Question

Can someone fix my coding please it's a calendar some days are placed wrong. I just want the Main menu to display first and second option and four option to end the program properly. Please remove option number 3 Find the number of days between 2 selected dates and the code that belongs to this function.

#include<stdio.h>

void days(int,int,int,int,int,int);

int month(int,int);

int mon[12]={31,28,31,30,31,30,31,31,30,31,30,31};

#define TRUE 1

#define FALSE 0

int days_in_month[]={0,31,28,31,30,31,30,31,31,30,31,30,31};

char *months[]=

{

" ",

"\n\n\nJanuary",

"\n\n\nFebruary",

"\n\n\nMarch",

"\n\n\nApril",

"\n\n\nMay",

"\n\n\nJune",

"\n\n\nJuly",

"\n\n\nAugust",

"\n\n\nSeptember",

"\n\n\nOctober",

"\n\n\nNovember",

"\n\n\nDecember"

};

int inputyear(void)

{

int year;

printf("Enter Year: ");

scanf("%d", &year);

return year;

}

int inputmonth(void)

{

int month;

printf("Enter a month (1-12): ");

scanf("%d", &month);

return month;

}

int determinedaycode(int year)

{

int daycode;

int d1, d2, d3;

d1 = (year - 1.)/ 4.0;

d2 = (year - 1.)/ 100.;

d3 = (year - 1.)/ 400.;

daycode = (year + d1 - d2 + d3) %7;

return daycode;

}

int determineleapyear(int year)

{

if(year% 4 == FALSE & year%100 != FALSE || year%400 == FALSE)

{

days_in_month[2] = 29;

return TRUE;

}

else

{

days_in_month[2] = 28;

return FALSE;

}

}

void calendar(int year, int daycode)

{

int month, day;

for ( month = 1; month <= 12; month++ )

{

printf("%s", months[month]);

printf("\n\nSun Mon Tue Wed Thu Fri Sat\n" );

// Correct the position for the first date

for ( day = 1; day <= 1 + daycode * 5; day++ )

{

printf(" ");

}

// Print all the dates for one month

for ( day = 1; day <= days_in_month[month]; day++ )

{

printf("%2d", day );

// Is day before Sat? Else start next line Sun.

if ( ( day + daycode ) % 7 > 0 )

printf(" " );

else

printf("\n " );

}

// Set position for next month

daycode = ( daycode + days_in_month[month] ) % 7;

}

}

void calendarm(int year, int daycode, int month)

{

int day;

printf("%s", months[month]);

printf("\n\nSun Mon Tue Wed Thu Fri Sat\n" );

// Correct the position for the first date

for ( day = 1; day <= 1 + daycode * 5; day++ )

{

printf(" ");

}

// Print all the dates for one month

for ( day = 1; day <= days_in_month[month]; day++ )

{

printf("%2d", day );

// Is day before Sat? Else start next line Sun.

if ( ( day + daycode ) % 7 > 0 )

printf(" " );

else

printf("\n " );

}

// Set position for next month

daycode = ( daycode + days_in_month[month] ) % 7;

}

int main(void)

{

int year, daycode, leapyear,month,c;

int a1,b1,c1,a2,b2,c2;

printf("\n MAIN MENU");

printf("\n 1.View calendar for specified month in selected year");

printf("\n 2.View the calendar for the specified year");

printf("\n 3.Find the number of days between 2 selected dates");

printf("\n 4.End the program \n");

scanf("%d",&c);

switch(c)

{

case 1:

year = inputyear();

month = inputmonth();

daycode = determinedaycode(year);

determineleapyear(year);

calendarm(year, daycode,month);

printf("\n");break;

case 2:

year = inputyear();

daycode = determinedaycode(year);

determineleapyear(year);

calendar(year, daycode);

printf("\n");break;

case 3:

printf("\n Enter first date(dd mm yyyy) : ");

scanf("%d%d%d",&a1,&b1,&c1);

printf("\n Enter second date(dd mm yyyy) : ");

scanf("%d%d%d",&a2,&b2,&c2);

if(c2>=c1)

days(c1,c2,b1,b2,a1,a2);

else

days(c2,c1,b2,b1,a2,a1);

break;

case 4: break;

default:printf("\n Enter correct value");break;

}

}

void days(int y1,int y2,int m1,int m2,int d1,int d2)

{

int count=0,i;

for(i=y1;i<y2;i++)

{

if(i%4==0)

count+=366;

else

count+=365;

}

count-=month(m1,y1);

count-=d1;

count+=month(m2,y2);

count+=d2;

if(count<0)

count=count*-1;

printf("\n The no. of days b/w the 2 dates = %d days",count);

}

int month(int a,int yy)

{

int x=0,c;

for(c=0;c<a-1;c++)

{

if(c==1)

{

if(yy%4==0)

x+=29;

else

x+=28;

}

else

x+=mon[c];

}

return(x);

}

Answer 1

PROGRAM CODE (Option-3 removed and also its functions code)

#include<stdio.h>

// Removing days and month function declarations (I am just commenting below line, so u can remove it)
// void days(int,int,int,int,int,int);
// int month(int,int);

int mon[12]={31,28,31,30,31,30,31,31,30,31,30,31};

