两个对象之间的日期比较
我不得不为我的大学实验室编写一个程序。在该程序中,我想比较格式为日/月/年的两个日期。我知道如何做到这一点,但不包括小时。现在我将该日期转换为0000年以来的日期,并简单地比较这两个值。问题是我的老师告诉我要添加几小时,现在我不知道如何比较这一点。有什么建议么?本次代码波纹管两个对象之间的日期比较
.h文件中
class timee
{
int day;
int month;
int year;
int hour;
long int count;
public:
timee();
timee(int,int,int,int);
long int daysCount();
bool operator>(const timee &);
bool operator>=(const timee &);
bool operator<=(const timee &);
bool operator==(const timee &);
timee & operator=(const timee &);
timee & operator+=(int);
timee & operator-=(int);
long int operator-(timee &);
friend ostream & operator<<(ostream &, const timee &);
friend istream & operator>>(istream &, timee &);
};
这里.cpp文件
timee::timee():day(0),month(0),year(0),hour(0),count(0){}
timee::timee(int day,int month,int year,int hour):day(day),month(month),year(year),hour(hour)
{
count = daysCount();
}
/*calculating the number of days that have passed since year 0000*/
long int timee::daysCount()
{
int month_days[] = {0,31,59,90,120,151,181,212,243,273,304,334};
// calculate number of leap years.
int leapyears = year/4;
if (isLeapYear(year) && month < 3)
{
// If this is a leap year
// And we have not passed Feburary then it does
// not count.....
leapyears --;
}
// convert year/month/day into a day count
count = year * 365 + month_days[month-1] + day + leapyears;
return count;
}
/*convering the date from days since year 0000 to year/month/day format */
timee timee::dateConversion()
{
int month_days[] = {0,31,59,90,120,151,181,212,243,273,304,334,365};
//calculate number of leap year
int leapyears = year/4;
if (isLeapYear(year) && month < 3)
{
// If this is a leap year
// And we have not passed Feburary then it does
// not count.....
leapyears --;
}
//calculating year
year = (count-leapyears)/365;
for(unsigned int i = 0; i <= 12; i++)
{
if((count-leapyears)%365 > month_days[i])
{
month = i+1;
}
}
day = ((count-leapyears)%365)-month_days[month-1];
return *this;
}
bool timee::operator>(const timee &obj)
{
return count>obj.count;
}
bool timee::operator>=(const timee &obj)
{
//if((count>=obj.count) && (hour>=obj.hour)) return true;
//else if((count<=obj.count) && (hour>obj.hour))return false;
}
bool timee::operator<=(const timee &obj)
{
return count<=obj.count;
}
bool timee::operator==(const timee &obj)
{
return count==obj.count;
}
timee & timee::operator=(const timee &obj)
{
day=obj.day;
month=obj.month;
year=obj.year;
hour=obj.hour;
count=obj.count;
return *this;
}
timee & timee::operator+=(int value)
{
count+=value;
this->dateConversion();
return *this;
}
timee & timee::operator-=(int value)
{
count-=value;
this->dateConversion();
return *this;
}
long int timee::operator-(timee &obj)
{
return count - obj.count;
}
ostream & operator<<(ostream &os, const timee &obj)
{
os << "Date: " << obj.day << "." << obj.month << "." << obj.year << " Hour: " << obj.hour << " " << obj.count << endl;
return os;
}
istream & operator>>(istream &is, timee &obj)
{
cout << "Type day, month and year" << endl;
is >> obj.day >> obj.month >> obj.year >> obj.hour;
obj.daysCount();
return is;
}
还有就是我试图重载> =运营商之一。请帮忙。
count
在你的算法是指自0年
虽然,你应该现在已经不是一天最小精度过去的天数,而是一个小时。所以,你应该简单地创建一个变量totalHours
正的小时数量自今年通过0
//Calculate number of days since year 0
count = year * 365 + month_days[month-1] + day + leapyears;
//Convert to number of HOURS since year 0, and add additional hour
totalHours = count*24 + hour;
感谢您的回复。这很有帮助,但现在我还有其他一些问题。我想创建一个反向工作的函数,这意味着从“从0年过去的小时”获得日/月/年和小时。现在我正在努力争取一年,但不知怎的,它不能像它应该那样工作,例如将日期1/1/2016小时1转换为“小时...”,然后返回给我2017年。它是因为使用int值? 'year =((count-leapyears)/ 24)/ 365;' – Bringer 2015-01-20 20:20:29
如果我说得对,现在想做的是从一个单一的组件(从0年开始的小时数),得到年数,几个月,几天还有剩余时间? 这比你到目前为止所做的还要复杂 – Angivare 2015-01-20 20:29:10
我使用这种格式的主要目的是将日期添加/移动x天更简单(至少我认为是这样)。但正如你所说,将其转换回来非常困难。 – Bringer 2015-01-20 20:33:56
有count
和obj.count
内operator >=
之间的3个可能的关系。 count < obj.count
或count == obj.count
或count > obj.count
。这同样适用于hours
和obj.hours
。这给出了3 * 3 = 9种可能的组合。写下每个组合的运算符结果,然后找到在代码中表达它的最简单方法。
请注意,您不需要为每个比较运算符执行此操作。通常你执行operator <
然后根据那个定义其他的。
小时会增加小数天数,所以如果您想保留一个类似的算法,您仍然不能返回'long int'。 – crashmstr 2015-01-20 18:54:27
http://*.com/questions/4196153/find-how-many-seconds-past-since-1-1-1970 – 2015-01-20 18:57:16