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.

#! /bin/bash
# some script here
shift
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?

Oliver Salzburg · Accepted Answer · 2012-03-13 10:27:32Z

8

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

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

answered Mar 13, 2012 at 10:27

Oliver Salzburg

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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!
– François ッ Vespa ت
Mar 13, 2012 at 10:30
It may be that you split these args later based on default IFS which default value is "<space><tab><newline>"
– Cougar
Mar 14, 2012 at 9:48
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.
– Marek Podyma
Nov 7, 2018 at 11:23

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