Escape shell arg from one script to another

Question

Given ./mysh0:

#!/bin/bash

exec ./mysh1 $*

And ./mysh1:

#!/bin/bash

echo $1
echo $2
echo $3

How do I call mysh0 such that the arguments to mysh1 and what's eventually printed are "A", "B 2" and "C"?

Calling this as ./mysh0 A "B 2" C does not work.

Using `"$@"` instead of `$*`, note the quotes. See also: http://unix.stackexchange.com/q/41571/38906 — cuonglm, Dec 10 '15 at 03:15
Hmm. Anyway working around it? I'm afraid `./mysh0` is an intermediate script that I do not own. — jvliwanag, Dec 10 '15 at 03:22
No, you can not. You must control the `mysh0`, because it decided how to pass argument to `mysh1`. — cuonglm, Dec 10 '15 at 03:27
Ok got it! Mind posting this as an answer so I can mark this as answered? — jvliwanag, Dec 10 '15 at 04:01

score 3 · Accepted Answer · edited Apr 13 '17 at 12:36

You must use "$@" instead of $*:

exec ./mysh1 "$@"

That's the right way to expand all positional arguments as separated words.

When you use $*, all positional arguments was concatenated in to one long string, with the first value of IFS as separator, which default to a whitespace, you got A B 2 C.

Now, because you use $* without double quote (which can lead to security implications and make your script choked), the shell perform split+glob on it. The long string you got above was split into four words, A, B, 2 and C.

Therefore, you actually passed four arguments to mysh1 instead of three.

Escape shell arg from one script to another

1 Answers1