Reputation: 1026
I want to make a program which takes no input and returns the date of the previous Monday. (I don't care about time zones. And I am only worried about Gregorian calendar). I am using date by Howard Hinnant. This is how I am doing it currently:
#include <iostream>
#include <date/date.h>
int main() {
auto todays_day = date::year_month_weekday(date::floor<date::days>(std::chrono::system_clock::now()));
auto todays_date = date::floor<date::days>(std::chrono::system_clock::now());
int offset = 0;
auto weekday = todays_day.weekday();
if(weekday == date::Tuesday)
offset = 1;
else if (weekday == date::Wednesday)
offset = 2;
else if (weekday == date::Thursday)
offset = 3;
else if (weekday == date::Friday)
offset = 4;
else if (weekday == date::Saturday)
offset = 5;
else if (weekday == date::Sunday)
offset = 6;
auto lastMonday = date::year_month_day(todays_date - date::days(offset));
std::cout << lastMonday;
}
Is there a better way to do this without boost::previous_weekday? (It's not a requirement not to use boost. I am just wondering if it is possible)
Upvotes: 3
Views: 1349
Reputation: 219428
The key to understanding how to do this more simply is knowing this one fact about Howard Hinnant's date library:
weekday
difference is circular (or modulo 7 if you prefer). That is, any weekday
subtracted from any weekday
results in a number of days
in the range [0, 6]. This effectively hides the underlying encoding of weekday
.
Thus there is no need to translate [Monday, Sunday] into [0, 6] (or any other encoding):
#include "date/date.h"
#include <iostream>
int
main()
{
auto todays_date = date::floor<date::days>(std::chrono::system_clock::now());
date::year_month_day lastMonday = todays_date -
(date::weekday{todays_date} - date::Monday);
std::cout << lastMonday << '\n';
}
Instead you just have to decide how many days you need to subtract from a sys_days
(todays_date
in this example). That number of days is today's weekday
minus Monday
. If today is Monday
, the result is days{0}
. If today is Sunday
, the result is days{6}
. We could just as well be talking about finding the previous Friday. The logic would not change.
Also, one can directly convert a sys_days
to a weekday
. No need to go though year_month_weekday
.
The code in the OP's question considers the "previous Monday" to be today if today happens to be a Monday. And that is fine. That is what is desired in many "previous weekday" algorithms. And it is the logic I have coded above.
But it is also common to want the previous-weekday-algorithm to result in last week if the weekday you are seeking is today. I.e. if today is Monday, compute a week ago instead of today. That too is easily doable, and by pretty much the same algorithm. One just has to subtract a day at the beginning of the algorithm if you desire this behavior:
auto todays_date = ...
todays_date -= date::days{1};
date::year_month_day lastMonday = ...
Upvotes: 7