如何知道Linux调度时间片? [英] How to know linux scheduler time slice?
问题描述
我在寻找我的Linux内核的时间片(或量子)的值。
I'm looking for the value of the time slice (or quantum) of my Linux kernel.
有没有这暴露出这样一个信息的/ proc
文件?
Is there a /proc
file which expose such an information ?
(或者)是我的分布的Linux头它明确定义?
(Or) Is it well-defined in the Linux header of my distributions ?
(或)是否有Linux的API(也许SYSINFO),揭露出这个值的C函数?
(Or) Is there a C function of the Linux API (maybe sysinfo) that expose this value ?
先谢谢了。
推荐答案
默认的Linux时间片的实时进程是在Linux内核为 RR_TIMESLICE
中的在include / linux /排程/ rt.h 。
The default Linux timeslice for realtime processes is defined in the Linux kernel as RR_TIMESLICE
in include/linux/sched/rt.h.
/*
* default timeslice is 100 msecs (used only for SCHED_RR tasks).
* Timeslices get refilled after they expire.
*/
#define RR_TIMESLICE (100 * HZ / 1000)
需要注意的是分配给特定进程实际的量子可能会有所不同比这个值:
您可以调整切片,通过调整 sched_latency_ns 并
sched_min_granularity_ns ,但请注意,切片是不是一个固定的
量子。还注意到,CFS preemption决定是根据
即时状态。任务可能已经收到一个完整的(变量)
CPU时间的切片,但preemption将被触发仅当更
值得的任务是可用的,所以一个切片是不是最大
不间断的CPU时间,你可能会想到它是..但它是
有些相似。
You can tune "slice" by adjusting sched_latency_ns and sched_min_granularity_ns, but note that "slice" is not a fixed quantum. Also note that CFS preemption decisions are based upon instantaneous state. A task may have received a full (variable) "slice" of CPU time, but preemption will be triggered only if a more deserving task is available, so a "slice" is not the "max uninterrupted CPU time" that you may expect it to be.. but it is somewhat similar.
不过,您可以使用 sched_rr_get_interval()
得到SCHED_RR间隔为给定的实时过程。
However, you can use sched_rr_get_interval()
to get the SCHED_RR interval for a given realtime process.
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