Calculate age in seconds - strange result - c++

Question

I'm new in programming and new in here.
Sorry for stupid question but i have problem with result in my "calculate your age in seconds" code. It gives me weird result like 6.17725e+10 or -6.17414e+10.
Program isn't finished yet but everything except results looks fine (i don't get any error.
Sorry again and I hope for your understanding:)
#include <iostream>
using namespace std;
void title()
{
cout << "Age Calculator" << endl << endl;
}
int byear()
{
cout << "Enter your birth year: ";
int by;
cin >> by;
return by;
}
int bmonth()
{
cout << "Enter your birth month: ";
int bm;
cin >> bm;
return bm;
}
int bday()
{
cout << "Enter your birth day: ";
int bd;
cin >> bd;
return bd;
}
int cyear()
{
int cy;
cout << "Enter current year ";
cin >> cy;
return cy;
}
int cmonth()
{
cout << "Enter current month: ";
int cm;
cin >> cm;
return cm;
}
int cday()
{
cout << "Enter current day: ";
int cd;
cin >> cd;
return cd;
}
void calculate(int by, int bm, int bd, int cy)
{
double y = 31104000;
long double cby = y * by;
long double cbm = 259200 * bm;
long double cbd = 8640 * bd;
long double ccy = 31104000 * cy;
cout << endl << cby << endl;
cout << endl << ccy << endl;
cout << endl << ccy - cby << endl;
}
int main()
{
title();
int by = byear();
int bm = bmonth();
int bd = bday();
int cy = cyear();
int cm = cmonth();
int cd = cday();
calculate(by, bm, bd, cy);
cin.get();
return 0;
}

Answer 1

Like Kenny Ostrom commented, the shown values may look strange due to the scientific notation used by cout. To show all digits, you can change cout's precision using cout.precision(your_precision_here). See question below.
How do I print a double value with full precision using cout?

Answer 2

First, the numeric format you are confused by is "scientific notation". That will be enough info to open up a world of google searches, or you can just force it not to print in scientific notation.
Second, you really want to use a time library for any calendar stuff. It will handle all kinds of calendar weirdness for you, including leap years. Fortunately we have time.h
Third, I recommend using an integer type for seconds, partly to avoid rounding errors and ugly decimals, but mainly because that's what time.h uses. Just make sure it is big enough. My compiler uses a 64 bit integer for time_t, so I used that:
#include <time.h>
#include <memory>
time_t get_age_in_seconds(int year, int month, int day)
{
struct tm birthday;
memset(&birthday, 0, sizeof(birthday));
birthday.tm_year = year - 1900; // years since 1900
birthday.tm_mon = month - 1; // months since January (0,11)
birthday.tm_mday = day; // day of the month (1,31)
time_t birthday_in_seconds = mktime(&birthday);
time_t now = time(NULL);
return now - birthday_in_seconds;
}

Answer 3

Don't use doubles to do the calculation. You're not going to have any fractional values since you're not doing any division.
More importantly, look into mktime(), time(), and difftime(). You should be using these to do your calculation.