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c++ - 重载 '-' 运算符

转载 作者:塔克拉玛干 更新时间:2023-11-03 02:10:59 26 4
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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。

我目前正盯着我的屏幕画一片空白。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。

main.cpp 中将发生的事情是将有两个变量,例如 beethovenDeathDatebeethovenBirthDate 将减去这两个变量以确定他的生命周期为了。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)
{
}

最佳答案

函数可以写成成员函数,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

免费功能看起来像这样:

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

TimeDuration 这里是一个完全独立的类型,表示时间长度。如果需要,您可以将其设为表示天数的 int,但在我看来,为此目的使用更具表现力的类型会更好。无论您对返回类型做出何种决定,类型为 Date(当然也不是 Date&)都没有任何意义。

假设您已经编写了一个向日期添加一天的函数,一个可能的(尽管不是非常有效)实现是这样的:

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

您可能需要的另一个函数(或者这可能是您最初真正想要的,但您的措辞并未表明)是一个可以从 Date 中减去时间长度的函数。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

关于c++ - 重载 '-' 运算符,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/18709604/

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