Is there a way to get only the no. of weekdays between 2 boost dates.
In the following, I'm only getting calendar days.
date begin_dt(2011,Aug,3);
date end_dt(day_clock::local_day());
days duration=end_dt-begin_dt;
std::cout<<"calendar days between begin & end date are: "<<duration<<std::endl;
Perhaps the simplest way is to run a day_iterator from start to finish:
#include <iostream>
#include <boost/date_time.hpp>
int main()
{
using namespace boost::gregorian;
date begin_dt(2011,Aug,3);
date end_dt(day_clock::local_day());
days duration=end_dt-begin_dt;
std::cout<<"calendar days between begin & end date are:" << duration << '\n';
int cnt=0;
for(day_iterator iter = begin_dt; iter!=end_dt; ++iter)
{
if( iter->day_of_week() != boost::date_time::Saturday
&& iter->day_of_week() != boost::date_time::Sunday)
++cnt;
}
std::cout << "of them " << cnt << " are weekdays\n";
}
You could start substracting the number of days to the next week start, and deleting either 1 or two depending if you're on saturday or before, or sunday. Then, you can divide the rest of the days remaining by 7, multiply that number by 2, and substract to the days. You have to make a case for the remainder too. If it is 6 (saturday), you have to remove one more. Not easy, but you get the idea.
Related
I was trying to make a program that calculated the fee of a membership at a swimming pool. The user would enter a date (the last time a member renewed the membership), and the program would calculate if their membership was overdue or not, using the current date.
The membership is meant to be overdue a week (or another arbitrary time period) before the start of the month they joined in a year's time. For example, if I joined in February 2016, I would have to pay on January 24 2017 or before to make sure the membership is overdue. As soon as it gets to January 25, a month fee should be charged ($15) and as soon as it reaches February 25, two months fee should be charged etc.
However, I do not know how to charge for subsequent months after the first one. For example, paying on February 3 should result in one month overdue but paying on February 26 should be two months but I do not know how to do this.
How can I fix my function because it doesn't seem to work?
E.g. I entered November 15 2016 and it should return 15 since the membership was due on October 24 2017 but it returns 0.
int membershipFine(int joinDay, int joinMonth, int joinYear, int currentDay, int currentMonth, int currentYear)
{
int dueDay[12] = {25, 22, 25, 24, 25, 24, 25, 25, 24, 25, 24, 25}; // the week before the end of each month in days
int correspondingMonth = joinMonth - 2; // finds the element position that corresponds
if (correspondingMonth == -1) // if they joined in january, the array will go to december
{
correspondingMonth = 11;
}
int differenceInMonths = currentMonth - joinMonth + 12 * (currentYear - joinYear);
if (differenceInMonths < 11)
{
return 0;
}
else if ((differenceInMonths == 11) && (joinDay < dueDay[correspondingMonth]))
{
return 0;
}
else if (differenceInMonths == 11)
{
return 15;
}
if (differenceInMonths > 11 && joinDay < dueDay[correspondingMonth]) // not sure about this if and else statement
{
return (differenceInMonths - 11) * 15;
}
else return (differenceInMonths - 10) * 15;
}
The best way to deal with dates and times is to use a library that raises the level of abstraction about integers, to dates and times. Howard Hinnant's free, open-source, header-only library is such a tool.
It has a {year, month, day} class called date::year_month_day that lends itself to year and month arithmetic. One could use this to change the API of membershipFine from taking 6 type-unsafe parameters to just two type-safe parameters:
int
membershipFine(date::year_month_day joinDate, date::year_month_day currentDate);
Your description of the due date appears to say that it is independent of the day of the month of the join date, and that it is 1 year, less 1 week from the first of the month of the join date. If this is true, this can be easily computed like this:
using namespace date;
year_month_day dueDate{
local_days{(joinDate.year()/joinDate.month() + years{1})/1} - weeks{1}};
The expression joinDate.year()/joinDate.month() creates a year_month object, which is just a year and month, neglecting the day-of-the-month from joinDate. I add 1 year to that year_month, which results in another year_month, exactly 1 year later.
To that sum, /1 is appended. This creates a year_month_day corresponding to the first day of the month of the aforementioned year_month.
Now even though year_month_day is great for year and month oriented arithmetic, it is not so great for day and week-oriented arithmetic. The best data structure for that is a {count-of-days} from some epoch. This library has such a data structure called local_days. So I convert to that, subtract 1 week, and then convert back to year_month_day.
All of this (to compute the due date) happens in the lines of code above.
Now I need to compute the fine based on the relationship between currentDate and dueDate. The fine is $0 if currentDate < dueDate, and otherwise is a function of the number of whole months (plus 1) currentDate is beyond dueDate (as I understand your problem statement):
int fine = 0;
if (currentDate >= dueDate)
{
auto differenceInMonths = currentDate.year()/currentDate.month() -
dueDate.year()/dueDate.month();
if (currentDate.day() >= dueDate.day())
++differenceInMonths;
fine = differenceInMonths.count() * 15;
}
The difference in months, neglecting the day-of-the-month, can be computed by converting to year_month objects and subtracting. Now if currentDate.day() < dueDate.day(), this is the correct answer. For example if the difference in months is 1, but the day of the month in currentDate has not yet exceeded the day of the month in dueDate, then we don't want to charge for a second month, else we do. If we do, differenceInMonths is incremented.
