I have two binary files that I want to execute one after the other, the thing is that I want to execute both for one minute. I have the following bash code:
./file_1
./file_2
but I do not know how to only run it for a minute.
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2 Answers
Here's a portable solution - run the binaries in the background, and kill them after a minute:
for file in "file_1" "file_2"; do
"./$file" &
file_pid=$!
sleep 60
kill "$file_pid"
done
The & operator starts a background job, and the special variable $! contains the PID of the last job. The loop is optional. We can use it to reduce duplicated code.
Comments
This is a one-liner using the timeout command
$ timeout 60 binary-1; timeout 60 binary-2
In this case, the format is:
timeout duration command
Where duration defaults to seconds, or also could be used 1m (1 minutes), from the man:
duration is a floating point number followed by an optional unit:
‘s’ for seconds (the default)
‘m’ for minutes
‘h’ for hours
‘d’ for days
