In Bash, how to find the lowest-numbered unused file descriptor?

Question

In a Bash-script, is it possible to open a file on \"the lowest-numbered file descriptor not yet in use\"?

I have looked around for how to do this, but it seems that Bas

Answer 1

I needed to support both bash v3 on Mac and bash v4 on Linux and the other solutions require either bash v4 or Linux, so I came up with a solution that works for both, using /dev/fd.

find_unused_fd() {
  local max_fd=$(ulimit -n)
  local used_fds=" $(/bin/ls -1 /dev/fd | sed 's/.*\///' | tr '\012\015' '  ') "
  local i=0
  while [[ $i -lt $max_fd ]]; do
    if [[ ! $used_fds =~ " $i " ]]; then
      echo "$i"
      break
    fi
    (( i = i + 1 ))
  done
}

For example to dup stdout, you can do:

newfd=$(find_unused_fd)
eval "exec $newfd>&1"