22

我正在尝试编写一个 Date 类以尝试学习 C++。

我正在尝试找到一种算法来添加或减去日期到日期，其中 Day 从 1 开始，Month 从 1 开始。事实证明它非常复杂，而且谷歌并没有出现太多，

有谁知道这样做的算法？

4

10 回答 10

21

最简单的方法是实际编写两个函数，一个将天转换为从给定开始日期开始的天数，另一个将转换回日期。一旦日期表示为天数，添加或减去它是微不足道的。

你可以在这里找到算法：http: //alcor.concordia.ca/~gpkatch/gdate-algorithm.html

于 2010-02-26T19:47:41.733 回答
16

你真的不需要这样的算法（至少不是名副其实的东西），标准库可以完成大部分繁重的工作；日历计算是出了名的棘手。只要您不需要早于 1900 年的日期，那么：

#include <ctime>

// Adjust date by a number of days +/-
void DatePlusDays( struct tm* date, int days )
{
    const time_t ONE_DAY = 24 * 60 * 60 ;

    // Seconds since start of epoch
    time_t date_seconds = mktime( date ) + (days * ONE_DAY) ;

    // Update caller's date
    // Use localtime because mktime converts to UTC so may change date
    *date = *localtime( &date_seconds ) ; ;
}

示例用法：

#include <iostream>

int main()
{
    struct tm date = { 0, 0, 12 } ;  // nominal time midday (arbitrary).
    int year = 2010 ;
    int month = 2 ;  // February
    int day = 26 ;   // 26th

    // Set up the date structure
    date.tm_year = year - 1900 ;
    date.tm_mon = month - 1 ;  // note: zero indexed
    date.tm_mday = day ;       // note: not zero indexed

    // Date, less 100 days
    DatePlusDays( &date, -100 ) ; 

    // Show time/date using default formatting
    std::cout << asctime( &date ) << std::endl ;
}
于 2010-02-26T21:06:31.030 回答
3

我假设这是为了某种练习，否则您将使用已经提供给您的时间课程。

您可以将时间存储为自某个日期以来的毫秒数。然后您可以添加适当的值并在调用类的访问器时将其转换为日期。

于 2010-02-26T19:46:20.177 回答
2

这是一个非常简单的方法的草图。为了简单起见，我假设d要添加的天数 是正数。很容易将以下内容扩展到d否定的情况。

要么d小于 365，要么d大于或等于 365。

如果d小于 365：

m = 1;
while(d > numberOfDaysInMonth(m, y)) {
    d -= numberOfDaysInMonth(m, y);
    m++;
}
return date with year = y, month = m, day = d;

如果d大于 365：

while(d >= 365) {
    d -= 365;
    if(isLeapYear(y)) {
        d -= 1;
    }
    y++;
}
// now use the case where d is less than 365

或者，您可以用Julian 形式表示日期，然后只需添加到 Julian 形式并转换为 ymd 格式。

于 2010-02-26T19:50:00.160 回答
1

一种方法是将日期映射到日期的儒略数，进行整数运算，然后转换回来。

您会发现大量有关 julian 函数的资源。

于 2010-02-26T19:46:58.527 回答
1

试试这个功能。它正确计算加法或减法。dateTime 参数必须采用 UTC 格式。

tm* dateTimeAdd(const tm* const dateTime, const int& days, const int& hours, const int& mins, const int& secs) {
    tm* newTime = new tm;
    memcpy(newTime, dateTime, sizeof(tm));

    newTime->tm_mday += days;
    newTime->tm_hour += hours;
    newTime->tm_min += mins;
    newTime->tm_sec += secs;        

    time_t nt_seconds = mktime(newTime) - timezone;
    delete newTime;

    return gmtime(&nt_seconds);
}

并且有使用示例：

time_t t = time(NULL);
tm* utc = gmtime(&t);
tm* newUtc = dateTimeAdd(utc, -5, 0, 0, 0); //subtract 5 days
于 2013-10-06T13:40:40.370 回答
1

我知道这是一个非常古老的问题，但在处理日期和时间时，这是一个有趣且常见的问题。所以我想分享一些代码来计算新日期，而不使用 C++ 中的任何内置时间功能。

#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
于 2017-06-22T12:10:43.317 回答
0

我建议首先编写一个例程，将年月日转换为自固定日期以来的天数，例如自 1.01.01 以来。还有一个对称的例程可以将其转换回来。

不要忘记正确处理闰年！

拥有这两个，您的任务将是微不足道的。

于 2010-02-26T19:47:49.167 回答
0

我知道这是近十年前提出的一个老问题。但是几天前我遇到了同样的任务，这是这里的答案

// C++ program to find date after adding 
// given number of days. 
#include<bits/stdc++.h> 
using namespace std; 

// Return if year is leap year or not. 
bool isLeap(int y) 
{ 
    if (y%100 != 0 && y%4 == 0 || y %400 == 0) 
        return true; 

    return false; 
} 

// Given a date, returns number of days elapsed 
// from the beginning of the current year (1st 
// jan). 
int offsetDays(int d, int m, int y) 
{ 
    int offset = d; 

    switch (m - 1) 
    { 
    case 11: 
        offset += 30; 
    case 10: 
        offset += 31; 
    case 9: 
        offset += 30; 
    case 8: 
        offset += 31; 
    case 7: 
        offset += 31; 
    case 6: 
        offset += 30; 
    case 5: 
        offset += 31; 
    case 4: 
        offset += 30; 
    case 3: 
        offset += 31; 
    case 2: 
        offset += 28; 
    case 1: 
        offset += 31; 
    } 

    if (isLeap(y) && m > 2) 
        offset += 1; 

    return offset; 
} 

// Given a year and days elapsed in it, finds 
// date by storing results in d and m. 
void revoffsetDays(int offset, int y, int *d, int *m) 
{ 
    int month[13] = { 0, 31, 28, 31, 30, 31, 30, 
                    31, 31, 30, 31, 30, 31 }; 

    if (isLeap(y)) 
        month[2] = 29; 

    int i; 
    for (i = 1; i <= 12; i++) 
    { 
        if (offset <= month[i]) 
            break; 
        offset = offset - month[i]; 
    } 

    *d = offset; 
    *m = i; 
} 

// Add x days to the given date. 
void addDays(int d1, int m1, int y1, int x) 
{ 
    int offset1 = offsetDays(d1, m1, y1); 
    int remDays = isLeap(y1)?(366-offset1):(365-offset1); 

    // y2 is going to store result year and 
    // offset2 is going to store offset days 
    // in result year. 
    int y2, offset2; 
    if (x <= remDays) 
    { 
        y2 = y1; 
        offset2 = offset1 + x; 
    } 

    else
    { 
        // x may store thousands of days. 
        // We find correct year and offset 
        // in the year. 
        x -= remDays; 
        y2 = y1 + 1; 
        int y2days = isLeap(y2)?366:365; 
        while (x >= y2days) 
        { 
            x -= y2days; 
            y2++; 
            y2days = isLeap(y2)?366:365; 
        } 
        offset2 = x; 
    } 

    // Find values of day and month from 
    // offset of result year. 
    int m2, d2; 
    revoffsetDays(offset2, y2, &d2, &m2); 

    cout << "d2 = " << d2 << ", m2 = " << m2 
        << ", y2 = " << y2; 
} 

// Driven Program 
int main() 
{ 
    int d = 14, m = 3, y = 2015; 
    int x = 366; 

    addDays(d, m, y, x); 

    return 0; 
}
于 2019-04-26T12:19:36.110 回答
0

不知道这是否有帮助。我正在研究一个调度系统，该系统（在第一个简单草案中）将开始日期计算为截止日期 - 提前期天数。我处理了经过的秒数（自纪元以来），以便在未来的代码草案中实现更高的精度。

#include <iostream>
#include <ctime>

int main() {
  // lead time in days
  int lead_time = 20;
  // assign a due_date of (midnight on) 3rd November 2020
  tm tm_due_date = { 0, 0, 0, 3, 11, 2020-1900};
  // convert due_date to seconds elapsed (since epoch)
  time_t tt_due_date = mktime(&tm_due_date);
  // subtract lead time seconds
  tt_due_date -= 60 * 60 * 24 * lead_time;
  // convert back (to local time)
  tm *start_date = localtime(&tt_due_date);
  // otput the result as something we can readily read
  std::cout << asctime(start_date) << "\n";
}
于 2020-04-30T14:32:50.960 回答