Question

使用以下代码通过监视器解决餐饮哲学家问题。我不明白为什么test（）方法需要

self[i].signal();

我理解的方式是，当一个人存在其关键部分时，就会调用信号，即。当哲学家吃完之前，而不是之前。在这种情况下，如果一个人调用resource.pickup（）然后将philosopher i设置为hungry，则调用test（）以查看BOTH chopstick现在是否可用（这与信号量解决方案相比是主要优势）。然而，在test（）中，如果两个邻居都没有进食而且我饿了，请拨打signal（）??！我理解的方式是，signal（）增加了资源计数器，并在一个人离开它的关键部分时被调用（我们在putdown（）中看到，我不明白为什么它在这里被称为

 // Dining-Philosophers Solution Using Monitors

monitor DP
{
    status state[5];
    condition self[5];

    // Pickup chopsticks
    Pickup(int i)
    {
        // indicate that I’m hungry
        state[i] = hungry;

        // set state to eating in test()
        // only if my left and right neighbors 
        // are not eating
        test(i);

        // if unable to eat, wait to be signaled
        if (state[i] != eating)
            self[i].wait;
    }

    // Put down chopsticks
    Putdown(int i)
    {

        // indicate that I’m thinking
        state[i] = thinking;

        // if right neighbor R=(i+1)%5 is hungry and
        // both of R’s neighbors are not eating,
        // set R’s state to eating and wake it up by 
        // signaling R’s CV
        test((i + 1) % 5);
        test((i + 4) % 5);
    }

    test(int i)
    {

        if (state[(i + 1) % 5] != eating
            && state[(i + 4) % 5] != eating
            && state[i] == hungry) {

            // indicate that I’m eating
            state[i] = eating;

            // signal() has no effect during Pickup(),
            // but is important to wake up waiting
            // hungry philosophers during Putdown()
            self[i].signal();
        }
    }

    init()
    {

        // Execution of Pickup(), Putdown() and test()
        // are all mutually exclusive,
        // i.e. only one at a time can be executing
for
    i = 0 to 4

        // Verify that this monitor-based solution is
        // deadlock free and mutually exclusive in that
        // no 2 neighbors can eat simultaneously
        state[i] = thinking;
    }
} // end of monitor

用餐的哲学家与监视器

0 个答案: