Shell Programming and Scripting

Code:

#!/bin/sh
exec 0</dev/null
exec 1>/dev/null

ls -l /proc/self/fd >&2

produces

Code:

total 0
lr-x------ 1 tyler users 64 Feb 18 10:38 0 -> /proc/7886/fd
lrwx------ 1 tyler users 64 Feb 18 10:38 1 -> /dev/pts/4
lrwx------ 1 tyler users 64 Feb 18 10:38 2 -> /dev/pts/4

I've verified the shell is actually redirecting stdout as expected, but when I run ls it is not inheriting this change. If I leave out the >&2, it produces no output, meaning it's only re-redirecting stdout to my console when I don't redirect stdout into stderr -- in other words, it only redirects stdout back to the console when I tell it not to use stdout at all! Why is it restoring stdout at all?

ls -l /proc/self/fd >&2
You are redirecting stdout (what ls is producing) into stderr (&2).

Unless you tell the script to dump stderr to /dev/null...:

Code:

./myscript 2>/dev/null

...it will be output to the shell you are running the script in.

Leaving the >&2 out of your script sends the output of ls to the default stdout (unless ls spits errors.. to stderr..)

Quite so. I have also already redirected /dev/null into stdin, and stdout into /dev/null. I don't want to redirect stderr, and have already redirected stdin and stdout to /dev/null. I expected ls to inherit stdin and stdout from the shell, but instead the shell for some copies stderr into stdout -- but only when I redirect ls's stdout. It doesn't just do this for terminals, either -- I've seen it do so for files, so I boiled the problem down to its minimum representation here. Which is /dev/null, of course, given the prior redirections. But when I redirect it into stderr, the shell takes the extraordinary step of restoring stdout to its previous state.

Code:

I don't want to redirect stderr, and have already redirected stdin and stdout to /dev/null.  I expected ls to inherit stdin and stdout from the shell

Yes but you haven't redirected stderr, which is why when to tell ls's stdout to redirect to stderr, you are getting output to your shell.

If you were to:

Code:

#!/bin/sh
exec 0>/dev/null
exec 1>/dev/null
exec 2>/dev/null

ls -l /proc/self/fd >&2

then your stderr would be redirected to /dev/null.. in which case you would get no output.

I'm still confused at what you are trying to do...??

I'm not really following this.

The default location for standard error is the screen. So it stands to reason that by redirecting ls to stderr (which you haven't redirected elsewhere) the output of ls will go to the screen.

In any case, is

Code:

exec 0</dev/null
exec 1>/dev/null

the same as

Code:

exec 0<&-
exec 1>&-

Code:

I'm not really following this.

I'm not either..
Can you tell me why you would pipe /dev/null into stdin?

Also,

Code:

exec 0<&-

What is this?

It closes standard input.