这里有很多答案推荐带引号或不带引号的$@或$*，但似乎没有人解释这些参数的真正作用以及为什么你应该这样做。所以让我从这个答案中偷取一个很好的总结:

+--------+---------------------------+
| Syntax |      Effective result     |
+--------+---------------------------+
|   $*   |     $1 $2 $3 ... ${N}     |
+--------+---------------------------+
|   $@   |     $1 $2 $3 ... ${N}     |
+--------+---------------------------+
|  "$*"  |    "$1c$2c$3c...c${N}"    |
+--------+---------------------------+
|  "$@"  | "$1" "$2" "$3" ... "${N}" |
+--------+---------------------------+

请注意，引号会造成所有的不同，如果没有引号，两者的行为是相同的。

出于我的目的，我需要将参数从一个脚本传递到另一个脚本，为此最好的选择是:

# file: parent.sh
# we have some params passed to parent.sh 
# which we will like to pass on to child.sh as-is

./child.sh $*

注意，在上述情况下，没有引号和$@也可以工作。

bash和其他类似bourne的炮弹:

bar "$@"

如果在带引号的字符串中包含$@和其他字符，当有多个参数时，行为是非常奇怪的，只有第一个参数包含在引号中。

例子:

#!/bin/bash
set -x
bash -c "true foo $@"

收益率:

$ bash test.sh bar baz
+ bash -c 'true foo bar' baz

但是先赋值给另一个变量:

#!/bin/bash
set -x
args="$@"
bash -c "true foo $args"

收益率:

$ bash test.sh bar baz
+ args='bar baz'
+ bash -c 'true foo bar baz'

#!/usr/bin/env bash
while [ "$1" != "" ]; do
  echo "Received: ${1}" && shift;
done;

只是认为在尝试测试args如何进入脚本时，这可能更有用

我的SUN Unix有很多限制，甚至“$@”也没有按预期解释。我的变通方法是${@}。例如,

#!/bin/ksh
find ./ -type f | xargs grep "${@}"

顺便说一下，我必须有这个特定的脚本，因为我的Unix也不支持grep -r

如果你确实希望传递相同的参数，请使用“$@”而不是普通的$@。

观察:

$ cat no_quotes.sh
#!/bin/bash
echo_args.sh $@

$ cat quotes.sh
#!/bin/bash
echo_args.sh "$@"

$ cat echo_args.sh
#!/bin/bash
echo Received: $1
echo Received: $2
echo Received: $3
echo Received: $4

$ ./no_quotes.sh first second
Received: first
Received: second
Received:
Received:

$ ./no_quotes.sh "one quoted arg"
Received: one
Received: quoted
Received: arg
Received:

$ ./quotes.sh first second
Received: first
Received: second
Received:
Received:

$ ./quotes.sh "one quoted arg"
Received: one quoted arg
Received:
Received:
Received: