Table of Contents
Project: dayofweek
A project for C/C++ programming, CSCS1320S14, by Alana Whittier during the Spring semester 2014.
This project was begun on March 7, 2014 and it took nearly 3 days to complete.
Objectives
The purpose of this project is to write a program that utilizes a mental math trick to determine what day of the week that January 1st of any given year falls on in the 21st century. A two digit year (between 2001 and 2099) is input from the user and is computed using a series of selection statements to derive the day of the week that Jan 1st falls on per calendar year, minus one day for leap years.
Prerequisites
In order to successfully accomplish/perform this project, the listed resources/experiences need to be consulted/achieved:
Resources:
- project assignment page
- consultation with C/C++ programming instructor, Matt Haas
Experiences:
- ability to obtain user input
- ability to obtain the day of the week January 1st falls on per given year.
- ability to use selection statements to achieve desired result
- ability to use Bitwise && to distinguish between leap year and non-leap year
Background
The purpose of this project is to use a mental math technique to programmatically compute the day on which January 1st falls per given calendar year in the 21st century. The “mental math” trick is used to simplify the process of computation without being dependent on a calculator. To see how this particular technique works for computing the week day that January 1st falls on, see the assignment link below. Project: MENTAL MATH (DAY OF WEEK) To implement the program to compile and execute successfully, it is required to know how to determine leap years from non-leap years. As we experience leap years as much as every four years, simple algorithms can be used to determine the leap year. However, every 100 years, a leap year is skipped to make up for the 11 minutes removed from the 365.25 day calendar year.
Scope
The scope of the project is to write a program that utilizes the described mental math technique (see assignment in Background section above) to compute the week day that January 1st happens to fall on for a given year in the 21st century (2000-2099). In doing so, the program should prompt the user for the last two digits of the year, perform the calculation of the day of the week that January 1st falls on for the year given by user input, and display the value. Applying pointers to enable the use of scanf() for obtaining input from the user, using if() statements, are used in the implementation of this program.
Attributes
Upon successful completion of this project, the following attributes are achieved.
- More familiarity with pointers and scanf: This project requires user input.
- Become more exposed to selection statements: if statements are used to compute the day of the week January 1st falls on per given calendar year.
- Get more comfortable with Bitwise && operators: used to differentiate between leap years and non-leap years to obtain the correct output.
Procedure
The actual steps taken to accomplish the project. Include images, code snippets, command-line excerpts; whatever is useful for intuitively communicating important information for accomplishing the project.
Code
</code> /*dayofweek.c- A program using a math technique to output the day of the week;
that January 1st falls on from calender years 2001 to 2099 based;
on what year the user inputs;
A C program Written by: Alana Whittier for CSCS1320S14 on March 7, 2014; */
#include <stdio.h> #include <stdlib.h>
int main() {
int i; int *p; p=&i; int year;
printf(“Type last two digits of year from 2001-2099;\n”); scanf(“%d”, p);
year=i; i=(float)1); i=(year+i)%7;
leap years! if(year%4==0&&(year/4!=0)){ printf(“year, %d, is a leap year\n”,2000+ year);} if(i==1&&(year%4==0&&(year/4!=0))){ printf(“Jan 1st,%d, is on : Sunday\n”, 2000+ year);} if(i==2&&(year%4==0&&(year/4!=0))){ printf(“Jan 1st, %d, is on: Monday\n”, 2000+year);} if(i==3&&(year%4==0&&(year/4!=0))){ printf(“Jan 1st, %d, is on: Tuesday\n”, 2000+year);} if(i==4&&(year%4==0&&(year/4!=0))){ printf(“Jan 1st, %d, is on: Wednesday\n”, 2000+year);} if(i==5&&(year%4==0&&(year/4!=0))){ printf(“Jan 1st, %d, is on: Thursday\n”, 2000+year);} if(i==6&&(year%4==0&&(year/4!=0))){ printf(“Jan 1st, %d, is on: Friday\n”, 2000+year);} if(i==7&&(year%4==0&&(year/4!=0))){ printf(“Jan 1st, %d, is on Saturday\n”, 2000+year);} if(i==0&&(year%4==0&&(year/4!=0))){ printf(“Jan 1st, %d, is on Saturday\n”, 2000+year);} not leap years!
else if(i==1&&(year%4!=0)){
printf("Jan 1st, %d, is on: Monday\n", 2000+year);}
else if(i==2&&(year%4!=0)){
printf("Jan 1st, %d, is on: Tuesday\n", 2000+year);}
else if(i==3&&(year%4!=0)){
printf("Jan 1st, %d, is on: Wednesday\n", 2000+year);}
else if(i==4&&(year%4!=0)){
printf("Jan 1st, %d, is on: Thursday\n", 2000+year);}
else if(i==5&&(year%4!=0)){
printf("Jan 1st, %d, is on: Friday\n", 2000+year);}
else if(i==6&&(year%4!=0)){
printf("Jan 1st, %d, is on: Saturday\n", 2000+year);}
else if(i==7&&(year%4!=0)){
printf("Jan 1st, %d, is on: Sunday\n", 2000+year);}
else if(i==0&&(year%4!=0)){
printf("Jan 1st, %d, is on: Sunday\n", 2000+year);}
return(0); }
</code>
Execution
lab46:~/src/cscs1320$ nano dayofweek.c lab46:~/src/cscs1320$ gcc -o dayofweek dayofweek.c lab46:~/src/cscs1320$ ./dayofweek Type last two digits of year from 2001-2099; 04 year, 2004, is a leap year Jan 1st, 2004, is on: Thursday lab46:~/src/cscs1320$ cal 01 2004 January 2004 Su Mo Tu We Th Fr Sa 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 lab46:~/src/cscs1320$ ./dayofweek Type last two digits of year from 2001-2099; 67 Jan 1st, 2067, is on: Saturday lab46:~/src/cscs1320$ cal 01 2067 January 2067 Su Mo Tu We Th Fr Sa 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
Reflection
Interestingly enough, the year 2000 would not be a leap year if only using the first 2 criteria. The criteria:
- If year is evenly divisible by 4, it is a leap year.
- If year is evenly divided by 100, it is not a leap year, except when
- the year is also evenly divisible by 400.
The year 2000 is a significant calendar year because it was the first instance when the third criteria was invoked since switching from the Julian calendar to the Gregorian one. The reason for leap years? This was necessary to keep our calendar in alignment with Earth's revolutions around the sun. If we did not add a day to our calendar ~every 4 years, we would lose nearly 6 hours every year, so after 100 years our calendar would be off by nearly 24 days as measured by earth's revolutions. Why is a leap year skipped every 100 years, except in special instances (like year 2000)? This is to account for the 11 minutes shaved off from the 365.25 calendar year.
While I wrote my code more simplistically saying if i divided by 4 equals zero, then it is a leap year and if i divided by 4 does not equal zero, not a leap year, the code worked. I could have added that i/100==0 || i/400==0 is a leap year to make my code more realistic to how the calendar interprets leap years from non-leap years, particularly in special instances such as 2000.
References
In performing this project, the following resources were referenced: