9.8 Header File:Chrono.h

namespace Chrono{
enum class Month{
//...
};
class Date{
//...
};
//helper functions..

File:Chrono.cpp

#include "Chrono.h"
// definitions

add_year function as a member function.

void Date::add_year(int n){
    y+=n;
}

How about leapyear? Ex)2016, Feb 29 => 2017, feb 29? Let's choose 2017, mar 1.

void Date::add_year(int n){
    if(m==feb && d==29 && !leapyear(y+n)){
        m==mar;
        d=1;
    }
    y+=n;
}

Make a leapyear

bool leapyear(int y){
//
}

Algorithm

1. if (year is not divisible by 4) then (it is a common year)
2. else if (year is not divisible by 100) then (it is a leap year)
3. else if (year is not divisible by 400) then (it is a common year)
4. else (it is a leap year)

textbook implementation

bool leapyear(int y)
{
    if (y%4) return false;
    if (y%100==0 && y%400) return false;
    return true;
}

Check: Make add_month

For simplicity, assume (n>0).

void Date::add_month(int n){
//?
}

Hint:

int days_in_month(int y, Month m){
    static int days[13]={0,31,28,31,30,31,30,31,31,30,31,30,31};
    if(int(m)==2 && leapyear(y)) return 29; //handle Feb separately.
    else return days[int(m)];
}

overloading

ostream & operator<<(ostream & os, const Month m){
    static string month_tbl[13]={"","jan","feb","mar","apr","may","jun","july", "aug", "sep", "oct", "nov", "dec"};
    os<<month_tbl[int(m)];
    return os;
}
ostream & operator<<(ostream & os, const Date & d){
    //Month mm=d.month();
    os<<d.month()<<"-"<<d.day()<<", "<<d.year();
    return os;
}

check!: add_day

Assume n>0.

int days_in_month(int y, Month m){
    static int days[13]={0,31,28,31,30,31,30,31,31,30,31,30,31};
    if(int(m)==2 && leapyear(y)) return 29; //handle feb separately.
    else return days[int(m)];
}
void Date::add_day(int n){
// Use days_in_month and add_month
}

Hint: Use add_month(1) after reset the day as 1.

n=n+d-1;
d=1;

Code!

Negative number cases

subtract month

Problem in add_month.

Feb 29, 2016
-12 mont
Feb 29, 2015 = March 1, 2015
March 1,2015
+12 month
March 1, 2016 (Not same)

In general,

d.add_month(1);
d.add_month(1);

is not equal to

d.add_month(2)

One implementation.

void Date::add_month(int n){ 
    int yy=(n/12);// extra years
    int mm=n%12;// extra month
    m=Month(int(m)+mm);
    y+=yy; //negative year is ok!
    if(int(m)>12){ 
        ++y;
        m=Month(int(m)-12); 
    }
    else if(m <= 0){  //negative month.
        m=Month(int(m)+12);
        --y;
    }
    if(d>days_in_month(y,Month(int(m)))){
        d-=days_in_month(y,Month(int(m)));//m<=days(m+1) since d=d-days<=3<days(m+1)
        m=Month(int(m)+1); //carry over. m<=12 since (m=12 && d>days=31) is always false.
    }
}

add_day with any number.

Code!

Hint

1. Split as n>0, n==0 and n<0
2. Use add_month(1) or add_month(-1) after reset the day as 1.

   n=n+d-1;
   d=1;

HW make a validation code for date.

bool Date::is_date(){
    //  1<=m<=12
    //  1<=d<=days_of_month
}