4

目前我正在编写一个程序,它有一个部分来确定两个日期之间的天数差异,但是通过重载减号运算符。

我目前正盯着我的屏幕画一个完全空白。我脑子里有一些转瞬即逝的想法,但它们就是那样,转瞬即逝。

在 main.cpp 中发生的事情是会有两个变量,例如beethovenDeathDatebeethovenBirthDate将减去这两个变量以确定他的寿命。如果我没记错的话,大约是 22000 天。

所以事不宜迟,这是我的代码:

日期.cpp

const std::string Date::MONTH_STRINGS[] = 
{
    "", //one based indexing
    "January",
    "February",
    "March",
    "April",
    "May", 
    "June",
    "July",
    "August",
    "September",
    "October",
    "November",
    "December"
};

const int Date::DAYS_PER_MONTH[] =
{
    0, //one based indexing
    31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31
};      

Date::Date(int day, int month, int year) : _year(year), _month(month), _day(day)
{
    isValid();
}

Date::Date()
{
    time_t t = time(0);   // get time now
    struct tm * now = localtime( & t );
    _year = now -> tm_year + 1900;
    _month = now -> tm_mon + 1;
    _day = now -> tm_mday;
}

int Date::maxDay(int month, int year)
{
    int ret = DAYS_PER_MONTH[month];
    if(isLeapYear(year) == true && month == 2)
    {
        ++ret;
    }
    return ret;
}

void Date::addDay(bool forward)
{
    if(forward)
    {
        if(_day < maxDay(_month, _year))
        {
            ++_day;
        }
        else
        {
            _day = MIN_DAY;
            ++_month;
            if(_month > MAX_MONTH)
            {
                _month = MIN_MONTH;
                ++_year;
            }    
        }
    }
    else
    {
        if(_day <= MIN_DAY)
        {
            --_month;
            if(_month < MIN_MONTH)
            {
                _month = MAX_MONTH;
                --_year;
            } 
            _day = maxDay(_month, _year);
        }
        else
        {
            --_day;
        }
    }    
}


std::string Date::toString() const
{
    if(isValid() == false)
    {
        return std::string();
    }
    std::stringstream ss;
    ss  << MONTH_STRINGS[_month] << " " << _day << ", " <<  _year;
    return ss.str();            
}    


bool Date::isValid() const
{
    if(_month < MIN_MONTH || _month > MAX_MONTH)
    {
        std::cerr << "Invalid date " << std::endl;
        return false;
    }
    int daysThisMonth = maxDay(_month, _year);

    if(_day < MIN_DAY || _day > daysThisMonth)
    {
        std::cerr << "Invalid date " << std::endl;            
        return false;   
    }
    return true;
}

bool Date::isLeapYear(int year)
{
    if(!(year % 4))
    {
        if(!(year % 100))
        {
            if(!(year % 400))
            {
                return true;
            }
            else
            {
                return false;
            }
        }
        else
        {
            return true;
        }
    }
    else
    {
        return false;
    }
}

bool Date::isLeapYear() const
{
    return isLeapYear(_year);
}

bool Date::isLeapDay() const
{
    return isLeapDay(_day, _month, _year);
}

bool Date::isLeapDay(int day, int month, int year) 
{
    if(day == 29 &&  month == 2 && isLeapYear(year) == true)
    {
        return true;
    }
    else
    {
        return false;
    }
}


void Date::addYears(int years)
{
    if(years == 0)
    {
        return;
    }
    if(isLeapDay() && !isLeapDay(_day, _month, _year + years))
    {
        _day = Date::DAYS_PER_MONTH[_month];
    }
    _year += years;
}

void Date::addMonths(int months)    
{
    if(months == 0)
    {
        return;
    }
    int deltayears = months / MAX_MONTH;
    int deltamonths = months % MAX_MONTH;
    int newMonth = 0;
    if(months > 0)
    {
        newMonth = (_month + deltamonths) % MAX_MONTH;
        if((_month + deltamonths) > MAX_MONTH)
        {
            ++deltayears;
        }
    }
    else
    {
        if((_month + deltamonths) < MIN_MONTH)
        {
            --deltayears;
            newMonth = _month + deltamonths + MAX_MONTH;
        }
        else
        {
            newMonth = _month + deltamonths;
        }       
    }
    if(_day > maxDay(newMonth, _year + deltayears))
    {
        _day = maxDay(newMonth, _year + deltayears);
    }
    _year += deltayears;
    _month = newMonth;

}

void Date::addDays(int days)
{
    if(days == 0)
    {
        return;
    }

    if(days < 0)
    {
        for(int i = 0; i > days; --i)
        {
            addDay(false);
        }
        return;
    }

    for(int i = 0; i < days; ++i)
    {
        addDay(true);
    }
}

std::ostream& operator<<(std::ostream& os, const Date& date)
{
    os << date.toString();
    return os;
}

Date Date::operator+(int days) const
{
    Date ret = *this;
    ret.addDays(days);
    return ret;
}

Date& Date::operator+=(int days)
{
    addDays(days);
    return *this;
}
//This is where I get stumped (the parameters was just one of my failed experiments
Date& Date::operator-(int day, int month, int year)
{
}
4

2 回答 2

2

该函数既可以写成成员,也可以写成自由函数。成员函数签名如下所示:

TimeDuration Date::operator-(Date const & rhs) const

免费功能如下所示:

TimeDuration operator-(Date const & lhs, Date const & rhs)

TimeDuration这是一个完全独立的类型,表示时间长度。如果您愿意,您可以int将其设为表示天数,但在我看来,为此目的使用更具表现力的类型会更好。无论您决定返回类型是什么,该类型都没有任何意义Date(当然不是Date&)。

鉴于您已经编写了一个将日期添加到日期的函数,一个可能的(尽管不是非常有效)的实现将是这样的:

if lhs_date comes before rhs_date
    add days to (a copy of) lhs_date until lhs_date == rhs_date
    return the negative of number of days added
if rhs_date comes before lhs_date
    add days to (a copy of) rhs_date until rhs_date == lhs_date
    return the number of days added
else
    return 0

您可能想要的另一个功能(或者这可能是您最初真正想要的,但您的措辞并未表明它)是一个可以从Date. 在这种情况下,返回值将是另一个Date对象(但不是Date&),并且可能的签名看起来像这样:

Date Date::operator-(TimeDuration rhs) const // member version
Date operator-(Date const & lhs, TimeDuration const & rhs) // non-member version
于 2013-09-10T01:59:31.510 回答
0

你应该做这个:

//This is where I get stumped (the parameters was just one of my failed experiments
TimeDuration& Date::operator-(Date const & d1)
{
    // ... processing ...
    // this - d1;
}

并将其称为:

Date d1 = new Date(20, 01, 2013);
TimeDuration duration = d1 - (new const Date(20, 01, 1922));
// Calculate no. of days or years using duration

逻辑如下:

  1. 将两个Date对象(第一个可能是隐式的)传递给重载函数并返回TimeDuration

  2. 要调用此运算符,您可以使用Date您拥有的数据创建一个对象,而不是单独传递每个值。

请检查确切的语法。

于 2013-09-10T02:35:21.300 回答