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?
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 
$* or "$*"may be required, & the purpose cannot be served by $@ or "$@"?
"$@" with the quotes.
for word ... loop, you could write: printf "%s\n" "$*" to be compared to printf "%s\n" "$@" and printf "%s\n" $@! (... or even printf %s\\n "$@" ;)A nice handy overview table from the Bash Hackers Wiki:
| Syntax | Effective result |
|---|---|
$* | $1 $2 $3 … ${N} |
$@ | $1 $2 $3 … ${N} |
"$*" | "$1c$2c$3c…c${N}" |
"$@" | "$1" "$2" "$3" … "${N}" |
where c in the third row is the first character of $IFS, the Input Field Separator, a shell variable.
If the arguments are to be stored, load them in an array variable.
"$@" (with quotes). None of the other variants work
$*
Expands to the positional parameters, starting from one. When the expansion occurs within double quotes, it expands to a single word with the value of each parameter separated by the first character of the IFS special variable. That is, "$*" 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
$@ 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.
This example may highlight the difference between @ and * when we use them. I declared two arrays fruits and vegetables
fruits=(apple pear plum peach melon) vegetables=(carrot tomato cucumber potato 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 plum peach melon Fruits: apple Fruits: pear Fruits: plum Fruits: peach Fruits: melon + --------------------------------------------- + Vegetables: carrot tomato cucumber potato onion Vegetables: carrot Vegetables: tomato Vegetables: cucumber Vegetables: potato Vegetables: onion 
echo "something $@"as an error