C++ compare dates to get age

Question

Hello I am new to C++ expecialy dates in C++.. how can I compare these numbers 29(day) 08(month) 86(year) with todays date to get age ?

Here is hwo I started my function:

std::string CodeP, day, month, year, age;

std::cout<<"Enter Permanent Code(example: SALA86082914) :\n "; //birthday first six numbers in code

std::cin>>CodeP;

year  = CodeP.substr (4,2);
month = CodeP.substr (6,2);
day = CodeP.substr (8,2);

std::cout<<"day :"<<day <<'\n';
std::cout<<"month :"<<month <<'\n';
std::cout<<"year :"<<year <<'\n';

//then get today's date to compare with the numbers of birthday to get age

Do you have access to C++11? If you don't know, what compiler are you using? What version? If this is for a class, there might be restrictions on what version of C++ you can use as well. — alecbz, Apr 02 '14 at 15:20
Look into [chrono](http://en.cppreference.com/w/cpp/chrono). (C++11) — user123, Apr 02 '14 at 15:20
@MohammadAliBaydoun: How will that help? `` doesn't have anything for dealing with calendars. — Mike Seymour, Apr 02 '14 at 15:23

score 1 · Accepted Answer · answered Apr 02 '14 at 15:22

1

First subtract the years to get the difference. Now compare the months; if today's month is greater than the birth month, you're done. If not, compare the days; if today's day is greater than the birth day, you're done. Otherwise subtract one from the difference and that's the age.

answered Apr 02 '14 at 15:22

Mark Ransom

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Age in what units? Beware February 29 and such fun. – Deduplicator Apr 02 '14 at 15:23
2

@Deduplicator age is generally specified in years unless a baby is less than a year old. And I think my formula still works for February 29, whether it's today or the birthday. – Mark Ransom Apr 02 '14 at 15:25
Hm, deal with less than 1 month, over two consecutive months? – Deduplicator Apr 02 '14 at 15:26
@MarkRansom Yes it works, but only for age in years. – HelloWorld123456789 Apr 02 '14 at 15:26
What is the best way to get today's date ? and any idea how to deal with the fact that I only have the 2 last digits of the year ? – Drew Salesse Apr 02 '14 at 15:39
@DrewSalesse, I thought 2 digit years went out of style 14 years ago? Anyway you can look for any Y2K advice for how to deal with it. And see http://stackoverflow.com/questions/997946/how-to-get-current-time-and-date-in-c for today's date. – Mark Ransom Apr 02 '14 at 15:43

score 1 · Answer 2 · answered Apr 02 '14 at 15:48

Try my code below, I already did the necessary code for the comparing and the processing of two dates that you will be needing. Take note that my code's role is just to compare two dates and give the resulting age based from those dates.

Sorry but I have to leave you the job of creating a code that will give you your expected output based from a six-digit birthday input. I guess you already know that you still have to come up with your own solution for your problem, we are only here to give you an idea on how you can tackle it. We could only do so much to help and support you. Hope my post was helpful!

