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

What does umask 077 mean?

990

---

How much ram do I have linux?

974

---

Which command is used to delete a group?

1186

---

How do you create a text file in linux?

1138

---

How do I check my cpu cores?

1011

---

What is cat command in linux?

1013

---

What is top command in linux?

992

---

What is a makefile in linux?

997

---

What commands are used to see all jobs running in the hadoop cluster and kill a job in linux?

963

---

What is linux pwd (print working directory) command?

972

---

How can I type in cmd?

930

---

How do you insert comments in the command line prompt?

1181

---

How to recover /etc/passwd file and /etc/shadow file?

990

---

What is umask 000?

945

---

What is the fastest way to enter a series of commands from the command-line?

1067

---