Issue
I have this script, but I do not know how to get the last element in the printout:
cat /proc/cpuinfo | awk '/^processor/{print $3}'
The last element should be the number of CPUs, minus 1.
Solution
grep -c ^processor /proc/cpuinfo
will count the number of lines starting with "processor" in /proc/cpuinfo
For systems with hyper-threading, you can use
grep ^cpu\\scores /proc/cpuinfo | uniq | awk '{print $4}'
which should return (for example) 8
(whereas the command above would return 16
)
Answered By - unbeli