XV6 中的 MLFQ 和 RR

Question

我试图了解 XV6 中的 MLFQ 和 RR 是如何工作的。我见过的一个实现如下：

void rr_scheduler(void){
struct proc *p;
struct cpu *c = mycpu();

c->proc = 0;
intr_on();

for (p = proc; p < &proc[NPROC]; p++)
{
    acquire(&p->lock);
    if (p->state == RUNNABLE)
    {
        // Switch to chosen process.  It is the process's job
        // to release its lock and then reacquire it
        // before jumping back to us.
        p->state = RUNNING;
        c->proc = p;
        swtch(&c->context, &p->context);

        // Process is done running for now.
        // It should have changed its p->state before coming back.
        c->proc = 0;
    }
    release(&p->lock);
}
void mlfq_scheduler(void){
struct proc *p;
struct cpu *c = mycpu();

c->proc = 0;
intr_on();

char high_avail = 0;
do
{
    high_avail = 0;
    for (p = proc; p < &proc[NPROC]; p++)
    {
        acquire(&p->lock);
        if (p->priority > 0)
        {

            release(&p->lock);
            continue;
        }

        if (p->state == RUNNABLE)
        {
            high_avail = 1;
            // Switch to chosen process.  It is the process's job
            // to release its lock and then reacquire it
            // before jumping back to us.
            p->state = RUNNING;
            c->proc = p;
            swtch(&c->context, &p->context);
            // check if we are still the right scheduler
            if (sched_pointer != &mlfq_scheduler)
            {
                release(&p->lock);
                return;
            }

            // Process is done running for now.
            // It should have changed its p->state before coming back.
            c->proc = 0;
        }
        release(&p->lock);
    }
} while (high_avail);

//  RR on low prio - break when high prio found
for (p = proc; p < &proc[NPROC]; p++)
{
    acquire(&p->lock);
    if (p->priority == 0 && p->state == RUNNABLE)
    {
        // found high prio - switch to high prio task
        release(&p->lock);
        break;
    }

    if (p->state == RUNNABLE)
    {
        // Switch to chosen process.  It is the process's job
        // to release its lock and then reacquire it
        // before jumping back to us.
        p->state = RUNNING;
        c->proc = p;
        swtch(&c->context, &p->context);
        // check if we are still the right scheduler
        if (sched_pointer != &mlfq_scheduler)
        {
            release(&p->lock);
            return;
        }

        // Process is done running for now.
        // It should have changed its p->state before coming back.
        c->proc = 0;
    }
    release(&p->lock);
}
}

我对 C 不太熟悉，但确实理解提供的代码。但是我不明白如何使用 MLFQ 更改优先级以及进程如何运行指定的时间直到 RR 中的切换。如果我认为这是在 trap.c 文件中实现的中断，那么我是否完全关闭了？在这种情况下，任何人都可以回答这是如何在 trap.c 文件中实现的吗？

感谢您的提前答复。

Answer 1

如果 MLFQ 正确实现，一些逻辑应该处理陷阱并更新进程的优先级。这可能位于

usertrap

中的

trap.c

内部，或者位于

yield

中的

proc.c

函数内部。前一个函数应该调用后者，并将进程产生的原因作为参数，这可以用于编辑优先级，例如，如果时间片已用完。