Given ./mysh0:
#!/bin/bash
exec ./mysh1 $*
And ./mysh1:
#!/bin/bash
echo $1
echo $2
echo $3
How do I call mysh0 such that the arguments to mysh1 and what's eventually printed are "A", "B 2" and "C"?
Calling this as ./mysh0 A "B 2" C does not work.
You must use "$@" 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 $* without double quote (which can lead to security implications and make your script choked), the shell perform split+glob on it. The long string you got above was split into four words, A, B, 2 and C.
Therefore, you actually passed four arguments to mysh1 instead of three.
"$@"instead of$*, note the quotes. See also: unix.stackexchange.com/q/41571/38906./mysh0is an intermediate script that I do not own.mysh0, because it decided how to pass argument tomysh1.