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al0ne
CPU 利用率计算逻辑问题
https://github.com/al0ne/LinuxCheck/blob/bebca8ca774963526e2e39d59bce95373e7aa2c8/LinuxCheck.sh#L135-L140
计算逻辑问题: /proc/stat 文件中，CPU 时间是以计时单位（jiffies）提供的，各列分别代表不同状态的累计时间。不能简单地加和然后除来计算 CPU 使用率。导致这种计算值偏差太大。
下面的代码是 GPT 写的，供参考 :)
cpu_use() {
echo "CPU Core | Usage % | Idle %"
echo "---------|---------|-------"
# 读取 CPU 数据前的延迟
sleep_duration=0.1
# 第一次读取 /proc/stat 并获取初始值
read prev_idle prev_total < <(awk '/^cpu /{print $5, $2+$3+$4+$5+$6+$7+$8+$9+$10+$11}' /proc/stat)
sleep $sleep_duration
# 第二次读取 /proc/stat 并获取新值
read idle total < <(awk '/^cpu /{print $5, $2+$3+$4+$5+$6+$7+$8+$9+$10+$11}' /proc/stat)
# 计算差异和 CPU 使用率
idle_diff=$((idle - prev_idle))
total_diff=$((total - prev_total))
usage=$(awk -v total_diff="$total_diff" -v idle_diff="$idle_diff" 'BEGIN {printf "%.2f", (100 * (total_diff - idle_diff) / total_diff)}')
idle_rate=$(awk -v idle_diff="$idle_diff" -v total_diff="$total_diff" 'BEGIN {printf "%.2f", (100 * idle_diff / total_diff)}')
echo "Overall | $usage% | $idle_rate%"
# 对每个 CPU 核心进行相同的操作
cpu_count=$(grep -c '^processor' /proc/cpuinfo)
for ((i=0; i<cpu_count; i++)); do
read prev_idle prev_total < <(awk -v cpu_id="$i" '/^cpu[0-9]+/{if (NR == cpu_id+2) print $5, $2+$3+$4+$5+$6+$7+$8+$9+$10+$11}' /proc/stat)
sleep $sleep_duration
read idle total < <(awk -v cpu_id="$i" '/^cpu[0-9]+/{if (NR == cpu_id+2) print $5, $2+$3+$4+$5+$6+$7+$8+$9+$10+$11}' /proc/stat)
idle_diff=$((idle - prev_idle))
total_diff=$((total - prev_total))
usage=$(awk -v total_diff="$total_diff" -v idle_diff="$idle_diff" 'BEGIN {printf "%.2f", (100 * (total_diff - idle_diff) / total_diff)}')
idle_rate=$(awk -v idle_diff="$idle_diff" -v total_diff="$total_diff" 'BEGIN {printf "%.2f", (100 * idle_diff / total_diff)}')
echo "CPU$i | $usage% | $idle_rate%"
done
}
