Avoiding busy waiting in bash, without the sleep command

In newer versions of bash (at least v2), builtins may be loaded (via enable -f filename commandname) at runtime. A number of such loadable builtins is also distributed with the bash sources, and sleep is among them. Availability may differ from OS to OS (and even machine to machine), of course. For example, on openSUSE, these builtins are distributed via the package bash-loadables.

Edit: fix package name, add minimum bash version.

Creating a lot of subprocesses is a bad thing in an inner loop. Creating one sleep process per second is OK. There's nothing wrong with

while ! test_condition; do
  sleep 1
done

If you really want to avoid the external process, you don't need to keep the fifo open.

my_tmpdir=$(mktemp -d)
trap 'rm -rf "$my_tmpdir"' 0
mkfifo "$my_tmpdir/f"

while ! test_condition; do
  read -t 1 <>"$my_tmpdir/f"
done

I recently had a need to do this. I came up with the following function that will allow bash to sleep forever without calling any external program:

snore()
{
    local IFS
    [[ -n "${_snore_fd:-}" ]] || { exec {_snore_fd}<> <(:); } 2>/dev/null ||
    {
        # workaround for MacOS and similar systems
        local fifo
        fifo=$(mktemp -u)
        mkfifo -m 700 "$fifo"
        exec {_snore_fd}<>"$fifo"
        rm "$fifo"
    }
    read ${1:+-t "$1"} -u $_snore_fd || :
}

NOTE: I previously posted a version of this that would open and close the file descriptor each time, but I found that on some systems doing this hundreds of times a second would eventually lock up. Thus the new solution keeps the file descriptor between calls to the function. Bash will clean it up on exit anyway.

This can be called just like /bin/sleep, and it will sleep for the requested time. Called without parameters, it will hang forever.

snore 0.1  # sleeps for 0.1 seconds
snore 10   # sleeps for 10 seconds
snore      # sleeps forever

There's a writeup with excessive details on my blog here