What does $@ mean in a shell script?

$@ 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