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According to this page, $@ and $* do pretty much the same thing: 

The $@ holds list of all arguments passed to the script. The $* holds list of all arguments passed to the script.

After searching all the top hits in google, I'm not able to find any explanation why there would be 2 seemingly duplicate syntaxes.

They appear to work the same in my scripts. 

cat foo.sh #!/bin/bash echo The parameters passed in are $@ echo The parameters passed in are $* ./foo.sh herp derp The parameters passed in are herp derp The parameters passed in are herp derp
  1. Is one preferred over the other?
  2. Why are there 2 builtin variables to do the exact same thing?

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They aren't the same. $* is a single string, whereas $@ is an actual array. To see the difference, execute the following script like so:

 > ./test.sh one two "three four"

The script:

#!/bin/bash echo "Using \"\$*\":" for a in "$*"; do echo $a; done echo -e "\nUsing \$*:" for a in $*; do echo $a; done echo -e "\nUsing \"\$@\":" for a in "$@"; do echo $a; done echo -e "\nUsing \$@:" for a in $@; do echo $a; done

The explanation and the results for the four cases are below.

In the first case, the parameters are regarded as one long quoted string:

Using "$*": one two three four

Case 2 (unquoted) - the string is broken into words by the for loop:

Using $*: one two three four

Case 3 - it treats each element of $@ as a quoted string:

Using "$@": one two three four

Last case - it treats each element as an unquoted string, so the last one is again split by what amounts to for three four:

Using $@: one two three four
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    Why make such subtle design? Commented Sep 20, 2021 at 8:30
  • I agree, it's a hacked, non-intuitive design predating Bash, the idea that "$@" should expand to multiple words. Some other syntax was needed. And that other syntax should have worked not only with $*, but also with $anyOtherVariable. Probably, unquoted $@ should have been words, and ${@otherVariable} should have worked similarly. Decades too late now! Commented Nov 1, 2023 at 21:26
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The difference comes in how they are expanded.

$* expands to a single argument with all the elements delimited by spaces (actually the first character of $IFS).
$@ expands to multiple arguments.

For example

#!/bin/bash echo "With *:" for arg in "$*"; do echo "<$arg>"; done echo echo "With @:" for arg in "$@"; do echo "<$arg>"; done

 

$ /tmp/test.sh 1 2 "3 4" With *: <1 2 3 4> With @: <1> <2> <3 4>
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    The OP want to know difference between $* and $@, not "$*" and "$@". Commented May 12, 2014 at 16:26
  • actually the first character of $IFS i find that annotation really critical , as if you've recklessly defined IFS as a long string ,bash won't raise and error and would just use the first characters of the string (even if it was a space preceding the desired character) Commented Dec 6, 2021 at 11:18
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You can review Bash Beginners Guide for more information. These to do pretty much the same thing with the difference of how it's separated:

$* - 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.

$@ - Expands to the positional parameters, starting from one. When the expansion occurs within double quotes, each parameter expands to a separate word.

But unless you set IFS to value other then default it could look identical.

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    This is the best explanation imo, because it notes that they are the same, and only behave differently when quoted. Commented Jul 8, 2020 at 23:38
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Is one preferred over the other?

A short answer is "$@".

If you use them without double quote, they are the same. There is no one preferred over the other in this case. But I suggest you to always use them with double quote, except you know what exactly you want.

Why are there 2 builtin variables to do the exact same thing?

There are no difference between $* and $@, but "$@" and "$*" have.

You can see my answer in this post to see how are they different.

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