Is it possible to manually set the epoch date/time to the January 1, 0000, so I might use the std::chrono::time_point::time_since_epoch to calculate the difference between a given date and January 1, 0000?
I tried the following:
#include <iostream>
#include <chrono>
#include <ctime>
int main(int argc, char*argv[])
{
std::tm epochStart = {};
epochStart.tm_sec = 0;
epochStart.tm_min = 0;
epochStart.tm_hour = 0;
epochStart.tm_mday = 0;
epochStart.tm_mon = 0;
epochStart.tm_year = -1900;
epochStart.tm_wday = 0;
epochStart.tm_yday = 0;
epochStart.tm_isdst = -1;
std::time_t base = std::mktime(&epochStart);
std::chrono::system_clock::time_point baseTp=
std::chrono::system_clock::from_time_t(base);
std::time_t btp = std::chrono::system_clock::to_time_t(baseTp);
std::cout << "time: " << std::ctime(&btp);
}
but this gives me
time: Thu Jan 1 00:59:59 1970
I would avoid std::time_t altogether. Using days_from_civil from chrono-Compatible Low-Level Date Algorithms, you can immediately compute any difference between std::chrono::system_clock::time_point, and any date in the proleptic Gregorian calendar1.
In addition to days_from_civil which takes a year/month/day triple and converts it into a count of days before/since 1970-01-01 (a chrono-compatible epoch), it is also convenient to create a custom chrono::duration to represent 24 hours:
typedef std::chrono::duration
<
int,
std::ratio_multiply<std::ratio<24>, std::chrono::hours::period>
> days;
Now you can create any epoch you want with just:
constexpr days epoch = days(days_from_civil(0, 1, 1)); // 0000-01-01
In C++1y this is even a compile-time computation!
And you can subtract this std::chrono::duration from any other std::chrono::duration:
auto delta = std::chrono::system_clock::now().time_since_epoch() - epoch;
delta is now a std::chrono::duration representing the amount of time between now, and 0000-01-01. You can then print that out however you want, or otherwise manipulate it. For example here is an entire working demo:
#include "../date_performance/date_algorithms"
#include <iostream>
#include <chrono>
typedef std::chrono::duration
<
int,
std::ratio_multiply<std::ratio<24>, std::chrono::hours::period>
> days;
int
main()
{
constexpr days epoch = days(days_from_civil(0, 1, 1));
auto delta = std::chrono::system_clock::now().time_since_epoch() - epoch;
days d = std::chrono::duration_cast<days>(delta);
std::cout << "It has been " << d.count() << " days, ";
delta -= d;
auto h = std::chrono::duration_cast<std::chrono::hours>(delta);
std::cout << h.count() << " hours, ";
delta -= h;
auto m = std::chrono::duration_cast<std::chrono::minutes>(delta);
std::cout << m.count() << " minutes, ";
delta -= m;
auto s = std::chrono::duration_cast<std::chrono::seconds>(delta);
std::cout << s.count() << " seconds ";
std::cout << " since 0000-01-01\n";
}
Which for me output:
It has been 735602 days, 19 hours, 14 minutes, 32 seconds since 0000-01-01
A word of warning about overflow:
The std::chrono::system_clock::time_point::duration is not guaranteed to have a range large enough to do this. It turns out that on my system it does. It is microseconds in a signed long long which will span +/- 292,000 years. If you need to avoid an overflow problem, you could truncate your std::chrono::system_clock::time_point::duration to courser units (e.g. seconds or days) to extend the range prior to subtracting 0000-01-01.
I got to thinking
And that usually leads to a disaster. However in this case I decided I should add to this post anyway. This:
constexpr days epoch = days(days_from_civil(0, 1, 1));
has type days, which is a duration. But it really isn't a duration. It is a point in time. It is a date. It is a time_point with a coarse precision. By introducing a new typedef, the code in this post can be cleaned up just a little bit more:
typedef std::chrono::time_point<std::chrono::system_clock, days> date_point;
Now instead of writing:
constexpr days epoch = days(days_from_civil(0, 1, 1));
One can write:
constexpr date_point epoch{days(days_from_civil(0, 1, 1))};
But even more importantly, instead of:
auto delta = std::chrono::system_clock::now().time_since_epoch() - epoch;
we can now write:
auto delta = std::chrono::system_clock::now() - epoch;
This delta still has exactly the same type and value as it did previously, and everything else in the demo still proceeds as exactly as it did before.
