Distribute NIC interrupts evenly among the first X cores

interrupt.sh

#!/bin/bash

mask() {
    local bytes=$((($2+3)/4))
    local remaining=$((bytes%8))
    local groups=$((bytes/8))
    local s=$(printf "%0${bytes}x" $((1<<$1)))
    local rc=${s::$remaining}
    s=${s:$remaining}
    while [ -n "$s" ]; do
        rc="$rc,${s::8}"
        s=${s:8}
    done
    echo $rc
}

if [ $1 == "show" ]; then
    for irq in $(grep -P 'enp3s0f(0|1)-' /proc/interrupts | awk '{print $1}' | tr -d ':' | sort -n | xargs); do
        echo $irq $(cat /proc/irq/$irq/smp_affinity)
    done
    exit 0
fi

ncpu=$(lscpu | grep -P '^CPU\(s\):' | awk '{print $2}')
inuse=${1:-$ncpu}

if [ $inuse -gt $ncpu ]; then
    echo "you've at most $ncpu CPUs, using $ncpu"
    inuse=$ncpu
fi

cpuid=0
for irq in $(grep -P 'enp3s0f(0|1)-' /proc/interrupts | awk '{print $1}' | tr -d ':' | sort -n | xargs); do
    mask $cpuid $ncpu >/proc/irq/$irq/smp_affinity
    cpuid=$(((cpuid+1)%inuse))
done