Bash – Timeout a command in bash without unnecessary delay

bashcommand-linetimeoututilities

This answer to Command line command to auto-kill a command after a certain amount of time

proposes a 1-line method to timeout a long-running command from the bash command line:

( /path/to/slow command with options ) & sleep 5 ; kill $!

But it's possible that a given "long-running" command may finish earlier than the timeout.
(Let's call it a "typically-long-running-but-sometimes-fast" command, or tlrbsf for fun.)

So this nifty 1-liner approach has a couple of problems.
First, the sleep isn't conditional, so that sets an undesirable lower bound on the time taken for the sequence to finish. Consider 30s or 2m or even 5m for the sleep, when the tlrbsf command finishes in 2 seconds — highly undesirable.
Second, the kill is unconditional, so this sequence will attempt to kill a non-running process and whine about it.

So…

Is there a way to timeout a typically-long-running-but-sometimes-fast ("tlrbsf") command that

  • has a bash implementation (the other question already has Perl and C answers)
  • will terminate at the earlier of the two: tlrbsf program termination, or timeout elapsed
  • will not kill non-existing/non-running processes (or, optionally: will not complain about a bad kill)
  • doesn't have to be a 1-liner
  • can run under Cygwin or Linux

… and, for bonus points

  • runs the tlrbsf command in the foreground
  • any 'sleep' or extra process in the background

such that the stdin/stdout/stderr of the tlrbsf command can be redirected, same as if it had been run directly?

If so, please share your code. If not, please explain why.

I have spent awhile trying to hack the aforementioned example but I'm hitting the limit of my bash skills.

Best Solution

You are probably looking for the timeout command in coreutils. Since it's a part of coreutils, it is technically a C solution, but it's still coreutils. info timeout for more details. Here's an example:

timeout 5 /path/to/slow/command with options