[英]How do I make virtual methods of classes that inherit the same base class figure out what to do with their argument?
[英]Q&A: How do I figure out what the last day of the month is?
我正在嘗試編寫一個自己滾動的時區轉換器,我需要一種方法來確定該月的最后一天是什么。 經過一番研究,我發現了尋找閏年的公式。
這是一個很小的貢獻,但也許我會節省其他人我花了 20 分鍾弄清楚並應用它的時間。
此代碼接受一個有符號的短月,索引為 0(0 是一月)和一個索引為 0 的 int 年(2012 年是 2012 年)。
它返回 1 索引日(27 日是 27 日,但在 SYSTEMTIME 結構等中,您通常需要 0 索引 - 只是抬頭)。
short _get_max_day(short month, int year) {
if(month == 0 || month == 2 || month == 4 || month == 6 || month == 7 || month == 9 || month == 11)
return 31;
else if(month == 3 || month == 5 || month == 8 || month == 10)
return 30;
else {
if(year % 4 == 0) {
if(year % 100 == 0) {
if(year % 400 == 0)
return 29;
return 28;
}
return 29;
}
return 28;
}
}
關於什么
#include <time.h>
#include <iostream>
int LastDay (int iMonth, int iYear)
{
struct tm when;
time_t lastday;
// Set up current month
when.tm_hour = 0;
when.tm_min = 0;
when.tm_sec = 0;
when.tm_mday = 1;
// Next month 0=Jan
if (iMonth == 12)
{
when.tm_mon = 0;
when.tm_year = iYear - 1900 + 1;
}
else
{
when.tm_mon = iMonth;
when.tm_year = iYear - 1900;
}
// Get the first day of the next month
lastday = mktime (&when);
// Subtract 1 day
lastday -= 86400;
// Convert back to date and time
when = *localtime (&lastday);
return when.tm_mday;
}
int _tmain(int argc, _TCHAR* argv[])
{
for (int m = 1; m <= 12; m++)
std::cout << "Last day of " << m << " is " << LastDay (m, 2002) << std::endl;
return 0;
}
打印出來(2002年)......
Last day of 1 is 31
Last day of 2 is 28
Last day of 3 is 31
Last day of 4 is 30
Last day of 5 is 31
Last day of 6 is 30
Last day of 7 is 31
Last day of 8 is 31
Last day of 9 is 30
Last day of 10 is 31
Last day of 11 is 30
Last day of 12 is 31
我使用一個簡單的函數,它返回(標准)COleDateTime的整個日期。 它可能不是那么快的其他選擇,但它非常有效,適用於閏年和非常簡單的證明。
這是我正在使用的代碼:
COleDateTime get_last_day_of_month(UINT month, UINT year)
{
if(month == 2)
{ // if month is feb, take last day of March and then go back one day
COleDateTime date(year, 3, 1, 0, 0, 0); // 1 March for Year
date -= 1; // go back one day (the standard class will take leap years into account)
return date;
}
else if(month == 4 || month == 6 || month == 9 || month == 11) return COleDateTime(year, month, 30, 0, 0, 0);
else return COleDateTime(year, month, 31, 0, 0, 0);
}
import datetime
from datetime import date
from dateutil.relativedelta import relativedelta
year = int((date.today()).strftime("%Y"))
month = list(range(1, 13, 1))
YearMonthDay = [(datetime.datetime(year, x, 1) + relativedelta(day=31)).strftime("%Y%m%d") for x in month]
print(YearMonthDay)
['20220131', '20220228', '20220331', '20220430', '20220531', '20220630', '20220731', '20220831', '20220930', '20221031', '2021'
在 C++20 中:
#include <chrono>
std::chrono::day
get_max_day(std::chrono::month m, std::chrono::year y)
{
return (y/m/std::chrono::last).day();
}
如果你真的需要一個類型不安全的 API:
int
get_max_day(int m, int y)
{
return unsigned{(std::chrono::last/m/y).day()};
}
Word Year, Month, Day;
TDateTime datum_tdatetime = Date();
// first day of actual month
datum_tdatetime.DecodeDate(&year, &month, &day);
day = 1;
datum_tdatetime = EncodeDate(year, month, day);
// last day of previous month
datum_tdatetime -= 1;
// first day of previous month
datum_tdatetime.DecodeDate(&year, &month, &day);
day = 1;
datum_tdatetime = EncodeDate(year, month, day);
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