This is both a small change, and a big change. By treating epoch as a time_point instead of a duration, the algebra of time_point's and duration's works for us, both simplifying and type-checking our expressions to help us write cleaner code with fewer mistakes.
For example one can add two duration's together. But it doesn't make any sense at all to:
epoch + epoch
By using time_point instead of duration for the type of epoch, the compiler catches such non-sensical expressions at compile time.
1The proleptic Gregorian calendar has a year 0. In the year 0 it is 2 days behind the Julian calendar. Using a year 0 is also consistent with ISO 8601. As long as all parties involved know what calendar you are using, then everything is fine. Conversion between non-positive years and "BC years" is trivial if desired.
It's possible, the code you've given (minus a small fix, tm_mday starts with 1) yields:
Sat Jan 1 00:00:00 0
Live example
The real problem is: Are you on 32-bit or 64-bit? With a 32-bit system, time_t is also only 32 bits and you are limited to 1970 +/- 68 years.
On a 64-bit system, the limits are given by std::mktime and std::strftime, in my own code I have unit test for those strings and the corresponding values:
"-2147481748-01-01 00:00:00" maps to -67768040609740800
"2147483647-12-31 23:59:59" maps to 67767976233532799
I should probably also mention that there are systems where the above does not work because the underlying OS functions are buggy. For the record: I'm on Linux.
No. mktime and friends are based on UNIX time, which starts on 1st January 1970.
There is in fact no such thing as 0th January, 0000, so it seems likely that you would be better off finding another way to solve whatever is your actual problem.
Related
There's many similar questions out there but I haven't found one specific to the GPS output data I am receiving. The data from my GPS is in decimal form:
GPS Week: 2145 and GPS Time: 330374.741371 (the manual says this is a double that represents the "time of week in seconds")
I'm trying to convert this time into human readable UTC time. I'm using old C++14, not 20, so I can't just use the to_utc() function I don't think. I'm mostly confused about the decimal. On this website: https://www.labsat.co.uk/index.php/en/gps-time-calculator it looks like the data is "secondsOfTheWeek.secondsOfTheDay. I'm not sure how to convert this to UTC time...
I believe this output data is the number of seconds since the GPS epoch time of midnight, Jan. 6 1980. And I know it doesn't count leap seconds so that has to be taken into account too. If I had some guidance on how to start getting this into UTC time I think I could figure out the rest, but I'm not really sure where to start...
Eventually I want to convert the time into a string to set an OS system w that time using "date -s "16 AUG 2021 13:51:00" or something like that. But first I just need to convert this GPS time.
There exists a free, open-source preview to the C++20 chrono bits which works with C++14.
#include "date/tz.h"
#include <chrono>
date::utc_seconds
convert(int gps_week, double gps_time)
{
using namespace date;
using namespace std::chrono;
int upper = static_cast<int>(gps_time);
auto gps_t = gps_seconds{} + weeks(gps_week) + seconds{upper};
return clock_cast<utc_clock>(gps_t);
}
This first forms a gps_time by adding the appropriate number of weeks to the gps epoch, and then the seconds.
Next you use clock_cast to transform this into utc_time (which does include leap seconds).
This can be used like so:
#include <iostream>
int
main()
{
using namespace date;
using namespace std;
cout << convert(2145, 330374.741371) << '\n';
}
Which outputs:
2021-02-17 19:45:56
The clock_cast changes the epoch from 1980-01-06 to 1970-01-01 and adds in the number of leap seconds that occur between the gps epoch and the utc time point. If the gps input happens to correspond to a leap second, this will properly print "60" in the seconds field. For example:
cout << convert(1851, 259216) << '\n'; // 2015-06-30 23:59:60
Some installation is required.