    class age
    {
    private:
    int day;
    int month;
    int year;
    public:
    age():day(1), month(1), year(1)
    {}
    void get()
    {
    cout<<endl;
    cout<<"enter the day(dd):";
    cin>>day;
    cout<<"enter the month(mm):";
    cin>>month;
    cout<<"enter the year(yyyy):";
    cin>>year;
    cout<<endl;
    }
    void print(age a1, age a2)
    {
    if(a1.day>a2.day && a1.month>a2.month)
    {
    cout<<"your age is DD-MM-YYYY"<<endl;
    cout<<"\t\t"<<a1.day-a2.day<<"-"<<a1.month-a2.month<<"-"<<a1.year-a2.year;
     cout<<endl<<endl;
     }
    else if(a1.day<a2.day && a1.month>a2.month)
    {
    cout<<"your age is DD-MM-YYYY"<<endl;
    cout<<"\t\t"<<(a1.day+30)-a2.day<<"-"<<(a1.month-1)-a2.month<<"-"<<a1.year-a2.year;?
    cout<<endl<<endl;
    }
    else if(a1.day>a2.day && a1.month<a2.month)
    {
   cout<<"your age is DD-MM-YYYY"<<endl;
   cout<<"\t\t"<<a1.day-a2.day<<"-"<<(a1.month+12)-a2.month<<"-"<<(a1.year-1)-a2.year;
   cout<<endl<<endl;
   }
   else if(a1.day<a2.day && a1.month<a2.month)
   {
   cout<<"your age is DD-MM-YYYY"<<endl;
    cout<<"\t\t"<<(a1.day+30)-a2.day<<"-"<<(a1.month+11)-a2.month<<"-"<<(a1.year-1)-a2.year;
   cout<<endl<<endl;
   }     
   }
   };
   int main()
   {
   age a1, a2, a3;
  cout<<"\t Enter the current date.";
  cout<<endl<<endl;
  a1.get();
  cout<<"\t enter Date of Birth.";
  cout<<endl<<endl;
  a2.get();
  a3.print(a1,a2);
  return 0;
  }

score 1 · Answer 3 · answered Apr 02 '14 at 16:03

this will calculate an aproximate of the age(years):

#include <ctime>
#include <iostream>
using namespace std;
int main(){
    struct tm date = {0};
    int day, month, year;
    cout<<"Year: ";
    cin>>year;
    cout<<"Month: ";
    cin>>month;
    cout<<"Day: ";
    cin>>day;
    date.tm_year = year-1900;
    date.tm_mon = month-1;
    date.tm_mday = day;
    time_t normal = mktime(&date);
    time_t current;
    time(&current);
    long d = (difftime(current, normal) + 86400L/2) / 86400L;
    cout<<"You have~: "<<d/365.0<<" years.\n";
    return (0);
}

Massa · Answer 4 · 2014-04-02T17:55:59.993

The best way to do that is using Boost.Gregorian:

Assuming

#include <boost/date_time/gregorian/gregorian.hpp>

using namespace boost::gregorian;

You would do:

date bday { from_undelimited_string(std::string("19")+a.substr(4,6)) };
date now = day_clock::local_day();

unsigned years = 0;
year_iterator y { bday };
while( *++y <= *year_iterator(now) )
  ++years;
--y;
std::cout << *y << "\n";

unsigned months = 0;
month_iterator m { *y };
while( *++m <= *month_iterator(now) )
  ++months;
--m;
std::cout << *m << "\n";

unsigned days = 0;
day_iterator d { *m };
while( *++d <= *day_iterator(now) )
  ++days;
--d;
std::cout << *d << "\n";

std::cout << years << " years, " << months << " months, " << days << " days\n";

score 0 · Answer 5 · answered Apr 02 '14 at 19:09

I found another (simpler?) answer, using Boost.Gregorian

#include <boost/date_time/gregorian/gregorian.hpp>

using namespace boost::gregorian;

template<typename Iterator, typename Date>
struct date_count : public std::pair<unsigned, Date> {
  typedef std::pair<unsigned, Date> p;
  operator unsigned() { return p::first; }
  operator Date() { return p::second; }
  date_count(Date begin, Date end) : p(0, begin) {
    Iterator b { begin };
    while( *++b <= end )
      ++p::first;
    p::second = *--b;
  }
};

And you just have to use it this way:

  date bday { from_undelimited_string(std::string("19")+a.substr(4,6)) };
  date now = day_clock::local_day();

  date_count<year_iterator, date> years { bday, now };
  date_count<month_iterator, date> months { years, now };
  date_count<day_iterator, date> days { months, now };

  std::cout << "From " << bday << " to " << now << " there are " <<
    std::setw(5) << years << " years, " <<
    std::setw(3) << months << " months, " <<
    std::setw(3) << days << " days\n";

If you just want the number of years, you can do:

date_count<year_iterator> years(bday, now);

and run with the number of years in the "years" variable. :D

C++ compare dates to get age

5 Answers5