根据撰写本文时的htop源代码，我的假设看起来是正确的：
（请参阅ProcessList.c中的
voidProcessList_scan(ProcessList*this)函数）
// Guest time is already accounted in usertimeusertime = usertime - guest; # As you see here, it subtracts guest from user timenicetime = nicetime - guestnice; # and guest_nice from nice time// Fields existing on kernels >= 2.6// (and RHEL's patched kernel 2.4...)idlealltime = idletime + ioWait; # ioWait is added in the idleTimesystemalltime = systemtime + irq + softIrq;virtalltime = guest + guestnice;totaltime = usertime + nicetime + systemalltime + idlealltime + steal + virtalltime;
因此，从
/proc/stat：第一行中列出的字段中（请参阅文档中的
1.8节）
user nice system idle iowait irq softirq steal guest guest_nicecpu 74608 2520 24433 1117073 6176 4054 0 0 0 0
通过算法，我们可以计算出CPU使用率百分比，如下所示：
PrevIdle = previdle + previowaitIdle = idle + iowaitPrevNonIdle = prevuser + prevnice + prevsystem + previrq + prevsoftirq + prevstealNonIdle = user + nice + system + irq + softirq + stealPrevTotal = PrevIdle + PrevNonIdleTotal = Idle + NonIdle# differentiate: actual value minus the previous onetotald = Total - PrevTotalidled = Idle - PrevIdleCPU_Percentage = (totald - idled)/totald