Further information
This Wikipedia article says that the time of week actually comes in units of 1.5 seconds, ranging in value from 0 to 403,199.
static_assert(403'200 * 1.5 == 7 * 24 * 60 * 60);
If one finds themself dealing with the data in this form, here is an alternate convert implementation which can deal with this input data directly:
using gps_tow = std::chrono::duration<int, std::ratio<3, 2>>;
auto
convert(date::weeks gps_week_num, gps_tow tow)
{
using namespace date;
return clock_cast<utc_clock>(gps_seconds{} + gps_week_num + tow);
}
The first step is to define a duration unit of 1.5 seconds. This type is called gps_tow above.
The convert function now takes two strictly typed parameters: a count of weeks, and a count of gps_tow. Then one simply adds these parts together, along with the gps epoch, and clock_cast's it to utc_clock.
It can be used like so:
cout << convert(weeks{1851}, gps_tow{172811}) << '\n';
The output for this example is:
2015-06-30 23:59:60.5
I am collecting GPS time (in ns) from a sensor and I am looking for a way to convert that to a UTC time in C++.
I have a working code before in python.
time_gps = time_gps * 10**(-9) # Converts ns -> s
gps_epoch = pd.datetime(year=1980, month=1, day=6)
delta = pd.to_timedelta(time_gps, unit='s')
time = gps_epoch + delta - pd.to_timedelta(19, unit='s')
Using the link "Using std::chrono / date::gps_clock for converting a double gps timestamp to utc/tai" helped me figure out how to convert from GPS time to UTC.
uint64_t gps_input_ns = 1281798087485516800;
date::gps_time<std::chrono::nanoseconds> gt_nano{date::round<std::chrono::nanoseconds>(std::chrono::duration<uint64_t, std::nano>{gps_input_ns})};
auto utc_nano = date::clock_cast<date::utc_clock>(gt_nano);
std::cout << utc_nano << " UTC\n";
Output: 2020-08-18 15:01:09.485516800 UTC
My next question is, how can I extract the date and time from the variable "utc_nano"? I'm not very familiar with chrono or the date library and therefore having problems trying to separate the date and time. Any help would be much appreciated.
I'm assuming that leap seconds are important to you since you're dealing with gps time which represents the physical seconds that are labeled leap seconds by UTC. It is fairly tricky to manipulate date/times with leaps seconds, which is why Unix Time is so popular in computer systems.
In the C++20 chrono preview library, Unix Time is modeled by sys_time, whereas true UTC is modeled by utc_time. The only difference between these two models is that sys_time doesn't count leap seconds and utc_time does.
The advantage of sys_time is that there exists a fast and efficient algorithm for translating the time duration since 1970-01-01 00:00:00 into fields: year, month, day, hour, minute, second, subseconds. So if you want to break utc_time into these fields, the trick is to first turn utc_time into sys_time, while remembering whether or not your utc_time is referencing a leap second. Indeed, this is exactly what the streaming operator for utc_time does.
There exists a helper function get_leap_second_info to aid in doing this. This function takes a utc_time and returns a {is leap second, count of leap seconds} struct. The first member is true if the argument refers to a leap second, the second argument tells you how many leap seconds there have been between the argument and 1970. So the first step is to get this information for utc_nano:
auto info = get_leap_second_info(utc_nano);
Now you can create a sys_time with this information. Since sys_time is just like utc_time excluding leap seconds, you can just subtract off the number of leap seconds that have occurred:
sys_time<nanoseconds> sys_nano{utc_nano.time_since_epoch() - info.elapsed};
Now you have a count of nanoseconds in Unix Time. Truncating to days precision gives you a count of days in Unix Time:
auto sys_day = floor<days>(sys_nano);
sys_day is a date. The time of day is simply the difference between the nanoseconds-precision time_point and the days-precision time_point:
auto tod = sys_nano - sys_day;
tod is a time. It is the duration since midnight. It may be short by a second. That information is in info.is_leap_second.
