What does $@ mean in a shell script?

Question

$@ is all of the parameters passed to the script.

For instance, if you call ./someScript.sh foo bar then $@ will be equal to foo bar.

If you do:

./someScript.sh foo bar

and then inside someScript.sh reference:

umbrella_corp_options "$@"

this will be passed to umbrella_corp_options with each individual parameter enclosed in double quotes, allowing to take parameters with blank space from the caller and pass them on.

$@ is nearly the same as $*, both meaning "all command line arguments". They are often used to simply pass all arguments to another program (thus forming a wrapper around that other program).

The difference between the two syntaxes shows up when you have an argument with spaces in it (e.g.) and put $@ in double quotes:

wrappedProgram "$@"
# ^^^ this is correct and will hand over all arguments in the way
#     we received them, i. e. as several arguments, each of them
#     containing all the spaces and other uglinesses they have.
wrappedProgram "$*"
# ^^^ this will hand over exactly one argument, containing all
#     original arguments, separated by single spaces.
wrappedProgram $*
# ^^^ this will join all arguments by single spaces as well and
#     will then split the string as the shell does on the command
#     line, thus it will split an argument containing spaces into
#     several arguments.

Example: Calling

wrapper "one two    three" four five "six seven"

will result in:

"$@": wrappedProgram "one two    three" four five "six seven"
"$*": wrappedProgram "one two    three four five six seven"
                             ^^^^ These spaces are part of the first
                                  argument and are not changed.
$*:   wrappedProgram one two three four five six seven

These are the command line arguments where:

$@ = stores all the arguments in a list of string
$* = stores all the arguments as a single string
$# = stores the number of arguments

The usage of a pure $@ means in most cases "hurt the programmer as hard as you can", because in most cases it leads to problems with word separation and with spaces and other characters in arguments.

In (guessed) 99% of all cases, it is required to enclose it in ": "$@" is what can be used to reliably iterate over the arguments.

for a in "$@"; do something_with "$a"; done

From the manual:

@

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).

Meaning.

In brief, $@ expands to the arguments passed from the caller to a function or a script. Its meaning is context-dependent: Inside a function, it expands to the arguments passed to such function. If used in a script (outside a function), it expands to the arguments passed to such script.

$ cat my-script
#! /bin/sh
echo "$@"

$ ./my-script "Hi!"
Hi!

$ put () { echo "$@"; }
$ put "Hi!"
Hi!

---

* Note: Word splitting.

The shell splits tokens based on the contents of the IFS environment variable. Its default value is \t\n; i.e., whitespace, tab, and newline. Expanding "$@" gives you a pristine copy of the arguments passed. Expanding $@ may not. More specifically, any arguments containing characters present in IFS might split into two or more arguments or get truncated.

Thus, most of the time what you will want to use is "$@", not $@.