Then the fine is simply the differenceInMonths, converted from months to integral, times 15.
<aside> If there are any <chrono> fans out there, the type of differenceInMonths is actually a std::chrono::duration with a period that is exactly the average month. Thus the .count() member function to access the underlying integral value.
I've added some print statements to the above code, and below I show the whole thing put together plus a driver with a few examples:
#include "date/date.h"
#include <iostream>
int
membershipFine(date::year_month_day joinDate, date::year_month_day currentDate)
{
using namespace date;
year_month_day dueDate{
local_days{(joinDate.year()/joinDate.month() + years{1})/1} - weeks{1}};
int fine = 0;
if (currentDate >= dueDate)
{
auto differenceInMonths = currentDate.year()/currentDate.month() -
dueDate.year()/dueDate.month();
if (currentDate.day() >= dueDate.day())
++differenceInMonths;
fine = differenceInMonths.count() * 15;
}
std::cout << "join Date is " << joinDate << '\n';
std::cout << "due Date is " << dueDate << '\n';
std::cout << "current Date is " << currentDate << '\n';
std::cout << "fine is $" << fine << '\n';
return fine;
}
int
main()
{
using namespace date::literals;
std::cout << membershipFine(feb/29/2016, jan/24/2017) << '\n';
std::cout << membershipFine(feb/29/2016, jan/25/2017) << '\n';
std::cout << membershipFine(feb/29/2016, feb/24/2017) << '\n';
std::cout << membershipFine(feb/29/2016, feb/25/2017) << '\n';
}
This outputs:
join Date is 2016-02-29
due Date is 2017-01-25
current Date is 2017-01-24
fine is $0
0
join Date is 2016-02-29
due Date is 2017-01-25
current Date is 2017-01-25
fine is $15
15
join Date is 2016-02-29
due Date is 2017-01-25
current Date is 2017-02-24
fine is $15
15
join Date is 2016-02-29
due Date is 2017-01-25
current Date is 2017-02-25
fine is $30
30
In summary, the use of a library such as this frees you from having to think in terms of ints so you can concentrate on the logic you have to implement in terms of dates and calendars. The result is compact and readable code that is far more likely to be correct.
Update
In the comments below the OP asks about how to parse a date from cin and how to get the current date. There are several options.
Here is how I recommend asking for a date:
date::year_month_day join;
while (true)
{
std::cout << "Enter join date as yyyy-mm-dd: ";
std::cin >> date::parse("%F", join);
if (!std::cin.fail())
break;
std::cin.clear();
std::string garbage;
std::getline(std::cin, garbage);
std::cout << "Please try again.\n";
}
If you prefer to ask for some other format, here is the complete list of parsing flags available for use.
And the OP asks how to get the current date. There are multiple answers. If you are content with the current date in UTC, that is the simplest:
using namespace std::chrono;
using namespace date;
year_month_day today = floor<days>(system_clock::now());
If you want the current date in your local time zone, you need to use "date/tz.h" (requires some installation) and this syntax:
year_month_day today{floor<days>(make_zoned(current_zone(),
system_clock::now()).get_local_time())};
If you want the current date in some time zone other than your current local time zone, that can be done with:
year_month_day today{floor<days>(make_zoned("America/Los_Angeles",
system_clock::now()).get_local_time())};
No matter how you parse your join and today, they can be used like this:
std::cout << membershipFine(join, today) << '\n';
how to find all sundays from given date between march 2016 - march 2018. with looping, there are 3 loops : looping for year, month, and date, then use if.
But im confuse about looping. can you help me ?
I'd use a good date library like this one. But that's probably not what your mentor is looking for.
#include "date.h"
#include <iostream>
int
main()
{
using namespace date;
for (sys_days start = mar/sun[1]/2016; start <= sys_days{mar/last/2018};
start += weeks{1})
std::cout << start << '\n';
}
2016-03-06
2016-03-13
2016-03-20
...
2018-03-18
2018-03-25
I have an exercice, which I am having a little trouble with.
I must create a calculator which takes two parameters: Start date and days to add (except saturday and sunday, only business days, from monday to friday). Another thing is that the sum has to include the start date.
E.g. let's take the start day July 12th 2016, and add 8 days, which correspond to July 21th 2016 (Saturday and Sunday excluded, and Tuesday, July 21th 2016 is counted as one day).
I hope I'm clear.
I tried to code something, but it is not working.
// rStringGridEd1->IntCells[3][row] is a custom stringgrid
// and correspond to the number of days to add, j is the
// counter for the loop
while (j < rStringGridEd1->IntCells[3][row])
{
if (DayOfWeek(date) != 1 || DayOfWeek(date) !=7)
{
// if current date (TDate date = "12/07/16") is not Saturday or Sunday increment date by one day
date++;
}
else if(DayOfWeek(date) == 1)
{
//If date correspond to sunday increment the date by one and j the counter by one
date=date+1;
j++;
}
else if(DayOfWeek(date) == 7)
{
//If date correspond to saturday increment the date by two days and j the counter by one
date=date+2;
j++;
}
j++;
}
Can anyone help me, please?