If you want these types as "field types", you could convert sys_day to type year_month_day:
year_month_day ymd = sys_days;
year_month_day has getters for year, month and day.
You can convert tod into a {hours, minutes, seconds, nanoseconds} struct with:
hh_mm_ss hms{tod};
This has getters: hours(), minutes(), seconds(), and subseconds(). The above syntax assumes C++17. If in C++11 or 14, the syntax is:
hh_mm_ss<nanoseconds> hms{tod};
hh_mm_ss doesn't directly support a count of 60s, but that information is still in info.is_leap_second. E.g.
std::cout << hms.seconds().count() + info.is_leap_second << '\n';
That will output 60 if and only if info.is_leap_second is true.
You can even try this code which makes use of C time related functions
uint64_t ns = 1281798087485516800ULL + 315964800000000000ULL; // offset between gps epoch and unix epoch is 315964800 seconds
struct timespec ts;
ts.tv_sec = ns / 1000000000ULL;
ts.tv_nsec = ns % 1000000000ULL;
struct tm stm;
gmtime_r(&ts.tv_sec, &stm);
std::cout << stm.tm_year + 1900 << "-" << stm.tm_mon + 1 << "-" << stm.tm_mday << " " << stm.tm_hour << ":" << stm.tm_min << ":" << stm.tm_sec << std::endl;
https://www.timeanddate.com/date/weekday.html computes various facts about a day of the year, for example:
Given an arbitrary date, how can these numbers be computed with the C++20 chrono specification?
This is remarkably easy with the C++20 chrono specification. Below I show a function which inputs an arbitrary date, and prints this information to cout. Though at the time of this writing, the C++20 chrono specification isn't yet shipping, it is approximated by a free, open-source library. So you can experiment with it today, and even include it in shipping applications as long as you adopt C++11 or later.
This answer will take the form of a function:
void info(std::chrono::sys_days sd);
sys_days is a day-precision time_point in the system_clock family. That means it is simply a count of days since 1970-01-01 00:00:00 UTC. The type alias sys_days is new with C++20, but the underlying type has been available since C++11 (time_point<system_clock, duration<int, ratio<86400>>>). If you use the open-source C++20 preview library, sys_days is in namespace date.
The code below assumes function-local:
using namespace std;
using namespace std::chrono;
to reduce verbosity. If you are experimenting with the open-source C++20 preview library, also assume:
using namespace date;
Heading
To output the first two lines is simple:
cout << format("{:%d %B %Y is a %A}\n", sd)
<< "\nAdditional facts\n";
Just take the date sd and use format with the familiar strftime/put_time flags to print out the date and text. The open-source C++20 preview library hasn't yet integrated the fmt library, and so uses the slightly altered format string "%d %B %Y is a %A\n".
This will output (for example):
26 December 2019 is a Thursday
Additional facts
Common intermediate results computed once
This section of the function is written last, because one doesn't yet know what computations will be needed multiple times. But once you know, here is how to compute them:
year_month_day ymd = sd;
auto y = ymd.year();
auto m = ymd.month();
weekday wd{sd};
sys_days NewYears = y/1/1;
sys_days LastDayOfYear = y/12/31;
We will need the year and month fields of sd, and the weekday (day of the week). It is efficient to compute them once and for all in this manner. We will also need (multiple times) the first and last days of the current year. It is hard to tell at this point, but it is efficient to store these values as type sys_days as their subsequent use is only with day-oriented arithmetic which sys_days is very efficient at (sub-nanosecond speeds).
Fact 1: day number of year, and number of days left in year
auto dn = sd - NewYears + days{1};
auto dl = LastDayOfYear - sd;
cout << "* It is day number " << dn/days{1} << " of the year, "
<< dl/days{1} << " days left.\n";
This prints out the day number of the year, with January 1 being day 1, and then also prints out the number of days remaining in the year, not including sd. The computation to do this is trivial. Dividing each result by days{1} is a way to extract the number of days in dn and dl into an integral type for formatting purposes.
