Accessing bash command line args $@ vs $*

Question

In many SO questions and bash tutorials I see that I can access command line args in bash scripts in two ways:

$ ~ >cat testargs.sh 
#!/bin/bash

echo "you passed me" $*
echo "you passed me" $@

Which results in:

$ ~> bash testargs.sh arg1 arg2
you passed me arg1 arg2
you passed me arg1 arg2

What is the difference between $* and $@?
When should one use the former and when shall one use the latter?

Answer 1

This example let may highlight the differ between "at" and "asterix" while we using them. I declared two arrays "fruits" and "vegetables"

fruits=(apple pear plumm peach melon)            
vegetables=(carrot tomato cucumber potatoe onion)

printf "Fruits:\t%s\n" "${fruits[*]}"            
printf "Fruits:\t%s\n" "${fruits[@]}"            
echo + --------------------------------------------- +      
printf "Vegetables:\t%s\n" "${vegetables[*]}"    
printf "Vegetables:\t%s\n" "${vegetables[@]}"    

See the following result the code above:

Fruits: apple pear plumm peach melon
Fruits: apple
Fruits: pear
Fruits: plumm
Fruits: peach
Fruits: melon
+ --------------------------------------------- +
Vegetables: carrot tomato cucumber potatoe onion
Vegetables: carrot
Vegetables: tomato
Vegetables: cucumber
Vegetables: potatoe
Vegetables: onion

Answer 2

$@ is same as $*, but each parameter is a quoted string, that is, the parameters are passed on intact, without interpretation or expansion. This means, among other things, that each parameter in the argument list is seen as a separate word.

Of course, "$@" should be quoted.

http://tldp.org/LDP/abs/html/internalvariables.html#ARGLIST

Answer 3

A nice handy overview table from the Bash Hackers Wiki:

where c in the third row is the first character of $IFS, the Internal Field Separator; an environment variable.

If the arguments are to be stored in a script variable and the arguments are expected to contain spaces, I wholeheartedly recommend employing a "$*" trick with the internal field separator $IFS set to tab.

Answer 4

The difference appears when the special parameters are quoted. Let me illustrate the differences:

$ set -- "arg  1" "arg  2" "arg  3"

$ for word in $*; do echo "$word"; done
arg
1
arg
2
arg
3

$ for word in $@; do echo "$word"; done
arg
1
arg
2
arg
3

$ for word in "$*"; do echo "$word"; done
arg  1 arg  2 arg  3

$ for word in "$@"; do echo "$word"; done
arg  1
arg  2
arg  3

---

one further example on the importance of quoting: note there are 2 spaces between "arg" and the number, but if I fail to quote $word:

$ for word in "$@"; do echo $word; done
arg 1
arg 2
arg 3

and in bash, "$@" is the "default" list to iterate over:

$ for word; do echo "$word"; done
arg  1
arg  2
arg  3

Answer 5

$*

Expands to the positional parameters, starting from one. When the expansion occurs within double quotes, it expands to a single word with the value of each parameter separated by the first character of the IFS special variable. That is, "$*" is equivalent to "$1c$2c...", where c is the first character of the value of the IFS variable. If IFS is unset, the parameters are separated by spaces. If IFS is null, the parameters are joined without intervening separators.

$@

Expands to the positional parameters, starting from one. When the expansion occurs within double quotes, each parameter expands to a separate word. That is, "$@" is equivalent to "$1" "$2" ... If the double-quoted expansion occurs within a word, the expansion of the first parameter is joined with the beginning part of the original word, and the expansion of the last parameter is joined with the last part of the original word. When there are no positional parameters, "$@" and $@ expand to nothing (i.e., they are removed).

Source: Bash man