bash shell 脚本中的 $@ 和 $* 有什么区别？
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What is the difference between $@ and $* in shell script?
提问by 0x90
in my script.sh:
在我的script.sh：
aa=$@
bb=$*
echo $aa
echo $bb
when running it:
运行时：
source script.sh a b c d e f g
I get:
我得到：
a b c d e f g
a b c d e f g
What is the difference between $@and $*?
$@和 和有什么不一样$*？
回答by Igor Chubin
There are no difference between $*and $@, but there is a difference between "$@"and "$*".
有没有什么区别$*和$@，但之间的差异"$@"和"$*"。
$ 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 "$*" they all were merged into one argument a b c d.
我们为脚本提供了三个参数（a,b c和d），但在“$*”中，它们都合并为一个参数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. "$@" is a special token which means "wrap each individual argument in quotes". So a "b c" dbecomes (or rather stays) "a" "b c" "d"instead of "a b c d"("$*") or "a" "b" "c" "d"($@or $*).
您可以在这里看到，这"$*"意味着始终只有一个参数，并且"$@"包含与脚本一样多的参数。"$@" 是一个特殊标记，意思是“将每个单独的参数用引号括起来”。所以a "b c" d变成（或更确切地说保持）"a" "b c" "d"而不是"a b c d"（"$*"）或"a" "b" "c" "d"（$@或$*）。
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