如何正确杀死bash中的进程

Question

I use the following script to kill process by timeout:

# $1 - name of program and its command line

#launch program and remember PID
eval "$1" &
PID=$!

echo "Program '"$1"' started, PID="$PID

i=1
while [ $i -le 300 ]
do
 ps -p $PID >> /dev/null
 if [ $? -ne 0 ]
  then
   wait $PID
   exit $? #success, return rc of program
  fi

 i=$(($i+1))
 echo "waiting 1 second..."
 sleep 1
done

#program does not want to exit itself, kill it
echo "killing program..."
kill $PID
exit 1 #failed

So far, it have worked excellent, but today, i've noticed a bunch of 'hanging' processes in htop, so i've checked out and it turns out, that $PID in this case is not ID of the program process but of the script itself, and all the times i checked, ID of the program is $PID+1 . Now, the question is, am i correct to assume, that it will always be $PID+1 and i won't kill something important by replacing kill $PID with something like kill $PID $($PID+1)

EDIT: $1 may have several urguments, like ./bzip2 -ds sample3.bz2 -k

Answer 1

You can solve the problem simply with the following change:

From:

eval "$1" &

To:

eval "$1 &"

The reason is explained in this answer .

Answer 2

I just started writing a script with this functionality. I was going to call it "timeout" but before I opened a blank file, I checked to see if there was already a command with the same name. There was...

timeout

edit

If you need "1" specifically as a return value on failure...

timeout 1 nano -w; `if [[ $? == 124 ]] ; then exit 1 ; fi ; exit $?`

Answer 3

What's wrong with plain

( eval "$1" ) &
sleep 300
kill %1

Answer 4

You are backgrounding eval , not the command it runs, and eval is a shell built-in, so you are forking a new shell; that's why (I think) $! is the PID of the current shell.

One simple solution is to avoid using eval (for this and the usual concerns over security).

$1 "$@" &
PID=$!

True, this doesn't allow you to pass an arbitrary bash command line (pipeline, && list, etc) to your script, but your use case may not need to support such generalization. What commands do you typically pass?

Answer 5

Also, here is some refactoring to your code, maybe you will learn something from it:

#launch program and remember PID
eval "$1" &
PID=$!

echo "Program '$1' started, PID=$PID" # you can safely use single quotes inside double quotes, your variables are going to work in  " " as well!

i=1
while (( i <= 300 )) # use (( )) for math operations!
do
    ps -p "$PID" >> /dev/null # it is a good rule to quote every variable, even if you're pretty sure that it doesn't contain spaces
    if [[ $? != 0 ]]; then # Try to use [[ ]] instead of [. It is modern bash syntax
        wait "$PID"
        exit "$?" #success, return rc of program
    fi
    ((i++))
    echo "waiting 1 second..."
    sleep 1
done

#program does not want to exit itself, kill it
echo "killing program..."
kill "$PID"
exit 1 #failed