mjhjmtu

Is there a way to close all the open file descriptors, without having an explicit list of them beforehand?

To answer literally, to close all open file descriptors for bash:

for fd in $(ls /proc/$$/fd); do
 eval "exec $fd>&-"
done

However this really isn't a good idea since it will close the basic file descriptors the shell needs for input and output. If you do this, none of the programs you run will have their output displayed on the terminal (unless they write to the tty device directly). If fact in my tests closing stdin (exec 0>&-) just causes an interactive shell to exit.

What you may actually be looking to do is rather to close all file descriptors that are not part of the shell's basic operation. These are 0 for stdin, 1 for stdout and 2 for stderr. On top of this some shells also seem to have other file descriptors open by default. In bash, for example, you have 255 (also for terminal I/O) and in dash I have 10, which points to /dev/tty rather than the specific tty/pts device the terminal is using. To close everything apart from 0, 1, 2 and 255 in bash:

for fd in $(ls /proc/$$/fd); do
 case "$fd" in
 0|1|2|255)
 ;;
 *)
 eval "exec $fd>&-"
 ;;
 esac
done

Note also that eval is required when redirecting the file descriptor contained in a variable, if not bash will expand the variable but consider it part of the command (in this case it would try to exec the command 0 or 1 or whichever file descriptor you are trying to close).

NOTE: Also using a glob instead of ls (eg /proc/$$/fd/*) seems to open an extra file descriptor for the glob, so ls seems the best solution here.

Update

For further information on the portability of /proc/$$/fd, please see Portability of file descriptor links. If /proc/$$/fd is unavailable, then a drop in replacement for the $(ls /proc/$$/fd), using lsof (if that is available) would be $(lsof -p $$ -Ff | grep f[0-9] | cut -c 2-).

In recent versions of Bash (4.1 and onward, year 2009 and later) you can specify the file descriptor to close using a shell variable:

for fd in $(ls /proc/$$/fd/); do
 [ $fd -gt 2 ] && exec fd<&-
done

The feature had been in Korn shell already (since 1993?) but apparently took some time to make its way into Bash.

Clear all file descriptors except i/o/e of the current shell, but also excludes the ones provided as arguments

clear_fds() 255)
 ;;
 *)
 eval "exec $fd>&-"
 ;;
 esac
 fi
done

No. The kernel can close only one FD at a time, and bash does not have "group commands" for FDs.

for fd in $(ls -1 /proc/27343/fd); do echo exec "$fd>&-"; done

Remove the echo after testing.

If this is not for the shell itself but for a command to be run then you can use nohup.

Another way without "eval" at all, is to use the form:

$ exec var_a>> file.txt
$ echo $var_a
10
$ ls -l /proc/self/fd/10
l-wx------ 1 0 0 64 Dec 11 18:32 /proc/self/fd/10 -> /run/user/0/tmp/file.txt
$ echo "aaaaa" >&$var_a
$ cat file.txt
aaaaa
$ exec var_a>&-
$ ls /proc/self/fd/10
ls: cannot access '/proc/self/fd/10': No such file or directory