Bash wait for process start

There are multiple issues with your code:

1. pidof doesn't print 0 or -1 if there's no process, so your logic is wrong
2. pidof can return multiple pids if there are multiple processes, which would break your comparison
3. > has to be escaped in [ ] otherwise your code is equivalent to [ $notepad ] > 0, directing to a file.
4. > isn't even the right operator. You wanted -gt, but as mentioned in point #1 and #2, you shouldn't compare numbers.
5. until runs the loop until the condition is true. it doesn't wait for the condition to become true, and then run the loop.

This is how you should do it:

# Wait for notepad to start until pids=$(pidof notepad) do sleep 1 done # Notepad has now started. # There could be multiple notepad processes, so loop over the pids for pid in $pids do taskset -p 03 $pid renice -n 5 -p $pid sleep 5 renice -n 0 -p $pid done # The notepad process(es) have now been handled 

If I understand correctly, you want to wait until notepad is launched. Then the pidof must be inside your loop.

while ! pidof notepad.exe >> /dev/null ; do sleep 1 done notepad=$(pidof notepad.exe) taskset -p 03 $notepad renice -n 5 -p $notepad sleep 5 renice -n 0 -p $notepad 

To answer your first question: '>' is not the mathematical/comparison operator 'greater than'; it's bash's way of letting you pipe output to a file (handle).

echo "some text" > myfile.txt 

will create a file named myfile.txt, and '>' sent the output of the command into that file. I would imagine that your file '0' has the pid in it, and nothing else.

Instead, try -gt (or related variants: -ge, -lt, -le, -eg, -ne) to test if one value is greater than (or: greater than or equal, less than, less than or equal, equal, not equal, respectively) to see if that helps. That is,

until [ $notepad -gt 0 ]