So I haven't seen here or anywhere else a way to get only the month hours and minutes using <ctime> or other library.
What I can do now is just get the full current date:
time_t now = time(0);
cout<<ctime(&now);
Any suggestions?
You can use the <ctime> standard library like this:
#include <ctime>
#include <iostream>
int main()
{
// get current time
std::time_t timer = std::time(0);
// convert to 'broken time'
std::tm bt = *std::localtime(&timer); // not thread safe
// extract month number from 'broken time' struct
std::cout << "month: " << (bt.tm_mon + 1) << '\n';
std::cout << "hours: " << (bt.tm_hour) << '\n';
std::cout << "mins : " << (bt.tm_min) << '\n';
}
The function std::localtime returns a pointer to an internal statc structure of type std::tm.
Because it returns a pointer to an internal struct it is best to copy it to a local version by dereferencing the pointer using *:
// copy what the returned pointer points to into `bt`.
std::tm bt = *std::localtime(&timer);
Since you are using ctime (C time), all C things should work. You could use strftime
char timestr[32];
strftime(timestr, sizeof(timestr), "%m:%H:%M", localtime(&now));
If boost is fine for you try this one :
boost::posix_time::to_simple_string(boost::posix_time::microsec_clock::local_time()).c_str()
try
std::put_time("%a %b %d %H:%M:%S %Y", &now);
Related
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(¤tTime);
// 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(¤tTime);
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);
}
}
I am struggling to execute a simple task. I want to take a date parameter from the command line argument and increment it several times by 1 day. The basic program should:
read the command line argument as the starting date, and
loop several times, incrementing that date by 1 day every time.
I convert the char* command line argument to a struct tm, then convert the struct tm to a time_t and add 60 * 60 * 24 = 1 day to it. I convert it back to struct tm to print it.
Here is the code:
#include <iostream>
#include <cstdlib>
#include <string>
#include "time.h"
int main(int argc, char *argv[])
{
char* start_date;
tm tm_start_date = {}; // solution: needs initialization
start_date = argv[1];
strptime(start_date, "%Y-%m-%d", &tm_start_date); // YYYY-MM-DD
char ch_stmt_date[11] = "";
time_t t_stmt_date = 0;
tm tm_stmt_date = {}; // solution: needs initialization;
tm_stmt_date = tm_start_date;
// time_t -> tm localtime_r(time_t, tm)
// tm -> time_t mktime(tm) returns time_t
std::cout << "start_date: " << start_date << " / tm_start_date: " << std::to_string(1900 + tm_start_date.tm_year) + std::to_string(tm_start_date.tm_mon + 1) +
std::to_string(tm_start_date.tm_mday) << std::endl;
// increment by 1 day per iteration
for (int i=0; i<5; i++)
{
// tm -> t_time
t_stmt_date = mktime(&tm_stmt_date);
std::cout << "t_stmt_date: " << t_stmt_date << std::endl;
// + 1 day
t_stmt_date += 60*60*24;
std::cout << "t_stmt_date: " << t_stmt_date << std::endl;
// time_t -> tm
localtime_r(&t_stmt_date, &tm_stmt_date);
strftime (ch_stmt_date, 11, "%Y-%m-%d", &tm_stmt_date);
std::cout << "ch_stmt_date: " << ch_stmt_date << std::endl;
}
return EXIT_SUCCESS;
}
The start date is correctly read and parsed into the struct tm.
However, subsequently I get one of two behaviors of the program:
Either I get a -1 on the first call of t_stmt_date = mktime(&tm_stmt_date); and a value of t_stmt_date: 86399 (1970-01-02) in the output. The rest of the loop then works correctly and iterates 5 times, incrementing 1970-01-02 by 1 day.
Or, the same code using the same command line parameter parses a nonsensical value on the first call of t_stmt_date = mktime(&tm_stmt_date); in the loop which is not a valid date, which, however is also correctly incremented by 60*60*24 on each of the 5 loops.
At this point I am desperate to understand the issue. I am working on Ubuntu 20.04 using gcc.
Thanks for your help.
Edit: Initializing the struct tm did the trick!
[NOTE]
You explicitly mention "using localtime and mktime" in the question's title, but I wasn't sure though after reading the rest of the text if that was mandatory, or you just needed to get a task done.
If you cannot use other libraries, just let me know and I'll remove this answer.
You could use std::chrono and Howard Hinnant's date library (C++11 onwards, header-only).
Or, should you be able to use a C++20 compiler, you would only need std::chrono.
