How to resolve the algorithm Leap year step by step in the Bash programming language
Published on 12 May 2024 09:40 PM
How to resolve the algorithm Leap year step by step in the Bash programming language
Table of Contents
Problem Statement
Determine whether a given year is a leap year in the Gregorian calendar.
Let's start with the solution:
Step by Step solution about How to resolve the algorithm Leap year step by step in the Bash programming language
Source code in the bash programming language
#!/bin/bash
is_leap_year () # Define function named is_leap_year
{
declare -i year=$1 # declare integer variable "year" and set it to function parm 1
echo -n "$year ($2)-> " # print the year passed in, but do not go to the next line
if (( $year % 4 == 0 )) # if year not dividable by 4, then not a leap year, % is the modulus operator
then
if (( $year % 400 == 0 )) # if century dividable by 400, is a leap year
then
echo "This is a leap year"
else
if (( $year % 100 == 0 )) # if century not divisible by 400, not a leap year
then
echo "This is not a leap year"
else
echo "This is a leap year" # not a century boundary, but dividable by 4, is a leap year
fi
fi
else
echo "This is not a leap year"
fi
}
# test all cases
# call the function is_leap_year several times with two parameters... year and test's expectation for 'is/not leap year.
is_leap_year 1900 not # a leap year
is_leap_year 2000 is # a leap year
is_leap_year 2001 not # a leap year
is_leap_year 2003 not # a leap year
is_leap_year 2004 is # a leap year
# Save the above to a file named is_leap_year.sh, then issue the following command to run the 5 tests of the function
# bash is_leap_year.sh
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