给定日期的c++星期几

第一:如果已经有可以处理相同问题的标准化函数，则不要编写自己的函数。问题是，您可能很容易犯错误(我已经在您的weekday()函数的第一行看到了现在的错误)，而标准化函数的实现已经经过了彻底的测试，您可以确信它们交付了您期望得到的结果。

话虽如此，这里有一种可能的方法，使用std::localtime和std::mktime:

#include <ctime>
#include <iostream>
int main()
{
  std::tm time_in = { 0, 0, 0, // second, minute, hour
      9, 10, 2016 - 1900 }; // 1-based day, 0-based month, year since 1900
  std::time_t time_temp = std::mktime(&time_in);
  //Note: Return value of localtime is not threadsafe, because it might be
  // (and will be) reused in subsequent calls to std::localtime!
  const std::tm * time_out = std::localtime(&time_temp);
  //Sunday == 0, Monday == 1, and so on ...
  std::cout << "Today is this day of the week: " << time_out->tm_wday << "n";
  std::cout << "(Sunday is 0, Monday is 1, and so on...)n";
  return 0;
}

老问题的新答案，因为它们是一种不断变化的工具…

c++ 20规范指出，以下代码将具有与问题中代码意图相同的功能:

#include <chrono>
#include <format>
#include <iostream>
int
main()
{
    using namespace std;
    using namespace std::chrono;
    year_month_day dmy;
    cin >> parse("%d %m %Y", dmy);
    cout << format("{:%A}", weekday{dmy}) << 'n';
}

今天可以使用这个免费的、开源的日期/时间库来试验这种语法，只是日期对象在namespace date而不是namespace std::chrono中，并且格式字符串的语法略有改变。

#include "date/date.h"
#include <iostream>
int
main()
{
    using namespace std;
    using namespace date;
    year_month_day dmy;
    cin >> parse("%d %m %Y", dmy);
    cout << format("%A", weekday{dmy}) << 'n';
}

你对闰年的理解是错误的:

闰年是每4年一次除了如果它能被100整除，但是即使如此如果它能被400整除它仍然是闰年

关于如何计算"天数"(dn)的清晰简明的解释可以在这里找到。

一旦您有了天数(dn)，只需执行模数7。结果将是星期几(下跌)。

下面是一个示例实现(不检查date是否是有效输入):

#include <iostream>
#include <iomanip>
typedef unsigned long ul;
typedef unsigned int ui;
// ----------------------------------------------------------------------
// Given the year, month and day, return the day number.
// (see: https://alcor.concordia.ca/~gpkatch/gdate-method.html)
// ----------------------------------------------------------------------
ul CalcDayNumFromDate(ui y, ui m, ui d)
{
  m = (m + 9) % 12;
  y -= m / 10;
  ul dn = 365*y + y/4 - y/100 + y/400 + (m*306 + 5)/10 + (d - 1);
  return dn;
}
// ----------------------------------------------------------------------
// Given year, month, day, return the day of week (string).
// ----------------------------------------------------------------------
std::string CalcDayOfWeek(int y, ul m, ul d)
{
  std::string day[] = {
    "Wednesday",
    "Thursday",
    "Friday",
    "Saturday",
    "Sunday",
    "Monday",
    "Tuesday"
  };
  ul dn = CalcDayNumFromDate(y, m, d);
  return day[dn % 7];
}
// ----------------------------------------------------------------------
// Program entry point.
// ----------------------------------------------------------------------
int main(int argc, char **argv)
{
  ui y = 2017, m = 8, d = 29; // 29th August, 2017.
  std::string dow = CalcDayOfWeek(y, m, d);
  std::cout << std::setfill('0') << std::setw(4) << y << "/";
  std::cout << std::setfill('0') << std::setw(2) << m << "/";
  std::cout << std::setfill('0') << std::setw(2) << d << ": ";
  std::cout << dow << std::endl;
  return 0;
}

您可以使用Boost c++库中的公历日期系统来查找给定日期的星期几。下面是一个简单的例子:

#include <boost/date_time.hpp>
#include <string>
#include <iostream>
const static std::string daysOfWeek[] = {
    "Sunday",
    "Monday",
    "Tuesday",
    "Wednesday",
    "Thursday",
    "Friday",
    "Saturday"
};
int getDayOfWeekIndex(int day, int month, int year) {
    boost::gregorian::date d(year, month, day);
    return d.day_of_week();
}
int main()
{
    const int index = getDayOfWeekIndex(30, 07, 2018);
    std::cout << daysOfWeek[index] << 'n';
}

此代码打印Monday。

我也遇到过同样的问题，但我找到了一个简单的解决方案。

根据这篇文章:
下面是Sakamoto, Lachman, Keith和Craver提出的计算日的简单函数。下面的函数星期天返回0，星期一返回1，等等。"

int dayofweek(int d, int m, int y)  
{  
    static int t[] = { 0, 3, 2, 5, 0, 3, 5, 1, 4, 6, 2, 4 };  
    y -= m < 3;  
    return ( y + y / 4 - y / 100 + y / 400 + t[m - 1] + d) % 7;  
}

当一个数能被7整除时会发生什么?

14/7 = 214 % 7 = 0

模运算符(% n)将返回一个从0到n的数字-1

如果n除以7，余数永远不可能是7所以

int b = (a - 2)  % 7;
    switch (b){
    case 1:
        return "Monday";
    case 2:
        return "Tuesday";
    case 3:
        return "Wednesday";
    case 4:
        return "Thursday";
    case 5:
        return "Friday";
    case 6:
        return "Saturday";
    case 7:
        return "Sunday";
    }
}

在这种情况下，它不可能是星期天

试试这个

int b = (a - 2)  % 7;
        switch (b){
        case 0:
            return "Sunday";
        case 1:
            return "Monday";
        case 2:
            return "Tuesday";
        case 3:
            return "Wednesday";
        case 4:
            return "Thursday";
        case 5:
            return "Friday";
        case 6:
            return "Saturday";
        default:
            return "Error";
        }

尝试使用CTime类

为例:

const CTime currTime = CTime::GetCurrentTime();
const int nWeekDay = currTime.GetDayOfWeek();
switch (nWeekDay)
{
    case 1:
        return "Monday";
    case 2:
        return "Tuesday";
    case 3:
        return "Wednesday";
    case 4:
        return "Thursday";
    case 5:
        return "Friday";
    case 6:
        return "Saturday";
    case 7:
        return "Sunday";
}

在上面的例子中，我使用的是当前时间，但你可以用不同的方式，用你想要的时间，例如:

const CTime currTime = CTime(year,month, day, hours, minutes, seconds );

int dayofweek(int day,int month,int year)
{
    int arr[] = {0,3,2,5,3,5,1,4,6,2,4};
    if(month<3)
        year--;
    return ((year+year/4-year/100+year/400+arr[month-1]+day)%7);
}
int main()
{
    int day,month,year;
    cout<<"Enter the Date for which day of the week need to be find (DD/MM/YYYY)."<<endl;
    cin>>day>>month>>year;
    int x = dayofweek(day,month,year);
    if(x==0)
        cout<<"Sunday"<<endl;
    else if(x==1)
        cout<<"Monday"<<endl;
    else if(x==2)
        cout<<"Tuesday"<<endl;
    else if(x==3)
        cout<<"Wednesday"<<endl;
    else if(x==4)
        cout<<"Thursday"<<endl;
    else if(x==5)
        cout<<"Friday"<<endl;
    else if(x==6)
        cout<<"Saturday"<<endl;
}