What does the following Bash script mean/do?

Question

What does the following script mean?

exec 4<&0 0</etc/XX/cfg
read line1
exec 0<&4

UPDATE

Thanks for the answer, just one part I'm not sure:

It redirects fd0 to fd4, and "/etc/XX/cfg" to fd0.

So why does read still work, shouldn't that be empty?

Answer 1

It redirects stdin (FD0) to FD4, redirects from /etc/XX/cfg to FD0, reads a line from FD0, and then moves FD4 back to FD0. In short, it saves, replaces, and restores stdin, while reading a line from a file in between.

read line1 < /etc/XX/cfg would be so much easier, but it's impossible to tell if it's a valid replacement based only on the code shown.

Answer 2

To recast this in syscalls (using C):

exec 4<&0 0</etc/XX/cfg

/* Duplicate fd0 as fd4.  */
dup2 (0, 4);

/* Open file on fd0.
   "open" always uses lowest available descriptor, so we don't need to check it. */
close (0);
open ("/etc/XX/cfg", O_RDONLY);

exec 0<&4

/* Close fd0 and duplicate fd4 as fd0.  */
dup2 (4, 0);

Answer 3

The way I read this - based on the bash manpage section on REDIRECTION - is that stdin (fd0) is redirected into fd4, then input is taken from /etc/XX/cfg into stdin - which will end up on fd4.

read line1 should then take from fd4 which is then put back to fd0.

A demo might show my meaning a little better and answer "why" you might do this. By adding another line and putting it in a wrapper:

$ vim ./test.sh
#!/bin/bash
exec 4<&0 0</etc/XX/cfg
read line1    # reads from fd4
exec 0<&4
read line2    # reads from fd0

echo $line1 
echo $line2

You can pipe or redirect stdin via test.sh but also read in configuration by redirection, so in the code above I've pulled in values from the "config" (assumption on my part based on the filename - bad :) ), but I can process via stdin as well.

e.g:

$ ./test.sh < somefile
$ cat somefile | ./test.sh

Hopefully this explains it.