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

[Peter Hofmann]

Hi all,

I stumbled upon some rather strange behaviour that I just can't explain.
Hopefully one of you can help me with that. :)

Let's start with a simple brace expansion:

 $ echo {1..3}
 1 2 3

Now add some quotes to prevent that expansion:

 $ echo "{1..3}"
 {1..3}

Adding command substitution:

 $ echo $(echo "${1..3}")
 {1..3}

So far, so good. It's what I expected. Let's add another level of
quotes:

 $ echo "$(echo "{1..3}")"
 1 2 3

Huh? Actually, I was expecting to get the same output as before.

Some debug output:

 $ set -x
 $ echo "$(echo "{1..3}")"
 ++ echo 1
 ++ echo 2
 ++ echo 3
 + echo 1 2 3
 1 2 3

Why's that?

a) As far as I understood, quotes inside of $(...) should not interfere
   with the outer quotes.
b) Why are there three subshells? Actually, that {1..3} doesn't get
   expanded. It's more like the call above is effectively equivalent to
   this:

$ echo "$(echo 1)" "$(echo 2)" "$(echo 3)"


To sum up my question: Why do I get

 $ echo "$(echo "{1..3}")"
 1 2 3

instead of

 $ echo "$(echo "{1..3}")"
 {1..3}

?


I saw this happening on every version of Bash I could find -- ranging
from Bash 4 in current Arch Linux to some old Bash 3 of msysgit[1] on
Windows. Tried it on the last two or three versions of Ubuntu. And so
on. To be honest, this almost convinces that I really missed something,
so any help is very much appreciated.

Below[2] is the information about my current system (Arch Linux) that
"bashbug" gave me.

Many thanks in advance,
Peter


[1] http://code.google.com/p/msysgit/
[2] bashbug-infos:

 Machine: i686
 OS: linux-gnu
 Compiler: gcc
 Compilation CFLAGS:  -DPROGRAM='bash' -DCONF_HOSTTYPE='i686' 
-DCONF_OSTYPE='linux-gnu' -DCONF_MACHTYPE='i686-pc-linux-gnu' 
-DCONF_VENDOR='pc' -DLOCALEDIR='/usr/share/locale' -DPACKAGE='bash' -DSHELL 
-DHAVE_CONFIG_H   -I.  -I. -I./include -I./lib   -march=i686 -mtune=generic -O2 
-pipe 
-DDEFAULT_PATH_VALUE='/usr/local/bin:/usr/bin:/bin:/usr/local/sbin:/usr/sbin:/sbin'
 -DSTANDARD_UTILS_PATH='/usr/bin:/bin:/usr/sbin:/sbin' 
-DSYS_BASHRC='/etc/bash.bashrc' -DSYS_BASH_LOGOUT='/etc/bash.bash_logout'
 uname output: Linux pinguin 2.6.37-ARCH #1 SMP PREEMPT Fri Feb 11 16:55:18 UTC 
2011 i686 AMD Athlon(tm) 64 X2 Dual Core Processor 4200+ AuthenticAMD GNU/Linux
 Machine Type: i686-pc-linux-gnu

 Bash Version: 4.1
 Patch Level: 9
 Release Status: release

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

[Peter Hofmann]

Hi,

On Fri, Feb 18, 2011 at 07:26:18PM -0500, Chet Ramey wrote:
> Brace expansion is strictly textual, is performed before all other
> expansions, and doesn't understand a whole lot of shell syntax.
> It does understand a little about quoted strings, so what you get is
> 
> echo "$(echo "1")" "$(echo "2")" "$(echo "3")"
> 
> The preamble is "$(echo ", the portion to be expanded is {1..3}, and the
> postscript is ")".

thank you for clearing that up!

Aha, I see. I've read that part about "strictly textual" and "performed
before all other expansions" in the manual, but I didn't realize all the
consequences. This means that my quotes get interpreted *after* the
brace expansion is done, right? As a result, a call like

 echo "$(echo "{1..3}')'

ends up as

 echo "$(echo "1')' "$(echo "2')' "$(echo "3')'

and that surely won't work.

This makes a lot more sense now.


Thanks again,
Peter