在bash shell脚本中传播所有参数

我正在写一个非常简单的脚本来调用另一个脚本，我需要将参数从当前脚本传播到正在执行的脚本。

例如，我的脚本名称是foo.sh，并调用bar.sh

foo.sh：

bar $1 $2 $3 $4

我怎样才能做到这一点，没有明确指定每个参数？

使用"[email protected]"，而不是简单的[email protected]，如果你真的希望你的参数传递相同。

注意：

$ cat foo.sh
#!/bin/bash
baz.sh [email protected]

$ cat bar.sh
#!/bin/bash
baz.sh "[email protected]"

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

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

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

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

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

对于bash和其他类似Bourne的炮弹：

java com.myserver.Program "[email protected]"

未经作者同意，本文严禁转载，违者必究！