Ytdyklly

Short answer: use "$@" (note the double quotes). The other forms are very rarely useful.

"$@" is a rather strange syntax. It is replaced by all the positional parameters, as separate fields. If there are no positional parameters ($# is 0), then "$@" expands to nothing (not an empty string, but a list with 0 elements), if there is one positional parameter then "$@" is equivalent to "$1", if there are two positional parameters then "$@" is equivalent to "$1" "$2", etc.

"$@" allows you to pass down the arguments of a script or function to another command. It is very useful for wrappers that do things like setting environment variables, preparing data files, etc. before calling a command with the same arguments and options that the wrapper was called with.

For example, the following function filters the output of cvs -nq update. Apart from the output filtering and the return status (which is that of grep rather than that of cvs), calling cvssm on some arguments behaves like calling cvs -nq update with these arguments.

cvssm () { cvs -nq update "$@" | egrep -v '^[?A]'; }

"$@" expands to the list of positional parameters. In shells that support arrays, there is a similar syntax to expand to the list of elements of the array: "${array[@]}" (the braces are mandatory except in zsh). Again, the double quotes are somewhat misleading: they protect against field splitting and pattern generation of the array elements, but each array element ends up in its own field.

_{Some ancient shells had what is arguably a bug: when there were no positional arguments, "$@" expanded to a single field containing an empty string, rather than into no field. This led to the workaround ${1+"$@"} (made famous via the Perl documentation). Only older versions of the actual Bourne shell and the OSF1 implementation are affected, none of its modern compatible replacements (ash, ksh, bash, …) are. /bin/sh is not affected on any system that was released in the 21st century that I know of (unless you count Tru64 maintenance release, and even there /usr/xpg4/bin/sh is safe so only #!/bin/sh script are affected, not #!/usr/bin/env sh scripts as long as your PATH is set up for POSIX compliance). In short, this is a historical anecdote that you don't need to worry about.}

"$*" always expands to one word. This word contains the positional parameters, concatenated with a space in between. (More generally, the separator is the first character of the value of the IFS variable. If the value of IFS is the empty string, the separator is the empty string.) If there are no positional parameters then "$*" is the empty string, if there are two positional parameters and IFS has its default value then "$*" is equivalent to "$1 $2", etc.

$@ and $* outside quotes are equivalent. They expand to the list of positional parameters, as separate fields, like "$@"; but each resulting field is then split into separate fields which are treated as file name wildcard patterns, as usual with unquoted variable expansions.

For example, if the current directory contains three files bar, baz and foo, then:

set --         # no positional parameters

for x in "$@"; do echo "$x"; done  # prints nothing

for x in "$*"; do echo "$x"; done  # prints 1 empty line

for x in $*; do echo "$x"; done    # prints nothing

set -- "b* c*" "qux"

echo "$@"      # prints `b* c* qux`

echo "$*"      # prints `b* c* qux`

echo $*        # prints `bar baz c* qux`

for x in "$@"; do echo "$x"; done  # prints 2 lines: `b* c*` and `qux`

for x in "$*"; do echo "$x"; done  # prints 1 lines: `b* c* qux`

for x in $*; do echo "$x"; done    # prints 4 lines: `bar`, `baz`, `c*` and `qux`

Here is a simple script to demonstrates the difference between $* and $@:

#!/bin/bash



test_param() {

  echo "Receive $# parameters"

  echo Using '$*'



  echo

  for param in $*; do

    printf '==>%s<==n' "$param"

  done;



  echo

  echo Using '"$*"'

  for param in "$*"; do

    printf '==>%s<==n' "$param"

  done;



  echo

  echo Using '$@'

  for param in $@; do

    printf '==>%s<==n' "$param"

  done;



  echo

  echo Using '"$@"';

  for param in "$@"; do

  printf '==>%s<==n' "$param"

  done

}



IFS="^${IFS}"



test_param 1 2 3 "a b c"

Output:

% cuonglm at ~

% bash test.sh

Receive 4 parameters



Using $*

==>1<==

==>2<==

==>3<==

==>a<==

==>b<==

==>c<==



Using "$*"

==>1^2^3^a b c<==



Using $@

==>1<==

==>2<==

==>3<==

==>a<==

==>b<==

==>c<==



Using "$@"

==>1<==

==>2<==

==>3<==

==>a b c<==

In array syntax, there is no difference when using $* or $@. It only make sense when you use them with double quotes "$*" and "$@".

