防止bash脚本挂起

Question

Suppose I have bash-script with following code:

function test() {
  some_code
  ...
  make
  some_code
}

test
some_other_code

test() could contain any code that might run unreasonably long.

I was trying to use something like:

function test() {
  cd $WORK_FOLDER
  make
}

run_timeout()
{
local timeout=$1
$2 &
local pid=$!
while ps $pid >/dev/null && [ $timeout -ne 0 ]; do
  sleep 1
  let timeout--  
done  
kill -9 $pid 2>/dev/null && echo "Process $pid killed because executed too long"
}

run_timeout 15 "test"
run_timeout 5 "test"

But make was still running after the estimated time.

Any suggestion how to solve this problem?

Is there any technique that prevents a bash script from hanging?

Answer 1

I guess the $2 & is where you run the long function right? I had the same problem and some time, depending on what is in the function you will have multiple process ... I don t know if my solution is the best way to do it, but it worked for me.

change :

$2 &

for :

awk '{system($2)}' &

this way, pid =$! will give you the awk pid and by killing the awk, you kill the whole process thing.