Execute a shell function with timeout

Question

Why would this work

timeout 10s echo \"foo bar\" # foo bar

but this wouldn\'t

function echoFooBar {
  echo \"foo bar\"
}

e         


        

Answer 1

This function uses only builtins

• Maybe consider evaling "$*" instead of running $@ directly depending on your needs

• It starts a job with the command string specified after the first arg that is the timeout value and monitors the job pid

• It checks every 1 seconds, bash supports timeouts down to 0.01 so that can be tweaked

• Also if your script needs stdin, read should rely on a dedicated fd (exec {tofd}<> <(:))

• Also you might want to tweak the kill signal (the one inside the loop) which is default to -15, you might want -9

## forking is evil
timeout() {
    to=$1; shift
    $@ & local wp=$! start=0
     while kill -0 $wp; do
        read -t 1
        start=$((start+1))
        if [ $start -ge $to ]; then
            kill $wp && break
        fi
    done
}