#define TRUE 1
#define FALSE 0

int days_in_month[]={0,31,28,31,30,31,30,31,31,30,31,30,31};

char *months[]=
{
   " ",  
   "\n\n\nJanuary",  
   "\n\n\nFebruary",  
   "\n\n\nMarch",  
   "\n\n\nApril",  
   "\n\n\nMay",  
   "\n\n\nJune",  
   "\n\n\nJuly",  
   "\n\n\nAugust",  
   "\n\n\nSeptember",  
   "\n\n\nOctober",  
   "\n\n\nNovember",
   "\n\n\nDecember"
};

int inputyear(void)
{
   int year;
  
   printf("Enter Year: ");
  
   scanf("%d", &year);
  
   return year;
}

int inputmonth(void)
{
   int month;
  
   printf("Enter a month (1-12): ");
  
   scanf("%d", &month);
  
   return month;
}

int determinedaycode(int year)
{
   int daycode;
  
   int d1, d2, d3;
  
   d1 = (year - 1.)/ 4.0;
  
   d2 = (year - 1.)/ 100.;
  
   d3 = (year - 1.)/ 400.;
  
   daycode = (year + d1 - d2 + d3) %7;
  
   return daycode;
}

int determineleapyear(int year)
{
   if(year% 4 == FALSE & year%100 != FALSE || year%400 == FALSE)
  
   {  
       days_in_month[2] = 29;
      
       return TRUE;
   }
   else
   {
       days_in_month[2] = 28;
      
       return FALSE;
   }
}

void calendar(int year, int daycode)
{
   int month, day;
  
   for ( month = 1; month <= 12; month++ )
   {
       printf("%s", months[month]);
      
       printf("\n\nSun Mon Tue Wed Thu Fri Sat\n" );
      
       // Correct the position for the first date
      
       for ( day = 1; day <= 1 + daycode * 5; day++ )
       {
           printf(" ");
       }
  
       // Print all the dates for one month
      
       for ( day = 1; day <= days_in_month[month]; day++ )
       {
           printf("%2d", day );
           // Is day before Sat? Else start next line Sun.
          
           if ( ( day + daycode ) % 7 > 0 )
               printf(" " );
          
           else
               printf("\n " );  
       }
       // Set position for next month
      
       daycode = ( daycode + days_in_month[month] ) % 7;
   }
}

void calendarm(int year, int daycode, int month)
{

   int day;
  
   printf("%s", months[month]);
  
   printf("\n\nSun Mon Tue Wed Thu Fri Sat\n" );
  
   // Correct the position for the first date
  
   for ( day = 1; day <= 1 + daycode * 5; day++ )
  
   {
  
   printf(" ");
  
   }
  
   // Print all the dates for one month
  
   for ( day = 1; day <= days_in_month[month]; day++ )
  
   {
  
   printf("%2d", day );
  
   // Is day before Sat? Else start next line Sun.
  
   if ( ( day + daycode ) % 7 > 0 )
  
   printf(" " );
  
   else
  
   printf("\n " );
  
   }
  
   // Set position for next month
  
   daycode = ( daycode + days_in_month[month] ) % 7;
}

int main(void)
{
   int year, daycode, leapyear,month,c;
  
   // Removing unwanted variables
   // int a1,b1,c1,a2,b2,c2;
  
   // Writing using a loop giving user choices as many times as he wants
  
   while(c != 3)
   {
       printf("\n MAIN MENU");
  
       printf("\n 1.View calendar for specified month in selected year");
      
       printf("\n 2.View the calendar for the specified year");
      
       // Removing option 3 so option-4 "End the program" will become optin-3
      
       printf("\n 3.End the program \n ");
      
       scanf(" %d",&c);
      
       switch(c)
      
       {
           case 1:
          
               year = inputyear();
              
               month = inputmonth();
              
               daycode = determinedaycode(year);
              
               determineleapyear(year);
              
               calendarm(year, daycode,month);
              
               printf("\n");break;
              
           case 2:
          
               year = inputyear();
              
               daycode = determinedaycode(year);
              
               determineleapyear(year);
              
               calendar(year, daycode);
              
               printf("\n");break;
              
           // Removing whole case-3 details and making case-4, case-3
          
           case 3: break;
          
           default:printf("\n Enter correct value");break;
       }  
   }  
}

// Also definitions of days and month functions are removed  

SAMPLE OUTPUT

MAIN MENU 1. View calendar for specified month in selected year 2.view the calendar for the specified year 3. End the program 2 Enter Year: 2020 January Sun Mon Tue Wed Thu Fri Sat 1 2 3 7 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 UNA ONU 00 February Sun Mon Tue Wed Thu Fri Sat 3 4 5 6 7 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 WOON NA LOMON March Sun Mon Tue Wed Thu Fri Sat 1 2 3 4 5 6 7 9 10 11 12 13 14 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 April Sun Mon Tue Wed Thu Fri Sat 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30

May Sun Mon Tue Wed Thu Fri Sat 1 2 3 4 5 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 NPP лою в June Sun Mon Tue Wed Thu Fri Sat 1 2 3 4 5 6 7 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 July Sun Mon Tue Wed Thu Fri Sat 1 2 3 4 5 7 8 9 10 11 12 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 como 1 August Sun Mon Tue Wed Thu Fri Sat 2 3 4 5 9 10 11 12 13 14 15 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 16

September Sun Mon Tue Wed Thu Fri Sat 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 3 October Sun Mon Tue Wed Thu Fri Sat 1 2 4 5 6 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 November Sun Mon Tue Wed Thu Fri Sat 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 December Sun Mon Tue Wed Thu Fri Sat 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 MAIN MENU 1. View calendar for specified month in selected year 2. view the calendar for the specified year 3.End the program 3 Process exited after 20.1 seconds with return value o Press any key to continue

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