what is the command To print script arguments
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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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