With the script below, I'd expect any fd pointing to /dev/null to be closed
when the second llfd() is executed. Surprisingly, fd 3 is closed, but fd 10 is
now open, pointing to /dev/null, as if eval copied it instead of closing it. Is
this a bug?
Thanks,
M
$ bash -c 'llfd () { ls -l /proc/$BASHPID/fd/; }; x=3; eval "exec
$x>/dev/null"; llfd; eval "llfd $x>&-"'
total 0
lrwx------ 1 matei matei 64 Feb 11 18:36 0 -> /dev/pts/2
lrwx------ 1 matei matei 64 Feb 11 18:36 1 -> /dev/pts/2
lrwx------ 1 matei matei 64 Feb 11 18:36 2 -> /dev/pts/2
l-wx------ 1 matei matei 64 Feb 11 18:36 3 -> /dev/null
lr-x------ 1 matei matei 64 Feb 11 18:36 8 -> /proc/4520/auxv
total 0
lrwx------ 1 matei matei 64 Feb 11 18:36 0 -> /dev/pts/2
lrwx------ 1 matei matei 64 Feb 11 18:36 1 -> /dev/pts/2
l-wx------ 1 matei matei 64 Feb 11 18:36 10 -> /dev/null
lrwx------ 1 matei matei 64 Feb 11 18:36 2 -> /dev/pts/2
lr-x------ 1 matei matei 64 Feb 11 18:36 8 -> /proc/4520/auxv
$ bash --version
GNU bash, version 4.2.24(1)-release (x86_64-pc-linux-gnu)
Copyright (C) 2011 Free Software Foundation, Inc.
License GPLv3+: GNU GPL version 3 or later <http://gnu.org/licenses/gpl.html>
This is free software; you are free to change and redistribute it.
There is NO WARRANTY, to the extent permitted by law.
$