[Demo]
#include <chrono>
#include <iostream> // cout
#include <sstream> // istringstream
#include <string>
#include "date/date.h"
int main()
{
namespace ch = std::chrono;
namespace dt = date;
const std::string start_date{"2022-01-31"}; // date
std::istringstream iss{ start_date }; // to string stream
dt::sys_days start_day{}; // to a time point with a day duration
dt::from_stream(iss, "%Y-%m-%d", start_day);
for (auto day{start_day}, end_day{start_day + dt::days{3}};
day < end_day;
day += dt::days{1}) // with which we can do date arithmetic
{
std::cout << dt::format("%Y-%m-%d\n", day);
}
}
// Outputs:
//
// 2022-01-31
// 2022-02-01
// 2022-02-02
How do I get a uint unix timestamp in C++? I've googled a bit and it seems that most methods are looking for more convoluted ways to represent time. Can't I just get it as a uint?
C++20 introduced a guarantee that time_since_epoch is relative to the UNIX epoch, and cppreference.com gives an example that I've distilled to the relevant code, and changed to units of seconds rather than hours:
#include <iostream>
#include <chrono>
int main()
{
const auto p1 = std::chrono::system_clock::now();
std::cout << "seconds since epoch: "
<< std::chrono::duration_cast<std::chrono::seconds>(
p1.time_since_epoch()).count() << '\n';
}
Using C++17 or earlier, time() is the simplest function - seconds since Epoch, which for Linux and UNIX at least would be the UNIX epoch. Linux manpage here.
The cppreference page linked above gives this example:
#include <ctime>
#include <iostream>
int main()
{
std::time_t result = std::time(nullptr);
std::cout << std::asctime(std::localtime(&result))
<< result << " seconds since the Epoch\n";
}
#include<iostream>
#include<ctime>
int main()
{
std::time_t t = std::time(0); // t is an integer type
std::cout << t << " seconds since 01-Jan-1970\n";
return 0;
}
The most common advice is wrong, you can't just rely on time(). That's used for relative timing: ISO C++ doesn't specify that 1970-01-01T00:00Z is time_t(0)
What's worse is that you can't easily figure it out, either. Sure, you can find the calendar date of time_t(0) with gmtime, but what are you going to do if that's 2000-01-01T00:00Z ? How many seconds were there between 1970-01-01T00:00Z and 2000-01-01T00:00Z? It's certainly no multiple of 60, due to leap seconds.
As this is the first result on google and there's no C++20 answer yet, here's how to use std::chrono to do this:
#include <chrono>
//...
using namespace std::chrono;
int64_t timestamp = duration_cast<milliseconds>(system_clock::now().time_since_epoch()).count();
In versions of C++ before 20, system_clock's epoch being Unix epoch is a de-facto convention, but it's not standardized. If you're not on C++20, use at your own risk.
#include <iostream>
#include <sys/time.h>
using namespace std;
int main ()
{
unsigned long int sec= time(NULL);
cout<<sec<<endl;
}
I created a global define with more information:
#include <iostream>
#include <ctime>
#include <iomanip>
#define __FILENAME__ (__builtin_strrchr(__FILE__, '/') ? __builtin_strrchr(__FILE__, '/') + 1 : __FILE__) // only show filename and not it's path (less clutter)
#define INFO std::cout << std::put_time(std::localtime(&time_now), "%y-%m-%d %OH:%OM:%OS") << " [INFO] " << __FILENAME__ << "(" << __FUNCTION__ << ":" << __LINE__ << ") >> "
#define ERROR std::cout << std::put_time(std::localtime(&time_now), "%y-%m-%d %OH:%OM:%OS") << " [ERROR] " << __FILENAME__ << "(" << __FUNCTION__ << ":" << __LINE__ << ") >> "
static std::time_t time_now = std::time(nullptr);
Use it like this:
INFO << "Hello world" << std::endl;
ERROR << "Goodbye world" << std::endl;
Sample output:
16-06-23 21:33:19 [INFO] main.cpp(main:6) >> Hello world
16-06-23 21:33:19 [ERROR] main.cpp(main:7) >> Goodbye world
Put these lines in your header file. I find this very useful for debugging, etc.
Windows uses a different epoch and time units: see
Convert Windows Filetime to second in Unix/Linux
What std::time() returns on Windows is (as yet) unknown to me (;-))
So say I set a time in tm to be 23:00:00
ptm->tm_hour = 23; ptm->tm_min = 0; ptm->tm_sec = 0;
And I want to allow a user to subtract time from this
ptm->tm_hour -= hourinput; ptm->tm_min -= minuteinput; ptm->tm_sec -= secondinput;
If the user subtracts 0 hours, 5 minutes, and 5 seconds, instead of showing up as 22:54:55, it will show up as 23:-5:-5.
I suppose I could do a bunch of if statements to check if ptm is below 0 and account for this, but is there a more efficient way of getting the proper time?
Yes, you can use std::mktime for this. It doesn't just convert a std::tm to a std::time_t, it also fixes the tm if some field went out of range. Consider this example where we take the current time and add 1000 seconds.
