Run Linux shell script with arguments

Question

Can somone help me with this: so i have this script

#!/bin/bash


echo -n "Enter a value for X:(999 to exit): "
read x

until [[ $x == 999 ]]
do

echo -n "Enter a value for Y: "
read y
echo "X="$x
echo "Y="$y
((a=y+x))
echo "X+Y="$a
((s=y-x))
echo "X-Y="$s
((m=y*x))
echo "X*Y="$m
((d=y/x))
echo "X/Y="$d
((m=y%x))
echo "X%Y="$m

echo -n "Enter a value for X:(999 to exit): "
read x
if [[ $x == 999 ]];
then
    exit 0
fi

done
exit 0

but i didnt know how to write the rest of it, the missing thing is: Use the two command line arguments when the script starts if the user supplied them, and then prompt for more numbers to continue in the loop.

Answer 1

Am guessing the arguments you are looking from the user are x and y values. The easiest way to check if user provided arguments is to use $# which gets you the number of arguments given by the user.

So use it like this:

if [ "$#" -eq 2 ];       #2 arguments provided by user
then
    x=$1
    ... 
fi