 16079 days after 1970-01-01.
Here is the full example but at milliseconds precision:
#include "date/date.h"
#include <chrono>
#include <iostream>
#include <sstream>
int
main()
{
using namespace std;
using namespace std::chrono;
using namespace date;
istringstream in{"Thu, 9 Jan 2014 12:35:34.123 +0000"};
sys_time<milliseconds> tp;
in >> parse("%a, %d %b %Y %T %z", tp);
auto tp_days = floor<days>(tp);
hh_mm_ss hms{tp - tp_days};
std::cout << tp << '\n';
std::cout << "Number of days = " << tp_days.time_since_epoch() << '\n';
std::cout << "Number of hours = " << hms.hours() << '\n';
std::cout << "Number of minutes = " << hms.minutes() << '\n';
std::cout << "Number of seconds = " << hms.seconds() << '\n';
std::cout << "Number of milliseconds = " << hms.subseconds() << '\n';
}
Output:
2014-01-09 12:35:34.123
Number of days = 16079d
Number of hours = 12h
Number of minutes = 35min
Number of seconds = 34s
Number of milliseconds = 123ms
This library is now part of C++20, but is in namespace std::chrono and found in the header <chrono>.
This is rather C-ish and not as elegant of a solution as Simple's answer, but I think it might work. This answer is probably wrong but I'll leave it up so someone can post corrections.
#include <iostream>
#include <ctime>
int main ()
{
struct tm timeinfo;
std::string buffer = "Thu, 9 Jan 2014 12:35:00";
if (!strptime(buffer.c_str(), "%a, %d %b %Y %T", &timeinfo))
std::cout << "Error.";
time_t now;
struct tm timeinfo2;
time(&now);
timeinfo2 = *gmtime(&now);
time_t seconds = difftime(mktime(&timeinfo2), mktime(&timeinfo));
time(&seconds);
struct tm result;
result = *gmtime ( &seconds );
std::cout << result.tm_sec << " " << result.tm_min << " "
<< result.tm_hour << " " << result.tm_mday;
return 0;
}
Cases covered (code is below):
since a give date until now
long int min0 = getMinutesSince( "2005-02-19 12:35:00" );
since the epoch until now
long int min1 = getMinutesSince1970( );
between two date+hours (since the epoch until a given date)
long int min0 = getMinutesSince1970Until( "2019-01-18 14:23:00" );
long int min1 = getMinutesSince1970Until( "2019-01-18 14:27:00" );
cout << min1 - min0 << endl;
Complete code:
#include <iostream>
#include <chrono>
#include <sstream>
#include <string>
#include <iomanip>
using namespace std;
// ------------------------------------------------
// ------------------------------------------------
long int getMinutesSince1970Until( string dateAndHour ) {
tm tm = {};
stringstream ss( dateAndHour );
ss >> get_time(&tm, "%Y-%m-%d %H:%M:%S");
chrono::system_clock::time_point tp = chrono::system_clock::from_time_t(mktime(&tm));
return
chrono::duration_cast<chrono::minutes>(
tp.time_since_epoch()).count();
} // ()
// ------------------------------------------------
// ------------------------------------------------
long int getMinutesSince1970() {
chrono::system_clock::time_point now = chrono::system_clock::now();
return
chrono::duration_cast<chrono::minutes>( now.time_since_epoch() ).count();
} // ()
// ------------------------------------------------
// ------------------------------------------------
long int getMinutesSince( string dateAndHour ) {
tm tm = {};
stringstream ss( dateAndHour );
ss >> get_time(&tm, "%Y-%m-%d %H:%M:%S");
chrono::system_clock::time_point then =
chrono::system_clock::from_time_t(mktime(&tm));
chrono::system_clock::time_point now = chrono::system_clock::now();
return
chrono::duration_cast<chrono::minutes>(
now.time_since_epoch()-
then.time_since_epoch()
).count();
} // ()
// ------------------------------------------------
// ------------------------------------------------
int main () {
long int min = getMinutesSince1970Until( "1970-01-01 01:01:00" );
cout << min << endl;
long int min0 = getMinutesSince1970Until( "2019-01-18 14:23:00" );
long int min1 = getMinutesSince1970Until( "2019-01-18 14:27:00" );
if ( (min1 - min0) != 4 ) {
cout << " something is wrong " << endl;
} else {
cout << " it appears to work !" << endl;
}
min0 = getMinutesSince( "1970-01-01 01:00:00" );
min1 = getMinutesSince1970( );
if ( (min1 - min0) != 0 ) {
cout << " something is wrong " << endl;
} else {
cout << " it appears to work !" << endl;
}
} // ()

