[SOLVED] CS代考计算机代写 chain /* The main thread set its priority to PRI_MIN and creates 7 threads

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/* The main thread set its priority to PRI_MIN and creates 7 threads
(thread 1..7) with priorities PRI_MIN + 3, 6, 9, 12, …
The main thread initializes 8 locks: lock 0..7 and acquires lock 0.

When thread[i] starts, it first acquires lock[i] (unless i == 7.)
Subsequently, thread[i] attempts to acquire lock[i-1], which is held by
thread[i-1], except for lock[0], which is held by the main thread.
Because the lock is held, thread[i] donates its priority to thread[i-1],
which donates to thread[i-2], and so on until the main thread
receives the donation.

After threads[1..7] have been created and are blocked on locks[0..7],
the main thread releases lock[0], unblocking thread[1], and being
preempted by it.
Thread[1] then completes acquiring lock[0], then releases lock[0],
then releases lock[1], unblocking thread[2], etc.
Thread[7] finally acquires & releases lock[7] and exits, allowing
thread[6], then thread[5] etc. to run and exit until finally the
main thread exits.

In addition, interloper threads are created at priority levels
p = PRI_MIN + 2, 5, 8, 11, … which should not be run until the
corresponding thread with priority p + 1 has finished.

Written by Godmar Back */

#include
#include “tests/threads/tests.h”
#include “threads/init.h”
#include “threads/lock.h”
#include “threads/thread.h”

#define NESTING_DEPTH 8

struct lock_pair {
struct lock *second;
struct lock *first;
};

static thread_func donor_thread_func;
static thread_func interloper_thread_func;

void
test_priority_donate_chain(void)
{
int i;
struct lock locks[NESTING_DEPTH – 1];
struct lock_pair lock_pairs[NESTING_DEPTH];

/* This test does not work with the MLFQS. */
ASSERT(!thread_mlfqs);

thread_set_priority(PRI_MIN);

for (i = 0; i < NESTING_DEPTH – 1; i++)lock_init(&locks[i]);lock_acquire(&locks[0]);msg(“%s got lock.”, thread_name());for (i = 1; i < NESTING_DEPTH; i++) {char name[16];int thread_priority;snprintf(name, sizeof name, “thread %d”, i);thread_priority = PRI_MIN + i * 3;lock_pairs[i].first = i < NESTING_DEPTH – 1 ? locks + i : NULL;lock_pairs[i].second = locks + i – 1;thread_create(name, thread_priority, donor_thread_func, lock_pairs + i);msg(“%s should have priority %d.Actual priority: %d.”,thread_name(), thread_priority, thread_get_priority());snprintf(name, sizeof name, “interloper %d”, i);thread_create(name, thread_priority – 1, interloper_thread_func, NULL);}lock_release(&locks[0]);msg(“%s finishing with priority %d.”, thread_name(), thread_get_priority());}static voiddonor_thread_func(void *locks_){struct lock_pair *locks = locks_;if (locks->first)
lock_acquire(locks->first);

lock_acquire(locks->second);
msg(“%s got lock”, thread_name());

lock_release(locks->second);
msg(“%s should have priority %d. Actual priority: %d”,
thread_name(), (NESTING_DEPTH – 1) * 3,
thread_get_priority());

if (locks->first)
lock_release(locks->first);

msg(“%s finishing with priority %d.”, thread_name(),
thread_get_priority());
}

static void
interloper_thread_func(void *arg_ UNUSED)
{
msg(“%s finished.”, thread_name());
}

// vim: sw=2

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[SOLVED] CS代考计算机代写 chain /* The main thread set its priority to PRI_MIN and creates 7 threads
30 $