Fact 2: Number of this weekday and total number of weekdays in year
sys_days first_wd = y/1/wd[1];
sys_days last_wd = y/12/wd[last];
auto total_wd = (last_wd - first_wd)/weeks{1} + 1;
auto n_wd = (sd - first_wd)/weeks{1} + 1;
cout << format("* It is {:%A} number ", wd) << n_wd << " out of "
<< total_wd << format(" in {:%Y}.\n}", y);
wd is the day of the week (Monday thru Sunday) computed at the top of this article. To perform this computation we first need the dates of the first and last wd's in the year y. y/1/wd[1] is the first wd in January, and y/12/wd[last] is the last wd in December.
The total number of wds in the year is just the number of weeks between these two dates (plus 1). The sub-expression last_wd - first_wd is the number of days between the two dates. Dividing this result by 1 week results in an integral type holding the number of weeks between the two dates.
The week number is done the same way as the total number of weeks except one starts with the current day instead of the last wd of the year: sd - first_wd.
Fact 3: Number of this weekday and total number of weekdays in month
first_wd = y/m/wd[1];
last_wd = y/m/wd[last];
total_wd = (last_wd - first_wd)/weeks{1} + 1;
n_wd = (sd - first_wd)/weeks{1} + 1;
cout << format("* It is {:%A} number }", wd) << n_wd << " out of "
<< total_wd << format(" in {:%B %Y}.\n", y/m);
This works just like Fact 2, except we start with the first and last wds of the year-month pair y/m instead of the entire year.
Fact 4: Number of days in year
auto total_days = LastDayOfYear - NewYears + days{1};
cout << format("* Year {:%Y} has ", y) << total_days/days{1} << " days.\n";
The code pretty much speaks for itself.
Fact 5 Number of days in month
total_days = sys_days{y/m/last} - sys_days{y/m/1} + days{1};
cout << format("* {:%B %Y} has ", y/m) << total_days/days{1} << " days.\n";
The expression y/m/last is the last day of the year-month pair y/m, and of course y/m/1 is the first day of the month. Both are converted to sys_days so that they can be subtracted to get the number of days between them. Add 1 for the 1-based count.
Use
info can be used like this:
info(December/26/2019);
or like this:
info(floor<days>(system_clock::now()));
Here is example output:
26 December 2019 is a Thursday
Additional facts
* It is day number 360 of the year, 5 days left.
* It is Thursday number 52 out of 52 in 2019.
* It is Thursday number 4 out of 4 in December 2019.
* Year 2019 has 365 days.
* December 2019 has 31 days.
Edit
For those who are not fond of the "conventional syntax", there is a complete "constructor syntax" that can be used instead.
For example:
sys_days NewYears = y/1/1;
sys_days first_wd = y/1/wd[1];
sys_days last_wd = y/12/wd[last];
can be replaced by:
sys_days NewYears = year_month_day{y, month{1}, day{1}};
sys_days first_wd = year_month_weekday{y, month{1}, weekday_indexed{wd, 1}};
sys_days last_wd = year_month_weekday_last{y, month{12}, weekday_last{wd}};
I wrote the following code using Howard Hinnants date.h library, to compute the fractional day of the year of the current time. I was wondering if there are shorter ways of doing it, because my code feels like an overkill of std::chrono and date calls. Can I directly calculate the number of fractional days since the start of the year (at microsecond precision) and avoid my two-step approach?
#include <iostream>
#include <chrono>
#include "date.h"
int main()
{
// Get actual time.
auto now = std::chrono::system_clock::now();
// Get the number of days since start of the year.
auto ymd = date::year_month_day( date::floor<date::days>(now) );
auto ymd_ref = date::year{ymd.year()}/1/1;
int days = (date::sys_days{ymd} - date::sys_days{ymd_ref}).count();
// Get the fractional number of seconds of the day.
auto microseconds = std::chrono::duration_cast<std::chrono::microseconds>(now - date::floor<date::days>(now));
double seconds_since_midnight = 1e-6*microseconds.count();
// Get fractional day number.
std::cout << "Fractional day of the year: " << days + seconds_since_midnight / 86400. << std::endl;
return 0;
}
Good question (upvoted).