#include <iostream>
#include <iomanip> // put_time
#include <ctime>
int main(int argc, char **argv) {
std::time_t t = std::time(nullptr);
std::tm tm = *std::localtime(&t);
std::cout << "Time: " << std::put_time(&tm, "%c %Z") << std::endl;
tm.tm_sec += 1000; // the seconds are now out of range
//std::cout << "Time in 1000 sec" << std::put_time(&tm, "%c %Z") << std::endl; this would crash!
std::mktime(&tm); // also returns a time_t, but we don't need that here
std::cout << "Time in 1000 sec: " << std::put_time(&tm, "%c %Z") << std::endl;
return 0;
}
My Output:
Time: 01/24/19 09:26:46 W. Europe Standard Time
Time in 1000 sec: 01/24/19 09:43:26 W. Europe Standard Time
As you can see, the time went from 09:26:46 to 09:43:26.
Here's another approach using Howard Hinnant's date library, which is on its way into C++2a.
#include <iostream>
#include "date/date.h"
using namespace std::chrono_literals;
// Time point representing the start of today:
auto today = date::floor<date::days>(std::chrono::system_clock::now());
auto someTp = today + 23h; // Today, at 23h00
auto anotherTp = someTp - 5min - 5s; // ... should be self-explanatory now :)
std::cout << date::format("%b-%d-%Y %T\n", anotherTp);
If you want to expose the manipulation of time points via a user interface, the compile-time constructs 23h, 5min and so on are of course not available. Those literals construct std::chrono::duration objects, so you need a mechanism to turn user input into equivalent instances.
I'm looking for a way to save the time in a HH::MM::SS fashion in C++. I saw here that they are many solutions and after a little research I opted for time and localtime. However, it seems like the localtime function is a little tricky, since it says:
All calls to localtime and gmtime use the same static structure, so
each call overwrites the results of the previous call.
The problem that this causes is shown in the next snippet of code:
#include <ctime>
#include <iostream>
using namespace std;
int main() {
time_t t1 = time(0); // get time now
struct tm * now = localtime( & t1 );
std::cout << t1 << std::endl;
sleep(2);
time_t t2 = time(0); // get time now
struct tm * now2 = localtime( & t2 );
std::cout << t2 << std::endl;
cout << (now->tm_year + 1900) << '-'
<< (now->tm_mon + 1) << '-'
<< now->tm_mday << ", "
<< now->tm_hour << ":" << now->tm_min << ":" << now->tm_sec
<< endl;
cout << (now2->tm_year + 1900) << '-'
<< (now2->tm_mon + 1) << '-'
<< now2->tm_mday << ", "
<< now2->tm_hour << ":" << now2->tm_min << ":" << now2->tm_sec
<< endl;
}
A typical output for this is:
1320655946
1320655948
2011-11-7, 9:52:28
2011-11-7, 9:52:28
So as you can see, the time_t timestamps are correct, but the localtime messes everything up.
My question is: how do I convert a timestamp ot type time_t into a human-readable time?
If you are worried about reentrancy in localtime and gmtime, there is localtime_r and gmtime_r which can handle multiple calls.
When it comes to formatting the time to your liking, check the function strftime.
the localtime() call stores the results in an internal buffer.
Every time you call it you overwrite the buffer.
An alternative solution would be to make a copy of the buffer.
time_t t1 = time(0); // get time now
struct tm* now = localtime( & t1 ); // convert to local time
struct tm copy = *now; // make a local copy.
// ^^^ notice no star.
But note: The only time you should be converting to local time is when you display the value. At all other times you should just keep the time as UTC (for storage and manipulation). Since you are only converting the objects for display convert then print immediately and then things will not go wrong.
localtime has what is best considered a legacy interface. It can't be
used in multithreaded code, for example. In a multithreaded
environment, you can use localtime_r under Posix or localtime_s
under Windows. Otherwise, all you have to do is save the results:
tm then = *localtime( &t1 );
// ...
tm now = *localtime( &t2 );
It would probably be more idiomatic, however, to only call localtime
immediately before formatting the output, e.g.:
std::string
timestampToString( time_t timeAndDate )
{
char results[100];
if ( strftime( results, sizeof( results ), "%Y-%m-%d, %H:%M:%S",
localtime( &timeAndDate) ) == 0 ) {
assert( 0 );
}
return results;
}
and then writing:
std::cout << formatTime( t1 ) << std::endl;
(You could also create a more generic formatting function, which took
the format as an argument.)
You can run continuous clock using following code. It works nicely.
#include<iostream>
#include <Windows.h>
#include<ctime>
using namespace std;
void main() {
while(true) {
system("cls"); //to clear screen
time_t tim;
time(&tim);
cout << ctime(&tim);
Sleep(1);
}
}