What is the difference between $@ and $* in shell scripts?
ShellScriptingShell Problem Overview
In shell scripts, what is the difference between $@ and $*?
Which one is the preferred way to get the script arguments?
Are there differences between the different shell interpreters about this?
Shell Solutions
Solution 1 - Shell
From here:
> $@ behaves like $* except that when quoted the arguments are broken up properly if there are spaces in them.
Take this script for example (taken from the linked answer):
for var in "$@"
do
echo "$var"
done
Gives this:
$ sh test.sh 1 2 '3 4'
1
2
3 4
Now change "$@" to $*:
for var in $*
do
echo "$var"
done
And you get this:
$ sh test.sh 1 2 '3 4'
1
2
3
4
(Answer found by using Google)
Solution 2 - Shell
A key difference from my POV is that "$@" preserves the original number of arguments. It's the only form that does. For that reason it is very handy for passing args around with the script.
For example, if file my_script contains:
#!/bin/bash
main()
{
echo 'MAIN sees ' $# ' args'
}
main $*
main $@
main "$*"
main "$@"
### end ###
and I run it like this:
my_script 'a b c' d e
I will get this output:
MAIN sees 5 args
MAIN sees 5 args
MAIN sees 1 args
MAIN sees 3 args
Solution 3 - Shell
With $@ each parameter is a quoted string. Otherwise it behaves the same.
See: http://tldp.org/LDP/abs/html/internalvariables.html#APPREF