Shell - difference between '\$@" and "\$*"

There are 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 c and d) but in "$*" they all were merged 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

You can see here, that "$*" means always one single argument, and "$@" contains as many arguments, as the script had. "$@" is a special token which means "wrap each individual argument in quotes". So a "b c" d becomes (or rather stays) "a" "b c" "d" instead of "a b c d" ("$*") or "a" "b" "c" "d" ($@ or $*).

Copied from: https://stackoverflow.com/questions/15596826/what-is-the-difference-between-and-in-shell-script