#!/bin/bash

# gcd.sh: greatest common divisor

# Uses Euclid's algorithm

# The "greatest common divisor" (gcd) of two integers

#+ is the largest integer that will divide both, leaving no remainder.

# Euclid's algorithm uses successive division.

# In each pass,

#+ dividend

#+ divisor

#+ until remainder = 0.

# The gcd = dividend, on the final pass.

#

# For an excellent discussion of Euclid's algorithm, see

#+ Jim Loy's site, http://www.jimloy.comumber/euclids.htm.

# ------------------------------------------------------

# Argument check

ARGS=2

E_BADARGS=85

if [ $# -ne "$ARGS" ]

then

echo "Usage: `basename $0` first-number second-number"

exit $E_BADARGS

fi

# ------------------------------------------------------

gcd ()

{

dividend=$1 # Arbitrary assignment.

divisor=$2 #! It doesn't matter which of the two is larger.

# Why not?

remainder=1 # If an uninitialized variable is used inside

#+ test brackets, an error message results.

until [ "$remainder" -eq 0 ]

do # ^^^^^^^^^^ Must be previously initialized!

let "remainder = $dividend % $divisor"

dividend=$divisor # Now repeat with 2 smallest numbers.

divisor=$remainder

done # Euclid's algorithm

} # Last $dividend is the gcd.

gcd $1 $2

echo; echo "GCD of $1 and $2 = $dividend"; echo

# Exercises :

# ---------

# 1) Check command-line arguments to make sure they are integers,

#+ and exit the script with an appropriate error message if not.

# 2) Rewrite the gcd () function to use local variables.

exit 0

2) Use this script as a prototype for a writing a script to determine if a year is a leap year. A year is a leap year if the year is evenly divisible by 4 unless it is evenly divisible by 100 in which case it is not a leap year unless it is evenly divisible by 400 in which case it is a leap year