阅读《java并发编程实战》第五章

阅读《java并发编程实战》第五章

Semaphore的应用举例

• Semaphore的应用举例：实现一个固定大小的Set。当容器满了之后，无法add，线程阻塞。

public class BoundedHashSet {
    // invariant: size of Set always less than or equal to given size
    private final Set<Integer> set;
    private final Semaphore semaphore;

    public BoundedHashSet(final int size) {
        if (size <= 0) {
            throw new IllegalArgumentException("size should be above 0.");
        }
        this.set = Collections.synchronizedSet(new HashSet<>());
        this.semaphore = new Semaphore(size);
    }

    public boolean add(int x) throws InterruptedException {
        semaphore.acquire();
        boolean wasAdded = false;
        try {
            wasAdded = set.add(x);
            return wasAdded;
        } finally {
            if (!wasAdded) {
                semaphore.release();
            }
        }
    }

    public boolean remove(int x) {
        boolean res = set.remove(x);
        if (res) {
            semaphore.release(); // release不会抛 InterruptedException
        }
        return res;
    }
}

构建可伸缩的结果缓存（实践同步的思路）

version 1: 线程安全，但并发度很差

public interface Computable <A, V> {
    V compute(A arg);
}

/**
 * ThreadSafe, but performance/flexibility is very bad.
 * Not Good
 */
@ThreadSafe
public class Memorizer1<A, V> implements Computable<A, V> {
    // 有点像静态代理模式，类似Collections.synchronizedMap()
    private final Computable<A, V> task; 

    // GuardedBy(this)
    private final Map<A, V> cache = new HashMap<>();

    public Memorizer1(Computable<A, V> task) {
        this.task = task;
    }

    @Override
    public synchronized V compute(A arg) {
        V ans = cache.get(arg);
        if (ans == null) {
            ans = task.compute(arg);
            cache.put(arg, ans);
        }
        return ans;
    }
}

version 2: 线程安全，但是task.compute()方法有可能被调用两次，因为compute

public class Memorizer2<A, V> implements Computable<A, V> {

    private final Computable<A, V> task;
    private final Map<A, V> cache = new ConcurrentHashMap<>();

    public Memorizer2(Computable<A, V> task) {
        this.task = task;
    }

    @Override
    public synchronized V compute(A arg) {
        V ans = cache.get(arg);
        if (ans == null) {
            ans = task.compute(arg);
            cache.put(arg, ans);
        }
        return ans;
    }
}

version 3: 线程安全，但是同样有version 2的问题，会存在多个线程进入到 if (ans == null) 里面并发执行，重复计算。

public class Memorizer3<A, V> implements Computable<A, V> {

    // have an idea: if the param needs to be passed from outter, then initialize it in constructor
    // otherwise initialize in declaration.
    private final Map<A, Future<V>> cache = new ConcurrentHashMap<>();
    private final Computable<A, V> task;

    public Memorizer3(Computable<A, V> task) {
        this.task = task;
    }

    @Override
    public V compute(A arg) {
        Future<V> ans = cache.get(arg);
        if (ans == null) {
            Callable<V> eval = () -> task.compute(arg);
            FutureTask<V> ft = new FutureTask<>(eval);
            cache.put(arg, ft);
            ans = ft;
            ft.run();
        }
        try {
            return ans.get();
        } catch (InterruptedException e) {
            throw new RuntimeException(e);
        } catch (ExecutionException e) {
            throw new RuntimeException(e);
        }
    }
}

version 4: 线程安全，并发度也高，也解决了同时计算的问题，主要是利用了并发容器ConcurrentHashMap的putIfAbsent原子操作，只有成功添加的线程才有机会执行计算。

public class Memorizer4<A, V> implements Computable<A, V> {
    private final Computable<A, V> task;
    private final Map<A, Future<V>> cache = new ConcurrentHashMap<>();

    public Memorizer4(Computable<A, V> task) {
        this.task = task;
    }

    @Override
    public V compute(A arg) {        
        while (true) {
            Future<V> ans = cache.get(arg);
            if (ans == null) {
                FutureTask<V> ft = new FutureTask<>(() -> task.compute(arg));
                // if already exists, it will return the current value, so that ans will not be null, thus not run task
                ans = cache.putIfAbsent(arg, ft);
                if (ans == null) { // it means put success
                    ans = ft;
                    ft.run();
                }
            }
            try {
                return ans.get();
            } catch (CancellationException e) {
                cache.remove(arg, ans);
            } catch (ExecutionException e) {
                throw new RuntimeException(e);
            } catch (InterruptedException e) {
                throw new RuntimeException(e);
            }
        }
    }
}

总结：通过这个例子，可以看到，充分利用并发容器类，以及并发工具类（FutureTask类，Task接口）可以简化并发编程的难度，以及提高并发的可伸缩性。