Here is what Lee Painton's excellent (and up-voted) answer would look like using this free, open-source C++11/14 date library which is built on top of <chrono>:
#include "date.h"
#include <iostream>
date::year_month_day
get_end_job_date(date::year_month_day start, date::days length)
{
using namespace date;
--length;
auto w = weeks{length / days{5}};
length %= 5;
auto end = sys_days{start} + w + length;
auto wd = weekday{end};
if (wd == sat)
end += days{2};
else if (wd == sun)
end += days{1};
return end;
}
You could exercise it like this:
int
main()
{
using namespace date::literals;
std::cout << get_end_job_date(12_d/jul/2016, date::days{8}) << '\n';
}
Which outputs:
2016-07-21
This simplistic calculator has a precondition that start is not on a weekend. If that is not a desirable precondition then you could detect that prior to the computation and increment start internally by a day or two.
The date library takes care of things like the relationship between days and weeks, and how to add days to a date. It is based on very efficient (non-iterative) algorithms shown and described here.
If you aren't required to use a loop then you might want to consider refactoring your solution with a simpler calculation. Consider, for example, that every five business days automatically adds seven days to the date. Thus using the quotient and remainder of the days to add should tell you how many total days to add to your date variable without resorting to a brute force loop.
Since it's an exercise I won't get into specifics of code, but a few things to consider might be how you can figure out what day of the week you end on knowing the day that you started on. Also, if you end on a friday what happens with the weekend that immediately follows it.
I'm writing a impala udf in c++ which gets week of year when provided with date in yyyyMMdd. But could not seem to find way to convert yyyyMMdd to week of year in c++. In java I can you Calendar, but how to go about it in c++.
TIA
You can just use std::mktime from <ctime>. Example:
std::tm date={};
date.tm_year=2014-1900;
date.tm_mon=9-1;
date.tm_mday=28;
std::mktime(&date);
After the call, date.tm_wday is adjusted (0=Sunday). date.tm_yday is also adjusted.
To get the week into the year, use: (date.tm_yday-date.tm_wday+7)/7
This calculation returns 1 for the first full week (namely, first week with a Sunday in it in the year, including Jan 1 in years that start with Sunday); and 0 for days in the first partial week.
I answered this here, but for the sake of completeness, I will repeat it here too:
Use iso_week.h from howardhinnant.github.io/iso_week.html :
#include <iostream>
#include "iso_week.h"
int main() {
using namespace iso_week;
using namespace std::chrono;
// Get the current time and floor to convert to the sys_days:
auto today = floor<days>(system_clock::now());
// Convert from sys_days to iso_week::year_weeknum_weekday format
auto yww = year_weeknum_weekday{today};
// Print current week number of the year
std::cout << "The current week of " << yww.year() << " is: "
<< yww.weeknum() << std::endl;
// Set any day
auto any_day = 2014_y/9/28;
// Get week of `any_day`
std::cout << "The week of " << any_day.year() << " on `any day` was: "
<< any_day.weeknum() << std::endl;
}
which gives the output:
The current week of 2019 is: W18
The week in 2014 on `any day` was: W09
Use boost::date library.It's simple and easy to use.
Hi to all thank all in advance to those who tried to answer or answer and part of this question.
Calculate the sum of the digits of the year.
Calculate the absolute value of the difference between the year and the ’reverse’ of the year.
Calculate the number of even factors of the day.
Calculate the greatest common divisor of the day, month and year.
Calculate the number of steps required to solve the Collatz problem for
the month
These are my tasks that I need to fulfill, as Engineering student this how far I went in this. In the following codes I expect something like this
19
90
0
1
0
T M B B
The answer that I get is
Please enter your birthdate (dd mm yyyy): 12 11 1981
19
8468304
Press any key to continue . . .
8468304
How to get it right I know that my equation is right or(formula, method). However this is what I know.
#include<iostream>
#include<cstdlib>
using namespace std;
int main()
{
cout << "Please enter your birthdate (dd mm yyyy): ";
int day, month, year, count,rev;
int sum = 0;
cin >> day>> month >>year;
while (year!=0)
{
int count = year%10;
sum +=count;
year /= 10;
}
while(year>0)
{
rev = year%10;
year=year/10;
}
cout<<sum<<endl;
cout << rev;
system ("pause");
return 0;
}//end main
Please help!
After your first loop, while (year != 0), you don't reset the value of year, so it remains at zero and the second loop doesn't execute at all.
You need to save the value of year and use it when you start the second loop.
Just a note on organisation: I'd suggest to write a subroutine/function for every task, like
int digit_sum(int year) {
/* ... */
return sum;
}
int reverse_difference(int year) {
/* ... */
return diff;
}
and so on. This way you'll also prevent errors like modifying the year variable during the first calculation without saving the original value (which you did, as David Winant already pointed out).