The code you provided will give the same result. To understand it better, try this:

foo () {

    for i in "$*"; do

        echo "$i"

    done

}



bar () {

    for i in "$@"; do

        echo "$i"

    done

}

The output should now be different. Here's what I get:

$ foo() 1 2 3 4

1 2 3 4

$ bar() 1 2 3 4

1

2

3

4

This worked for me on bash. As far as I know, ksh should not differ much. Essentially, quoting $* will treat everything as one word, and quoting $@ will treat the list as separate words, as can be seen in the example above.

As an example of using the IFS variable with $*, consider this

fooifs () {

    IFS="c"            

    for i in "$*"; do

        echo "$i"

    done

    unset IFS          # reset to the original value

}

I get this as a result:

$ fooifs 1 2 3 4

1c2c3c4

Also, I've just confirmed it works the same in ksh. Both bash and ksh tested here were under OSX but I can't see how that would matter much.

The difference is important when writing scripts that should use the positional parameters in the right way...

Imagine the following call:

$ myuseradd -m -c "Carlos Campderrós" ccampderros

Here there are just 4 parameters:

$1 => -m

$2 => -c

$3 => Carlos Campderrós

$4 => ccampderros

In my case, myuseradd is just a wrapper for useradd that accepts the same parameters, but adds a quota for the user:

#!/bin/bash -e



useradd "$@"

setquota -u "${!#}" 10000 11000 1000 1100

Notice the call to useradd "$@", with $@ quoted. This will respect the parameters and send them as-they-are to useradd. If you were to unquote $@ (or to use $* also unquoted), useradd would see 5 parameters, as the 3rd parameter which contained a space would be split in two:

$1 => -m

$2 => -c

$3 => Carlos

$4 => Campderrós

$5 => ccampderros

(and conversely, if you were to use "$*", useradd would only see one parameter: -m -c Carlos Campderrós ccampderros)

So, in short, if you need to work with parameters respecting multi-word parameters, use "$@".

   *      Expands  to  the positional parameters, starting from one.  When

          the expansion occurs within double quotes, it expands to a  sin‐

          gle word with the value of each parameter separated by the first

          character of the IFS special variable.  That is, "$*" is equiva‐

          lent to "$1c$2c...", where c is the first character of the value

          of the IFS variable.  If IFS is unset, the parameters are  sepa‐

          rated  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  begin‐

          ning  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).

// man bash . is ksh, afair, similar behaviour.

Talking about differences between zsh and bash:

With quotes around $@ and $*, zsh and bash behave the same, and I guess the result is quite standard among all shells:

 $ f () { for i in "$@"; do echo +"$i"+; done; }; f 'a a' 'b' ''

 +a a+

 +b+

 ++

 $ f () { for i in "$*"; do echo +"$i"+; done; }; f 'a a' 'b' ''

 +a a b +

Without quotes, results are the same for $* and $@, but different in bash and in zsh. In this case zsh shows some odd behaviour:

bash$ f () { for i in $*; do echo +"$i"+; done; }; f 'a a' 'b' ''

+a+

+a+

+b+

zsh% f () { for i in $*; do echo +"$i"+; done; }; f 'a a' 'b' ''  

+a a+

+b+

(Zsh usually don't split textual data using IFS, unless explicitly requested, but notice that here the empty argument is unexpectedly missing in the list.)

One of the answers says $* (which I think of as a "splat") is rarely useful.

I search google with G() { IFS='+' ; w3m "https://encrypted.google.com/search?q=$*" ; }

Since URL’s are often split with a +, but my keyboard makes   easier to reach than +, $* + $IFS feel worthwhile.