2

I have wrote a simple script that takes in any number of parameters to demonstrate the difference between $@ and $*: 

#!/bin/bash                                                                
echo "double quoted $*     $@"
echo 'single quoted $*     $@'

On the CLI I did 

$./stuff.sh a b c d e f dfs

And this is what prints out 

double quoted a b c d e f dfs     a b c d e f dfs
single quoted $*     $@

Since they are identical does that mean $@ equals to $*? Or is there a point I am missing?

Stefan van den Akker
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cyc115
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  • See [this](http://stackoverflow.com/questions/3008695/what-the-difference-between-and-in-bash), [this](http://stackoverflow.com/questions/2761723/what-is-the-difference-between-and-in-shell-scripts), [this](http://stackoverflow.com/questions/17967074/difference-between-and-in-bash-scripting), [this](http://stackoverflow.com/questions/15596826/what-is-the-difference-between-and-in-shell-script). – devnull Sep 27 '13 at 14:27

3 Answers3

6

From Special Parameters in Bash Reference Manual

*

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).

Better with an example:

$ d=(a b c)
$ for i in "${d[*]}"; do echo $i; done <---- it is a field all together
a b c
$ for i in "${d[@]}"; do echo $i; done <---- each item is a different field
a
b
c
fedorqui
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  • Correct info, but quoting back to a really awful source. (One of the constant annoyances in #bash is helping folks unlearn bugs the ABS taught them to write). – Charles Duffy Sep 27 '13 at 14:32
  • Precisely I was just looking for a better reference. Updated :) – fedorqui Sep 27 '13 at 14:33
4
$ set -- "hello world" foo bar
$ printf '<%s> ' "$@"; echo
<hello world> <foo> <bar>     # "$@" keeps original separation
$ printf '<%s> ' "$*"; echo
<hello world foo bar>         # "$*" combines everything into one string
$ printf '<%s> ' $*; echo
<hello> <world> <foo> <bar>   # $* (no quotes) loses delimitation entirely.
$ printf '<%s> ' $@; echo
<hello> <world> <foo> <bar>   # unquoted, $@ is identical to unquoted $*.

Technically, $* isn't limited to space-delimiting; it uses the first character of $IFS. This can be used to your advantage:

jars=( *.jar )
IFS=:
export CLASSPATH=${jars[*]}
Charles Duffy
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  • Could you provide an example for `$@` (without double quotes) like you did for the others please? Your examples help me to quickly understand the differences. – Elie G. Jan 08 '20 at 17:14
  • When used without quotes, `$@` is exactly identical to `$*` without quotes. I'll gladly add an example to that effect. – Charles Duffy Jan 08 '20 at 17:23
3

From http://linuxsig.org/files/bash_scripting.html:

$* Stores all the arguments that were entered on the command line

so

$1 $2 ...

"$@" Stores all the arguments that were entered on the command line, individually quoted

so

"$1" "$2" ...
Alberto Zaccagni
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