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Question

这个问题已经有答案了:

Semaphore solution to reader-writer: order between updating reader count and waiting or signaling on read/write binary semaphore? (1 个回答)

已关闭 5 年前。

我正在练习readers and writers problem并提出以下解决方案。但是，在打印出以下结果后，程序会自行阻塞。

$ Read content: planets

我的想法是 first solution 的修改版本。使用 2 个信号量，一个称为 readMutex，确保一次只有一个读取线程更新 numOfReaders，而另一个信号量称为 accessToResource，确保当读取器正在阅读内容，作者应该等待。这是我的代码。

import java.util.concurrent.Semaphore;

public class ReadersAndWriters {
    public static final Semaphore accessToResource = new Semaphore(1, true);
    public static final Semaphore readMutex = new Semaphore(1, true);
    public static String content = "planets";
    public static int numOfReaders;

    static class Reader extends Thread {
        void read() {
            try {
                readMutex.acquire();
                numOfReaders++;
                if (numOfReaders == 1) {
                    accessToResource.acquire();
                }
                readMutex.release();

                // read content, not a critical section
                System.out.println("Read content:\t" + content);

                readMutex.acquire();
                numOfReaders--;
                if (numOfReaders <= 0) {
                    accessToResource.release();
                }
                readMutex.release();
            } 
            catch (InterruptedException e) {
                e.printStackTrace();
            }
        }

        @Override
        public void run() {
            read();
        }
    }

static class Writer extends Thread {
    String text;

    Writer(String text) {
        this.text = text;
    }

    void write() {
        try {
            readMutex.acquire();
            accessToResource.acquire();

            // critical section
            content = text;
            System.out.println("Content changed:\t" + content);
            // end of critical section

            accessToResource.release();
            readMutex.release();
        }
        catch (InterruptedException e) {
            e.printStackTrace();
        }
    }

    @Override
    public void run() {
        write();
    }
}

public static void main(String[] args) {
    Reader t1 = new Reader();
    Reader t2 = new Reader();
    Writer t3 = new Writer("stars");
    Reader t4 = new Reader();
    Writer t5 = new Writer("restaurant at the end of universe");
    Reader t6 = new Reader();
    Reader t7 = new Reader();
    t1.start();
    t2.start();
    t3.start();
    t4.start();
    t5.start();
    t6.start();
    t7.start();
}
}

Answer 1

当一个Reader已获取accessToResource并尝试获取readMutex时:

if (numOfReaders == 1) {
    accessToResource.acquire();
}
readMutex.release();

System.out.println("Read content:\t" + content);

readMutex.acquire();  // -------> here

当另一个Writter已获取readMutex并尝试获取accessToResource时:

readMutex.acquire();
accessToResource.acquire(); // ------> here

发生死锁。

由于 accessToResource 和 readMutex 的许可为 1，因此 Reader 和 都没有权限作家可以继续前进。