Question

C++ code:

Days in current month:

Write a program that can determine the number of days in a month for a specified month and year. The program should allow a user to enter two integers responding a month and a year, and it should determine how many days are in the specified month. The integers 1 through 12 will be used to identify the months of January through December. The user indicates the end of days in the current month and terminates.

Use the following criteria to identify leap year:

1. A year Y is divisible by 100. Then Y is a leap year if and only if it is divisible by 400. For example, 2000 is a leap year but 2100 is not.

2. A year Y is not divisible by 100. Then Y is a leap year if and only if it is divisible by 4. For example, 2008 is a leap year but 2009 is not.

Here is sample run of the program:

Enter month and year: 2 2008 [Enter]

29 days

Enter month and year: 0 0[Enter]

The current month, September 2009, has 30 days.

Answer 1

#include<iostream>
#include<string>
#include <time.h>

using namespace std;

string getMonth(int a){
   switch(a){
       case 1 : return "January";
                break;
       case 2 : return "February";
                break;
       case 3 : return "March";
                break;
       case 4 : return "April";
                break;
       case 5 : return "May";
                break;
       case 6 : return "June";
                break;
       case 7 : return "July";
                break;
       case 8 : return "August";
                break;
       case 9 : return "September";
                break;
       case 10 : return "October";
                break;
       case 11 : return "November";
                break;
       case 12 : return "December";
                break;
   }
}

int getDays(int m, int y){

   if (m == 2){
      if ((y % 100 == 0 && y % 400 == 0) || (y % 100 != 0 && y % 4 == 0)){
         return 29;
      }
      else {
        return 28;
      }
   }
   switch(m) {
      case 1:
      case 3:
      case 5:
      case 7:
      case 8:
      case 10:
      case 12: return 31;
               break;
      case 4:
      case 6:
      case 9:
      case 11: return 30;
               break;
             
     
   }
}

void getCurrentData(){

   time_t theTime = time(NULL);
   struct tm *aTime = localtime(&theTime);

   int day = aTime->tm_mday;
   int month = aTime->tm_mon + 1;
   int year = aTime->tm_year + 1900;

   int n = getDays(month,year);
   string name = getMonth(month);
   cout << "The current month, " << name << " " << year << ", has " << n << " days" << endl;
}

int main(){

   int m,y;

   cout << "Enter month and year:";
   cin >> m >> y;
   if (m == 0 && y == 0){
       getCurrentData();
   }
   else {
     int n = getDays(m,y);
     cout << n << " days" << endl;
   }
   return 0;
}