Re: Brace expansion inside of command substitution - broken or is it me?

[Greg Wooledge]

On Fri, Feb 18, 2011 at 10:53:31PM +0100, Andreas Schwab wrote:
> Greg Wooledge <wooledg@eeg.ccf.org> writes:
> 
> > On Fri, Feb 18, 2011 at 10:32:13PM +0100, Peter Hofmann wrote:
> >> 
> >>  $ echo "$(echo "{1..3}")"
> >>  1 2 3
> >> 
> >> Huh?
> >
> > Brace expansion is a funny thing.  My belief at the moment -- I'm sure
> > someone will correct me if I'm wrong -- is that because you've got
> > everything quoted up, it's all seen as one "word" by the parser.  And
> > it's a word that just happens to have a brace expansion in it.  So,
> > the parser expands it out something like this:
> >
> > $ echo "$(echo "1")" "$(echo "2")" "$(echo "3")"
> >
> > Counting PIDs on my sequentially-generating-PIDs OS seems to confirm that
> > it's running three child processes, so that lends a tiny bit of evidence
> > to my theory.
> 
> $ set -x
> $ echo "$(echo "{1..3}")"
> ++ echo 1
> ++ echo 2
> ++ echo 3
> + echo 1 2 3
> 1 2 3
> 
> Andreas.

So... what I said was correct?

If I expand it out the way I said, and do your set -x trick, I get the
same output:

imadev:~$ set -x
imadev:~$ echo "$(echo "1")" "$(echo "2")" "$(echo "3")"
++ echo 1
++ echo 2
++ echo 3
+ echo 1 2 3
1 2 3

So I think you're saying I'm correct.