bash - Is this the faster way to test cpu load using shell scripting? -

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i'm relatively new shell scripting , i'm in process of writing own health checking scripts using bash.

is following script test cpu load best can have in terms of performance, readability , maintainability?

#!/bin/sh  getloadavg5 () {   echo $(cat /proc/loadavg | cut -f2 -d' ') }  getnumcpus () {   echo $(cat /proc/cpuinfo | grep '^processor' | wc -l) }  awk \   -v failthold=0.8 \   -v warnthold=0.7 \   -v loadavg=$(getloadavg5) \   -v numcpus=$(getnumcpus) \   'begin {     ratio=loadavg/numcpus     if (ratio >= failthold) exit 2     if (ratio >= warnthold) exit 1     exit 0   }'

this might more suitable code review stackexchange, without condoning use of load averages in way, here ideas:

#!/bin/sh read -r 1 5 fifteen rest < /proc/loadavg cpus=$(grep -c '^processor' /proc/cpuinfo)  awk \   -v failthold=0.8 \   -v warnthold=0.7 \   -v loadavg="$five" \   -v numcpus="$cpus" \   'begin {     ratio=loadavg/numcpus     if (ratio >= failthold) exit 2     if (ratio >= warnthold) exit 1     exit 0   }'

it doesn't have of unnecessary cats/echos.

it happens run faster forking 1 or 2 times (depending on shell) instead of ~10, if performance issue shell scripts should avoided in general.


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