一道经常考的面试题

fox_lin · 2019-09-06 10:02:56

runtime2.go 中关于 p 的定义: 其中 runnext 指针决定了下一个要运行的 g,根据英文的注释大致意思是说:

  • 1 . 如果 runnext 不为空且是就绪的,那么应该优先运行该 runnext 指向的 g 而不是运行队列中到达运行时间的 g。
  • 2 . 此时要被运行的 g 将继承当前 g 剩余的时间片继续执行。
  • 3 . 之所以这样做是为了消除将最后的 g 添加到队列末尾导致的潜在的调度延迟。

所以当设置 runtime.GOMAXPROCS(1) 时,此时只有一个 P,创建的 g 依次加入 P, 当最后一个即 i==9 时,加入的最后 一个 g 将会继承当前主 goroutinue 的剩余时间片继续执行,所以会先输出 9, 之后再依次执行 P 队列中其它的 g。

个人看源码找到的答案,如有错误请立刻指出。

type p struct {
    lock mutex

    id          int32
    status      uint32 // one of pidle/prunning/...
    link        puintptr
    schedtick   uint32     // incremented on every scheduler call
    syscalltick uint32     // incremented on every system call
    sysmontick  sysmontick // last tick observed by sysmon
    m           muintptr   // back-link to associated m (nil if idle)
    mcache      *mcache
    racectx     uintptr

    deferpool    [5][]*_defer // pool of available defer structs of different sizes (see panic.go)
    deferpoolbuf [5][32]*_defer

    // Cache of goroutine ids, amortizes accesses to runtime·sched.goidgen.
    goidcache    uint64
    goidcacheend uint64

    // Queue of runnable goroutines. Accessed without lock.
    runqhead uint32
    runqtail uint32
    runq     [256]guintptr
    // runnext, if non-nil, is a runnable G that was ready'd by
    // the current G and should be run next instead of what's in
    // runq if there's time remaining in the running G's time
    // slice. It will inherit the time left in the current time
    // slice. If a set of goroutines is locked in a
    // communicate-and-wait pattern, this schedules that set as a
    // unit and eliminates the (potentially large) scheduling
    // latency that otherwise arises from adding the ready'd
    // goroutines to the end of the run queue.
    runnext guintptr
        ...
}
#4
更多评论

“每个goroutine生成后,在P的本地G队列未满的时候,是依次加入到P的本地G队列里的” 这句话中的P、G含义?

#1

P是逻辑处理器P0 G是程序中开的协程吧?

#2