Thread: Help with day of the week program

Hi,

Let me apologize in advance if this is in the wrong section or if this question has already been asked and answered. I've looked around the site, but have probably overlooked the answer if it's already here.

So, for my computer programming class (I'm a beginner), we are to create a program that "determines the day of the week a person was born on based on the birth date he or she enters". I've written the program and showed the teacher. It was perfect until he tested it with the date 12 - 7 - 1941. It printed out the Monday, instead of Sunday. So there is something wrong with the way I did the leapyear.

Here is my code:

Code:

 #include <iostream>

using namespace std;

int main() {

int birthyear;

int year;

int day;

int month;

int leapyear;

int total;

int birthday;

cout << "Enter your year of birth (yyyy): " << endl;

cin >> birthyear;

cout << "Enter your day of birth (dd): " << endl;

cin >> day;

cout << "Enter your month of birth (mm): " << endl;

cin >> month;

year = birthyear - 1900;

leapyear = year % 4;

total = (year / 4) + year + day;

 

switch (leapyear) {

case 0:

 switch (month){

  case 1:

  total = total + 1;

  break;

 case 2:

  total = total + 4;

  break;

 case 3:

  total = total + 4;

  break;

 case 4:

  total = total + 0;

  break;

 case 5:

  total = total + 2;

  break;

 case 6:

  total = total + 5;

  break;

 case 7:

  total = total + 0;

  break;

 case 8:

  total = total + 3;

  break;

 case 9:

  total = total + 6;

  break;

 case 10:

  total = total + 1;

  break;

 case 11:

  total = total + 4;

  break;

 case 12:

  total = total + 6;

  break;

}

default:

 switch (month){

  case 1:

  case 2:

  total = total - 1;

  break;

}

}

birthday = total % 7;

switch (birthday){

 case 1:

  cout << "Sunday" << endl;

  break;

  case 2:

  cout << "Monday" << endl;

  break;

 case 3:

  cout << "Tuesday" << endl;

  break;

 case 4:

  cout << "Wednesday" << endl;

  break;

 case 5:

  cout << "Thursday" << endl;

  break;

 case 6:

  cout << "Friday" << endl;

  break;

 case 7:

  cout << "Saturday" << endl;

  break;

}

return 0;

}

Now, my teacher said that I had to fix the way I did the switch statement that is determining when a year is a leap year (leap years had to be taken into consideration). Also, we are only doing dates from 1900 and on. So any year that is entered from before 1900 will not print anything out.

What I am not understanding is how I can fix this. The teacher didn't do a good job explaining what he found wrong, just that I needed to fix it. So I would appreciate if someone could explain what is wrong.

Thanks in advance,
Tiffany

I have no idea what your algorithm is, but just as a guess you should probably have twelve cases in your non-leap year, just like you have twelve cases in your leap year?

Or to make CodeMonkey's even nicer looking...

Code:

switch (leapyear) 
    {
        case 0:
        {
             switch (month)
             {
              case 9: case 12: ++total;
              case 6: ++total;
              case 2: case 11: case 3: ++total;
              case 8: ++total;
              case 5: ++total;
              case 1: ++total;
              case 4: case 7: break;
             }
        }break;
        default:
        {
             switch (month)
             {
              case 1: case 2: { total -= 1; }break;
             }
        }
    }//leapyear

To be honest, I don't even know what <ctime> functions are . . . ^_^
We were just given this problem right after learning about switch statements, which is what we're suppose to use in this particular program. And the spacing does look better, but the way that I have mine spaced is the way that was taught to the class.

So would tabstop's suggestion be a solution? Having 12 cases in the non leap year?

Thanks for the replies so far.

Originally Posted by Punkakitty

To be honest, I don't even know what <ctime> functions are . . . ^_^
We were just given this problem right after learning about switch statements, which is what we're suppose to use in this particular program. And the spacing does look better, but the way that I have mine spaced is the way that was taught to the class.

So would tabstop's suggestion be a solution? Having 12 cases in the non leap year?

Thanks for the replies so far.

Looking at it again, I'm believing that the last two cases are just supposed to be an adjustment -- which means that the first case should not have a break so that we "fall through" to the default -- except then you have it backwards, since case 0 is the leap year case, and you want to do these settings in all cases and the adjustment only in the leap years.

In other words, you need to look at what it is you're supposed to be computing.