How to parse a date string into a c++11 std::chron

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.

回答1:

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));

回答2:

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.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. For this part I'm also going to use a second header-only library "chrono_io.h":

#include "chrono_io.h"
#include "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);
    auto tp_days = floor<days>(tp);
    auto hms = make_time(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 factory function make_time takes any duration (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.

Finally, using "chrono_io.h", one can just print out all of these durations. This example outputs:

Number of days    = 16079[86400]s
Number of hours   = 12h
Number of minutes = 35min
Number of seconds = 34s

The [86400]s represents the units of the duration that has days-precision. So 2014-01-09 is 16079 days after 1970-01-01.

回答3:

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;
}