传播 Bash shell 脚本中的所有参数

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

例如，我的脚本名称是foo.sh并打电话bar.sh.

foo.sh:

bar $1 $2 $3 $4

在不明确指定每个参数的情况下如何做到这一点？

Use "$@"而不是普通的$@如果您确实希望传递相同的参数。

Observe:

$ 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: