目前我正在编写一个程序,其中有一个部分用于确定两个日期之间的天数差异,但是通过重载减号运算符.
我正在盯着我的屏幕画一个完整的空白.我头脑中有一些短暂的想法,但它们就是那个,转瞬即逝.
在main.cpp中发生的事情是,将会有两个变量,例如beethovenDeathDate和beethovenBirthDate,它们将被减去以确定他的生存时间.如果我没记错的话,这大约是22000天.
所以不用多说,这是我的代码:
Date.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)
{
}
最佳答案 该函数既可以作为成员编写,也可以作为自由函数编写.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)
这里的TimeDuration是一个完全独立的类型,代表一段时间.如果你愿意,你可以把它变成一个表示天数的int,但在我看来,为了这个目的,有一个更具表现力的类型会更好.无论你决定什么类型的返回类型,它都没有任何意义,因为类型是Date(当然不是Date&).
一个可能的(尽管不是非常有效)实现,假设你已经编写了一个函数来添加一天的日期,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想要的功能,但您的措辞并未表明它)是一个可以从Date中减去一段时间的功能.在这种情况下,返回值将是另一个Date对象(但不是Date&),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