Given ./mysh0:


exec ./mysh1 $*

And ./mysh1:


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.

  • 2
    Using "$@" instead of $*, note the quotes. See also: unix.stackexchange.com/q/41571/38906
    – cuonglm
    Dec 10 '15 at 3:15
  • Hmm. Anyway working around it? I'm afraid ./mysh0 is an intermediate script that I do not own.
    – jvliwanag
    Dec 10 '15 at 3:22
  • 2
    No, you can not. You must control the mysh0, because it decided how to pass argument to mysh1.
    – cuonglm
    Dec 10 '15 at 3:27
  • Ok got it! Mind posting this as an answer so I can mark this as answered?
    – jvliwanag
    Dec 10 '15 at 4:01

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.

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.