I think first we need to decide on what the right answer is. There's your answer, and currently the only other answer is Matteo's. For demonstration purposes, I've modified both answers to substitute in a "fake now" so that we can compare apples to apples:
using namespace std::chrono_literals;
auto now = date::sys_days{date::March/27/2019} + 0h + 32min + 22s + 123456us;
(approximately now at the time I'm writing this)
Chiel's code gives:
Fractional day of the year: 85.0225
Matteo's code gives:
Fractional day of the year: 85.139978280740735
They are close, but not close enough to both be considered right.
Matteo's code works with "average years":
auto this_year = date::floor<date::years>(now);
The length of a date::years is 365.2425 days, which is exactly right if you average all civil years over a 400 year period. And working with the average year length can be very useful, especially when dealing with systems that don't care about human made calendars (e.g. physics or biology).
I'm going to guess that because of the way Chiel's code is written, he would prefer a result that refers more precisely to this specific year. Therefore the code presented below is Chiel's's algorithm, resulting in exactly the same result, only slightly more efficient and concise.
// Get actual time.
auto now = std::chrono::system_clock::now();
// Get the number of days since start of the year.
auto sd = date::floor<date::days>(now);
auto ymd = date::year_month_day( sd );
auto ymd_ref = ymd.year()/1/1;
std::chrono::duration<double, date::days::period> days = sd - date::sys_days{ymd_ref};
// Get the fractional number of seconds of the day.
days += now - sd;
// Get fractional day number.
std::cout << "Fractional day of the year: " << days.count() << std::endl;
The first thing I noted was that date::floor<date::days>(now) was being computed in 3 places, so I'm computing it once and saving it in sd.
Next, since the final answer is a double-based representation of days, I'm going to let <chrono> do that work for me by storing the answer in a duration<double, days>. Any time you find yourself converting units, it is better to let <chrono> do it for you. It probably won't be faster. But it definitely won't be slower, or wrong.
Now it is a simple matter to add the fractional day to the result:
days += now - sd;
using whatever precision now has (microseconds or whatever). And the result is now simply days.count().
Update
And with just a little bit more time to reflect ...
I noticed that with the simplified code above, one can more easily see the entire algorithm as a single expression. That is (removing namespace qualification in order to get everything on one line):
duration<double, days::period> days = sd - sys_days{ymd_ref} + now - sd;
And this clearly algebraically simplifies down to:
duration<double, days::period> days = now - sys_days{ymd_ref};
In summary:
using namespace std::chrono;
using namespace date;
// Get actual time.
auto now = system_clock::now();
// Get the start of the year and subract it from now.
using ddays = duration<double, days::period>;
ddays fd = now - sys_days{year_month_day{floor<days>(now)}.year()/1/1};
// Get fractional day number.
std::cout << "Fractional day of the year: " << fd.count() << '\n';
In this case, letting <chrono> do the conversions for us, allowed the code to be sufficiently simplified such that the algorithm itself could be algebraically simplified, resulting in cleaner and more efficient code that is provably equivalent to the original algorithm in the OP's question.
Depending on the amount of input data, I have a program that runs in seconds or in days. At the end of my program, I want to print the elapsed "clock wall" time: in seconds if it is less then one minute, in min and sec if it is less than one hour, in hour-min-sec if it is less than one day, and in day-hour-min-sec otherwise. Here is the code I am using:
#include <cstdio>
#include <ctime>
#include <unistd.h> // for sleep
int main (int argc, char ** argv)
{
time_t startRawTime, endRawTime;
time (&startRawTime);
printf ("%s", ctime (&startRawTime));
sleep (3); // any preprocessing of input data
time (&endRawTime);
printf ("%s", ctime (&endRawTime));
printf ("%.0fs\n", difftime (endRawTime, startRawTime));
time_t elapsed = static_cast<time_t>(difftime (endRawTime, startRawTime));
struct tm * ptm = gmtime (&elapsed);
printf ("%id %ih %im %is\n", ptm->tm_mday, ptm->tm_hour, ptm->tm_min, ptm->tm_sec);
return 0;
}
Here is what it prints:
Mon Apr 9 14:43:16 2012
Mon Apr 9 14:43:19 2012
3s
1d 0h 0m 3s
Of course the last line is wrong (it should be "0d"). It seems it can be solved easily by printing ptm->tm_mday - 1. However, ptm->tm_mday will also be "1" when there really was one day elapsed between the two dates. And so in that case, I don't want to make it appear as "0d".
So is there a way to handle this properly? Or should I get the result of difftime as a double (that is, as a number of seconds) and then calculate myself the number of sec/min/hours/days?
Remark: my code is used only on Linux, compiled with gcc -lstdc++.
A time_t value represents a particular moment in time. The result of difftime is the interval, in seconds, between two moments. That's a very different thing.
In your code, difftime() returns 3.0, since there are 3 seconds between the two specified times. Converting that to time_t gives you a moment 3 seconds after the epoch; on most systems, that's going to be 3 seconds past midnight GMT on January 1, 1970. The tm_mday value is 1 because that was the first day of the month.
You might be able to make this work by subtracting 1 from the tm_mday value, since tm_mday is 1-based rather than 0-based. But you'll still get meaningless results for longer intervals. For example, an interval of 31.5 days will give you noon on February 1, because January has 31 days; that's not relevant to the information you're trying to get.
Just treat the result of difftime() as a double (because that's what it is) and compute the number of days, hours, minutes, and seconds by simple arithmetic.
(With some loss of portability, you can just subract the time_t values directly rather than using difftime(). That will make some of the arithmetic a little easier, but it will break on systems where a time_t value is something other than an integer count of seconds since some epoch. difftime() exists for a reason.)
Of course the last line is wrong (it should be "0d"). It seems it can
be solved easily by printing "ptm->tm_mday - 1". However, ptm->tm_mday
will also be "1" when there really was one day elapsed between the two
dates. And so in that case, I don't want to make it appear as "0d".
That's not correct; if the time interval is just over 1 day, ptm->tm_mday will be 2. You can verify this with a small modification to your code:
time (&endRawTime);
endRawTime += 86400; // add this line
printf ("%s", ctime (&endRawTime));
When I make this change, I get this output:
Mon Apr 9 13:56:49 2012
Tue Apr 10 13:56:52 2012
86403s
2d 0h 0m 3s
which could be corrected by subtracting 1 from ptm->tm_mday. But again, that's not the right approach.
Here's an example using the <chrono> library, a typesafe timing library that will prevent you from making the kind of mistake you're making. In chrono time_points and durations are not interchangeable, and if you try to use them that way then you get compiler errors.
#include <chrono>
#include <iostream>
#include <thread>
#include <cassert>
template<typename Rep,typename Period>
void print_duration(std::chrono::duration<Rep,Period> t) {
assert(0<=t.count() && "t must be >= 0");
// approximate because a day doesn't have a fixed length
typedef std::chrono::duration<int,std::ratio<60*60*24>> days;
auto d = std::chrono::duration_cast<days>(t);
auto h = std::chrono::duration_cast<std::chrono::hours>(t - d);
auto m = std::chrono::duration_cast<std::chrono::minutes>(t - d - h);
auto s = std::chrono::duration_cast<std::chrono::seconds>(t - d - h - m);
if(t>=days(1))
std::cout << d.count() << "d ";
if(t>=std::chrono::hours(1))
std::cout << h.count() << "h ";
if(t>=std::chrono::minutes(1))
std::cout << m.count() << "m ";
std::cout << s.count() << "s";
}
int main() {
auto start = std::chrono::steady_clock::now();
std::this_thread::sleep_for(std::chrono::seconds(3));
auto finish = std::chrono::steady_clock::now();
print_duration(finish-start);
std::cout << '\n';
}
Notes for GCC
Older versions of GCC have monotonic_clock instead of steady_clock. 4.7 has steady_clock.
In order to access std::this_thread::sleep_for you may have to define _GLIBCXX_USE_NANOSLEEP for some reason.