main returns int, and you should validate the input somewhere or all hell could break loose. Good idea though, and a nice start.

Thats a good.......

excuse me... can you explain the int weekday(int month, int day, int year) fuction at the buttom part?
i can't understand the purpose of the variables ix, tx, and vx
and the if statement at the very bottom... pls,,,help!

can't we use a string function pls?

can't we use a string function pls?

Um...what?

can't we use a string function pls?

Do'et.

This is a very nostalgic problem for me.
Several years ago (so long ago that Dani was probably not born yet) I had to solve this problem. I did all sorts of pre-internet researc but could not find an answer.

Then - the aha moment

I realized that there is no information in a single date that can tell you what day of the week it fell on but if you do know the day of the week for a certain day (e.g. today) you can

1. Calculate the difference in days between your known date and the date in question.
2. Find the modulo 7 difference of that difference.
3. Since the days of the week continuously repeat in a 7 day cycle, the modulo 7 value will determine the day of the week you want.

There are 2 caveats.

1. Leap years (i.e. an extra day) occur every four years EXCEPT if the year is 0 mod 400 (e.g. 2000 was not a leap year).
2. Several centuries ago (I don't recall when) a few months were dropped from the calendar (by papal decree?) to bring it back into synch with the seasons.

I wrote code to do this calculation but I can't find it, so I'll leave implementation as
"an exercise for the reader". Have fun!

I wrote code to do this calculation but I can't find it, so I'll leave implementation as "an exercise for the reader". Have fun!

Piece of cake considering I've already written this for my standard C library implementation. Here's the meat of it, slightly modified:

#include <stdio.h>

#define _LEAP_YEAR(year)  (((year) > 0) && !((year) % 4) && (((year) % 100) || !((year) % 400)))
#define _LEAP_COUNT(year) ((((year) - 1) / 4) - (((year) - 1) / 100) + (((year) - 1) / 400))

const int yeardays[2][13] = {
    { -1, 30, 58, 89, 119, 150, 180, 211, 242, 272, 303, 333, 364 },
    { -1, 30, 59, 90, 120, 151, 181, 212, 243, 273, 304, 334, 365 }
};

const int monthdays[2][13] = {
    { 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31 },
    { 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31 }
};

int weekday(int year, int month, int day)
{
    int ydays, mdays, base_dow;

    /* Correct out of range months by shifting them into range (in the same year) */
    month = (month < 1) ? 1 : month;
    month = (month > 12) ? 12 : month;

    mdays = monthdays[_LEAP_YEAR(year)][month - 1];

    /* Correct out of range days by shifting them into range (in the same month) */
    day = (day < 1) ? 1 : day;
    day = (day > mdays) ? mdays : day;

    /* Find the number of days up to the requested date */
    ydays = yeardays[_LEAP_YEAR(year)][month - 1] + day;

    /* Find the day of the week for January 1st */
    base_dow = (year * 365 + _LEAP_COUNT(year)) % 7;

    return (base_dow + ydays) % 7;
}

Dates are tricky though, and this only works within the confines of ISO C library requirements.

can't we use a string function pls?

Not functions in <string>, but a couple of functions in <ctime>

#include <iostream>
#include <ctime>

int main()
{
    int  month, day, year ;
    std::cin >> month >> day >> year ;
    // validate

    std::tm tm ;
    tm.tm_mon = month - 1 ;
    tm.tm_mday = day ;
    tm.tm_year = year - 1900 ;
    tm.tm_hour = tm.tm_min = tm.tm_sec = 0 ;
    tm.tm_isdst = -1 ;

    static const char* const wdays[] =
    { "Sunday", "Monday", "Tuesday", "Wednesday", "Thursday", "Friday","Saturday" } ;

    std::time_t t = std::mktime( &tm ) ;
    const std::tm* ptm = std::localtime( &t ) ;

    if( ptm ) std::cout << wdays[ ptm->tm_wday ] << '\n' ;
    else std::cout << "invalid date\n" ;
}

Also see: Zeller's congruence and Doomsday Rule.

// On what day were you born?
#include <stdio.h>
#include <conio.h>
#include <iostream>
#include <fstream>

using namespace std; 

long jd(int y, int m, int d);

void main() 
{
    int Year, Month, Day, DayOfWeek;
    long Julian;

    Year = 2001;  // If only one time doing might I hard code the date
    Month = 10;
    Day = 02;

    Julian = jd(Year, Month, Day); // Convert the date to Julian
    DayOfWeek = Julian % 7         // The remainder should give us the day of week  
    switch(DayOWeek)
    {

        case 0:
            cout << "Your birthday occurred on Monday"
            break;
        case 1:
            cout << "Your birthday occurred on Tuesday"
            break;
        case 2:
            cout << "Your birthday occurred on Wednesday"
            break;
        case 3:
            cout << "Your birthday occurred on Thursday"
            break;
        case 4:
            cout << "Your birthday occurred on Friday"
            break;
        case 5:
            cout << "Your birthday occurred on Saturday"
            break;
        case 6:
            cout << "Your birthday occurred on Sunday"
            break;
        default:
            cout << " You must not have been born"
            break
    }
}

// Convert date to julian
 long jd(int y, int m, int d)
 {
    y+=8000;
    if(m<3) { y--; m+=12; }
     return (y*365) +(y/4) -(y/100) +(y/400) -1200820 +(m*153+3)/5-92 +d-1;
 }