Conversion of times in C++

This is what I am trying to do:
Get the local time (from the system);
Convert that time to the UTC format and associate it with some member variable of current object.
Later on, given the timezone of the user, I wish to convert it into the correct local time and display it to the user.
Looking up few things on SO and CppReference, I could come up with the following snippet:
#include <iostream>
#include <iomanip>
#include <ctime>
#include <string>
using namespace std;
int main()
{
time_t lt = std::time(0);
//time(0) gives current time, but LTime shows the UTC time (not local time)
string LTime = std::ctime(&lt); //localtime() gives error
cout<<LTime;
//timestamp = mktime(&tm) - timezone;
//time_t timestamp = mktime(&tm) - _timezone;
//std::cout << "timestamp: " << std::put_time(timestamp, "%c %Z") << '\n';
return 0;
}
The example on cppreference.com illustrates how the value can be printed using put_time(); but how to store it in a variable?
How to convert the UTC time format to current timezone (given some timezone as the input)? I tried using the commented code above as per this link but it does not take any parameter.
You can use local time get the local time and gmt time for UTC
You can set the Time zone using the list Time zone wiki
#include <iostream>
#include <iomanip>
#include <ctime>
int main()
{
std::time_t result = std::time(nullptr);
auto local = std::asctime(std::localtime(&result));
std::cout <<local;
std::cout << "UTC: " << std::put_time(std::gmtime(&result), "%c %Z") << '\n';
putenv("TZ=Asia/Singapore");
local = std::asctime(std::localtime(&result));
std::cout <<"Asia/Singapore Time "<<local;
}
Output
Thu Sep 14 21:59:37 2017
UTC: Fri Sep 15 01:59:37 2017 UTC
Asia/Singapore Time Fri Sep 15 09:59:37 2017
Program ended with exit code: 0

Valid Date function using <ctime> mktime

I wanted a function that would take three inputs of day, month, year and tell me whether it is valid or not. Then using the example on http://www.cplusplus.com/reference/ctime/mktime/
I tried to implement my function:
bool ValidDate(int d, int m, int y)
{
struct tm *timeinfo;
time_t rawtime;
time (&rawtime);
timeinfo = localtime(&rawtime);
timeinfo->tm_year = y - 1900;
timeinfo->tm_mon = m - 1;
timeinfo->tm_mday = d;
if (mktime(timeinfo) == -1 )
return false;
else return true;
}
The problem is that the function is returning not as i want it to.
e.g im checking like
if (ValidDate(4,13,2010)) // out put is valid
std::cout << "valid\n";
else std::cout << "Invalid\n";
ValidDate(4,22,2010) // valid
ValidDate(344,13,2010) //valid
ValidDate(4,133,2010) //valid
ValidDate(31,12, 1920) //invalid
ValidDate(31,9,2010) //valid
ValidDate(4,9,2010) //valid
Why? thanks.
EDIT:
all dates entered were invalid except 31,12,1920 and 4,9,2010 and non of the outputs were correct.
mktime return is as follow :
Time since epoch as a std::time_t object on success or -1 if time cannot be represented as a std::time_t object.
std::time_t is defined as follow :
Arithmetic type capable of representing times.
Although not defined, this is almost always a integral value holding the number of seconds (not counting leap seconds) since 00:00, Jan 1 1970 UTC, corresponding to POSIX time.
So 31/12/1920 cannot be represented into a std::time_t as it is before the epoch.
As for the other invalid dates that are reported as valid, mktime also states :
The values in [the parameter] are permitted to be outside their normal ranges.
Here is the example taken from cppreference :
#include <iostream>
#include <iomanip>
#include <ctime>
int main()
{
std::time_t t = std::time(NULL);
std::tm tm = *std::localtime(&t);
std::cout << "Today is " << std::put_time(&tm, "%c %Z") <<'\n';
tm.tm_mon -= 100; // tm_mon is now outside its normal range
std::mktime(&tm);
std::cout << "100 months ago was " << std::put_time(&tm, "%c %Z") << '\n';
}
Output is :
Today is Wed Dec 28 09:56:10 2011 EST
100 months ago was Thu Aug 28 10:56:10 2003 EDT

Convert the GMT date string to CTime or COLEDatetime

How to Convert the GMT date string to CTime or COLEDatetime
"2011/06/01 3:20:20 GMT + 5:30"
COleDateTime::ParseDateTime
The lpszDate parameter can take a variety of formats. For example, the following strings contain acceptable date/time formats:
"25 January 1996"
"8:30:00"
"20:30:00"
"January 25, 1996 8:30:00"
"8:30:00 Jan. 25, 1996"
"1/25/1996 8:30:00" // always specify the full year,
// even in a 'short date' format
#include <iostream>
#include <ctime>
using namespace std;
int main( )
{
// current date/time based on current system
time_t now = time(0);
// convert now to string form
char* dt = ctime(&now);
cout << "The local date and time is: " << dt << endl;
// convert now to tm struct for UTC
tm *gmtm = gmtime(&now);
dt = asctime(gmtm);
cout << "The UTC date and time is:"<< dt << endl;
}