How to preserve white space in Bash arguments

Question

I use the command:

cm1 cm2 arg1 arg2 'argument 3'

It first goes to cm1, which will then redirect arg1 arg2 'argument 3' to another file.

/usr/bin/cm1:

#! /bin/bash
# some script here
shift
cm2 $@

/usr/bin/cm2:

echo $#
# This returns 4 in lieu of 3 because the white space in 'argument 3' causes the argument to be split into two arguments.

So, how can I pass arguments from one script to another and make sure white space won't be read as an argument separator?

Solution

I assume you have to re-wrap it into quotes, like so:

#! /bin/bash
# some script here
shift
cm2 "$@"

Comments

[François ッ Vespa ت]

Thanks! It works on the simple example I gave. Now on the real script I am coding and that has more complex rerouting, arguments get broken at some point. But that means the error is coming from somewhere else... Thanks!

[Cougar]

It may be that you split these args later based on default IFS which default value is "<space><tab><newline>"

[Marek Podyma]

I was looking for opposite behavior - how to force my script to iterate over arguments separated with spaces. In my script I had quoted reference ("$@") so all arguments were treated as one concatenated string. I replaced: >for path in "$@"< with: >for path in $@< and it worked. Thank you.