Answer Posted / shabab

$* and $@ are the right answers

$# - prints out the number of arguments passed

Consider the below code
########################
for i in "$*"
do
print $i
done

for i in "$@"
do
print $i
done
########################

and you call the script by saying

#samp.sh hai welcome to "Unix Forum"
hai welcome to Unix Forum
hai
welcome
to
Unix Forum


The first line is the output by printing out $*
the next four lines are with the help of $@.

So
$* will combine all arguments to a single string
$@ will have each arguments as a seperate string

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