Getting the day info only from a date int C/C++ [closed] - c++

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I have a program that is getting a date from some protocol, in format of DD/MM/YYYY.
The Problem is that I need to know the info of that day(day of the week, day in the year...) and I don't know how to do it.
Usually, when I want to get a day info, I am using time(time_t*) and convert the result to tm struct using localltime_r(tm*, time_t*) and then I have everything i need.
But in this case, this is not the current time(so I can not use time(time_t*)) and I don't have nothing except the date.
I can create a new tm struct and fill only tm_year, tm_mon, tm_mday and use mktime(tm*), but I am not sure if this will give me the right detail's of the desired date.

Answer 1

You might consider using Howard Hinnant's free, open-source date/time library. It works with C++11 and later, ported to gcc, clang and VS. It has been accepted into the C++20 draft specification.
Example code:
#include "date/date.h"
#include <iostream>
#include <sstream>
int
main()
{
using namespace std;
using namespace date;
istringstream in{"09/07/2018"};
sys_days sd;
in >> parse("%d/%m/%Y", sd);
cout << "Day of the week is " << weekday{sd} << '\n';
cout << "Day of the year is " << sd - sys_days{year_month_day{sd}.year()/1/0} << '\n';
}
Output:
Day of the week is Mon
Day of the year is 190d
If you would rather do the computations yourself, here are extremely efficient non-iterative public domain algorithms for calendrical computations. The library referenced above is nothing but a type-safe wrapper around these algorithms with more pleasant syntax.

Answer 2

In case you wanted to get write methods which get the day of week/year as opposed to using a library, here's what I would do.
You're going to need to take leap years into account. For any given year, determine whether or not it is a leap year. Something like this would work (using a formula found here: https://en.wikipedia.org/wiki/Leap_year#Algorithm:
bool isLeapYear(short year)
{
bool leapYear = false;
if (((year % 4 == 0 && year % 100 != 0)) || (year % 400 == 0))
leapYear = true;
return leapYear;
}
From this, calculating the day of the year is straightforward. Simply add up all the days for each month, if it is a leap year, add 29 days onto your tally for day of year if you come across February.
As for finding the day of the week, it really helps if you start with some lower bound for the year (in this case LOWYEAR = 1760) and start with the first day of that year (STARTDAYOFWEEK = 2). Each day of the week is corresponds to a number (0-6) where Sunday is 0, Monday is 1, etc.
int DayOfWeek(void)
{
//jan 1 1760 was tuesday (2)
short dayCount = STARTDAYOFWEEK;
for (unsigned i = LOWYEAR; i < year; i++)
{
if (isLeapYear(i))
dayCount += 2;
else
dayCount++;
}
return (dayCount + dayOfYear) - 1) % DAYSINWEEK;
}
Finding the day of week is now really easy, after calculating dayCount, the day of week is found by adding dayCount with dayOfYear and modding by DAYSINWEEK (6).
The resulting number will correspond to the day of week (0-6).