Redirecting only stderr to a pipe

Neither ls nor grep need anything open on their fd 3 (they don't use that fd), so it's better practice to close it (release the resources you don't need). We only use that fd 3 to be able to restore ls stdout to the original one (before executing ls).

Remember that in a pipe line, all commands run concurrently in their own process. Redirections apply for each separately, so closing the fd 3 for one cannot close it for the other one.

It won't make much difference in practice in this case to close or not to close other than you might reach a limit on the number of file descriptors sooner if you don't.

In other cases (commands themselves starting other processes for instance), that may hold resources and cause problems. Think for instance if stdout is a pipe for instance, a background process may end up inheriting that fd, preventing any reader from that pipe to see EOF until that process has returned.

A better way to write it would be:

{ ls -l 2>&1 >&3 3>&- | grep bad 3>&-; } 3>&1 

That way, the fd 3 is only redirected temporarily for the duration of that command group (and restored afterwards or closed).

See the difference:

$ { ls -l /proc/self/fd 2>&1 >&3 3>&- | grep bad 3>&-; } 3>&1 total 0 lrwx------ 1 stephane stephane 64 Apr 2 09:29 0 -> /dev/pts/4 lrwx------ 1 stephane stephane 64 Apr 2 09:29 1 -> /dev/pts/4 l-wx------ 1 stephane stephane 64 Apr 2 09:29 2 -> pipe:[575886] lr-x------ 1 stephane stephane 64 Apr 2 09:29 3 -> /proc/20918/fd/ $ { ls -l /proc/self/fd 2>&1 >&3 | grep bad 3>&-; } 3>&1 total 0 lrwx------ 1 stephane stephane 64 Apr 2 09:29 0 -> /dev/pts/4 lrwx------ 1 stephane stephane 64 Apr 2 09:29 1 -> /dev/pts/4 l-wx------ 1 stephane stephane 64 Apr 2 09:29 2 -> pipe:[575900] lrwx------ 1 stephane stephane 64 Apr 2 09:29 3 -> /dev/pts/4 lr-x------ 1 stephane stephane 64 Apr 2 09:29 4 -> /proc/20926/fd/ 

In the second invocation, ls had its fd 3 also open to the terminal (which was open on stdout at the time I ran the pipeline) for no good reason.

Note that in ksh93, with exec, you don't need to close those fds as fds other than 0, 1, 2 are automatically closed upon executing a command.

$ ksh93 -c 'exec 3>&1; ls -l /dev/fd/' total 0 lrwx------ 1 stephane stephane 64 Apr 2 09:34 0 -> /dev/pts/16 lrwx------ 1 stephane stephane 64 Apr 2 09:34 1 -> /dev/pts/16 lrwx------ 1 stephane stephane 64 Apr 2 09:34 2 -> /dev/pts/16 lr-x------ 1 stephane stephane 64 Apr 2 09:34 3 -> /proc/21105/fd $ ksh93 -c 'exec 3>&1; ls -l /dev/fd/ 3>&3' total 0 lrwx------ 1 stephane stephane 64 Apr 2 09:34 0 -> /dev/pts/16 lrwx------ 1 stephane stephane 64 Apr 2 09:34 1 -> /dev/pts/16 lrwx------ 1 stephane stephane 64 Apr 2 09:34 2 -> /dev/pts/16 lrwx------ 1 stephane stephane 64 Apr 2 09:34 3 -> /dev/pts/16 lr-x------ 1 stephane stephane 64 Apr 2 09:34 4 -> /proc/21108/fd 

I don't know why it says (but not 'ls') above, sounds like a mistake (possibly mine ;-).

Close fd3 twice because you fork two subshell in script, each subshell is inherited and copy parent's file descriptor. Close fd3 in each subshell doesn't affect others (and the parent, too).

So the comment line is very unclear and causes misleading.

If you want to redirect only stder to pipe, you can use process substitution:

ls -l 2> >(grep bad) 

or swap stderr and stdout:

ls -l 3>&1 1>&2 2>&3 | grep bad 

For anyone else arriving here because of the question summary, as opposed understanding the code snippet from "Advanced Bash Scripting Guide", I found the other answers hard to reason through. The following approach worked for my purposes and I find it easier to read and understand its intention:

$ ls -l 2>&1 1>&- | grep bad 

To me this says that I want to ignore stdout by closing it, then I want to redirect stderr to stdout so I can pipe it onto the next command.