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What is the difference between $ @ and $ * in a shell script?

in script.sh:

aa=$@
bb=$*
echo $aa
echo $bb

at startup:

 source script.sh a b c d e f g

I get:

a b c d e f g
a b c d e f g

What is the difference between $@and $*?

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1 answer

There is no difference between $*and $@, but there is a difference between "$@"and "$*".

$ cat 1.sh
mkdir "$*"

$ cat 2.sh
mkdir "$@"

$ sh 1.sh a "b c" d

$ ls -l
total 12
-rw-r--r-- 1 igor igor   11 mar 24 10:20 1.sh
-rw-r--r-- 1 igor igor   11 mar 24 10:20 2.sh
drwxr-xr-x 2 igor igor 4096 mar 24 10:21 a b c d

We gave three arguments to the script ( a, b cand d), but in "$ *" they were all combined into one argument a b c d.

$ sh 2.sh a "b c" d

$ ls -l
total 24
-rw-r--r-- 1 igor igor   11 mar 24 10:20 1.sh
-rw-r--r-- 1 igor igor   11 mar 24 10:20 2.sh
drwxr-xr-x 2 igor igor 4096 mar 24 10:21 a
drwxr-xr-x 2 igor igor 4096 mar 24 10:21 a b c d
drwxr-xr-x 2 igor igor 4096 mar 24 10:21 b c
drwxr-xr-x 2 igor igor 4096 mar 24 10:21 d

, "$*" , "$@" , script. "$ @" - , " ". , a "b c" d (, , ) "a" "b c" "d" "a b c d" ("$*") "a" "b" "c" "d" ($@ $*).

, :

http://tldp.org/LDP/abs/html/internalvariables.html#ARGLIST

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