Я знаю, что это очень старый вопрос, но это интересный и некоторые общие один, когда дело доходит до работы с датами и временем. Поэтому я подумал о совместном использовании кода, который вычисляет новую дату без использования встроенных функций времени на C++.
#include <iostream>
#include <string>
using namespace std;
class Date {
public:
Date(size_t year, size_t month, size_t day):m_year(year), m_month(month), m_day(day) {}
~Date() {}
// Add specified number of days to date
Date operator + (size_t days) const;
// Subtract specified number of days from date
Date operator - (size_t days) const;
size_t Year() { return m_year; }
size_t Month() { return m_month; }
size_t Day() { return m_day; }
string DateStr();
private:
// Leap year check
inline bool LeapYear(int year) const
{ return year % 4 == 0 && (year % 100 != 0 || year % 400 == 0); }
// Holds all max days in a general year
static const int MaxDayInMonth[13];
// Private members
size_t m_year;
size_t m_month;
size_t m_day;
};
// Define MaxDayInMonth
const int Date::MaxDayInMonth[13] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
//===========================================================================================
/// Add specified number of days to date
Date Date::operator + (size_t days) const {
// Maximum days in the month
int nMaxDays(MaxDayInMonth[m_month] + (m_month == 2 && LeapYear(m_year) ? 1 : 0));
// Initialize the Year, Month, Days
int nYear(m_year);
int nMonth(m_month);
int nDays(m_day + days);
// Iterate till it becomes a valid day of a month
while (nDays > nMaxDays) {
// Subtract the max number of days of current month
nDays -= nMaxDays;
// Advance to next month
++nMonth;
// Falls on to next year?
if (nMonth > 12) {
nMonth = 1; // January
++nYear; // Next year
}
// Update the max days of the new month
nMaxDays = MaxDayInMonth[nMonth] + (nMonth == 2 && LeapYear(nYear) ? 1 : 0);
}
// Construct date
return Date(nYear, nMonth, nDays);
}
//===========================================================================================
/// Subtract specified number of days from date
Date Date::operator - (size_t days) const {
// Falls within the same month?
if (0 < (m_day - days)) {
return Date(m_year, m_month, m_day - days);
}
// Start from this year
int nYear(m_year);
// Start from specified days and go back to first day of this month
int nDays(days);
nDays -= m_day;
// Start from previous month and check if it falls on to previous year
int nMonth(m_month - 1);
if (nMonth < 1) {
nMonth = 12; // December
--nYear; // Previous year
}
// Maximum days in the current month
int nDaysInMonth = MaxDayInMonth[nMonth] + (nMonth == 2 && LeapYear(nYear) ? 1 : 0);
// Iterate till it becomes a valid day of a month
while (nDays >= 0) {
// Subtract the max number of days of current month
nDays -= nDaysInMonth;
// Falls on to previous month?
if (nDays > 0) {
// Go to previous month
--nMonth;
// Falls on to previous year?
if (nMonth < 1) {
nMonth = 12; // December
--nYear; // Previous year
}
}
// Update the max days of the new month
nDaysInMonth = MaxDayInMonth[nMonth] + (nMonth == 2 && LeapYear(nYear) ? 1 : 0);
}
// Construct date
return Date(nYear, nMonth, (0 < nDays ? nDays : -nDays));
}
//===========================================================================================
/// Get the date string in yyyy/mm/dd format
string Date::DateStr() {
return to_string(m_year)
+ string("/")
+ string(m_month < 10 ? string("0") + to_string(m_month) : to_string(m_month))
+ string("/")
+ string(m_day < 10 ? string("0") + to_string(m_day) : to_string(m_day));
}
int main() {
// Add n days to a date
cout << Date(2017, 6, 25).DateStr() << " + 10 days = "
<< (Date(2017, 6, 25) /* Given Date */ + 10 /* Days to add */).DateStr() << endl;
// Subtract n days from a date
cout << Date(2017, 6, 25).DateStr() << " - 10 days = "
<< (Date(2017, 6, 25) /* Given Date */ - 10 /* Days to subract */).DateStr() << endl;
return 0;
}
Output
2017/06/25 + 10 days = 2017/07/05
2017/06/25 - 10 days = 2017/06/15
Я удивлен, что этот вопрос существует без сопроводительного ответа «Использовать Boost» со ссылкой на документацию. – jww