Calender Program in C - Beginners / Lab Assignments Source Code in C

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Home » C Home » Beginners / Lab Assignments Home » Calender Program in C

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Title	Calender Program in C
Author	Leya
Author Email	leyajacob [at] hotmail.com
Description	This was an assignment given to me last semester. (S3 in B.Tech). This is a very simple, down to earth program to generate the calendar for any given year from the 21st century onwards. It has no extra trimmings and is very easy to understand.
Category	C » Beginners / Lab Assignments
Hits	385189
Code	Select and Copy the Code
//Program to display calender for a given year #include <stdio.h> #include <conio.h> #include <process.h> #include <ctype.h> unsigned long days=0; //stores the days elapsed since 01.01.1899 void display(int n) //contains the number of days to display { int i, column, k, flag=0, j; printf("Sun Mon Tues Wed Thur Fri Sat "); for(i=1; i<=n; i++) { k=days%7;//remainder gives the starting day of each month if(flag==0) { for(j=1; j<=k; j++)//controls tabs of first week printf(" "); flag=1;//ensures that block is only executed once column=k; } printf("%d ", i); column++; if(column%7==0)//prints new line at the end of each week printf(" "); } printf(" Press any key to continue "); getch(); } void calculate(int year) //function calculates no. of days elapsed since 1899 { int i, month; for(i=1899; i<year; i++) //1899 chosen because Jan 1, 1899 is a Sunday { if((i%400==0)?1:((i%100==0)?0:((i%4==0)?1:0))) /This is because a leap year does not strictly fall on every fourth year. If a year is divisible by 4, then it is a leap year, but if that year is divisible by 100, then it is not a leap year. However, if the year is also divisible by 400, then it is a leap year. Eg: 1900 is not a leap year/ days+=366; else days+=365; } for(month=1; month<=12; month++) { printf(" "); switch(month) /switch case used to display each month and increment no. of days/ { case 1: printf(" JANUARAY %d ", year); display(31); days+=31; break; case 2: printf(" FEBURARY %d ", year); if((year%400==0)?1:((year%100==0)?0:((year%4==0)?1:0))) { display(29); days+=29; } else { display(28); days+=28; } break; case 3: printf(" MARCH %d ", year); display(31); days+=31; break; case 4: printf(" APRIL %d ", year); display(30); days+=30; break; case 5: printf(" MAY %d ", year); display(31); days+=31; break; case 6: printf(" JUNE %d ", year); display(30); days+=30; break; case 7: printf(" JULY %d ", year); display(31); days+=31; break; case 8: printf(" AUGUST %d ", year); display(31); days+=31; break; case 9: printf(" SEPTEMBER %d ", year); display(30); days+=30; break; case 10: printf(" OCTOBER %d ", year); display(31); days+=31; break; case 11: printf(" NOVEMBER %d ", year); display(30); days+=30; break; case 12: printf(" DECEMBER %d ", year); display(31); days+=31; break; } } } void main() { char ch[10]; int i, year, choice; do { clrscr(); days=0; printf("Enter the year in 'yyyy' format: "); scanf("%s", ch);//stores input first as a string for(i=0; i<strlen(ch); i++) if(ch[i]<'0' \|\| ch[i]>'9')//checks for invalid inputs { printf(" Invalid Year!"); printf(" END OF PROGRAM"); getch(); exit(0); } year = atoi(ch); //converts the year from string to integer datatype clrscr(); printf(" Calender for Year %d", year); printf(" ******************** "); calculate(year); //calls function to calculate no. of days elapsed printf(" ***************************************** "); printf(" Press 1 to continue, 2 to exit "); scanf("%d", &choice); }while(choice==1); clrscr(); printf(" END OF PROGRAM"); getch(); }

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