Java 面试总结

1. Java 并发volatile 问题代码

class NumberDemo {
//private AtomicInteger count = new AtomicInteger(0);
    private volatile int count = 0;

    public void add() {
        this.count++;
    }

    public int getCount() {
        return this.count;
    }
}

public class ThreadDemo {
    public static void main(String[] args) {
        NumberDemo d = new NumberDemo();
        //CountDownLatch latch = new CountDownLatch(10);
        for (int i = 1; i <= 10; i++) {
        new Thread(()-> {
              for (int j = 1; j <= 100; j++) {
                    d.add();
              }
        }).start();
    }
    if (Thread.activeCount() > 1) {
        Thread.yield();
    }
    System.out.println("count is :" + d.getCount());
}

 1. 问题这个count 虽然添加了volatile ，但是count++ 不是原子的.

2.主线程在启动所有子线程后，通过检查 Thread.activeCount() > 1 并调用 Thread.yield() 尝试等待子线程完成。这种方式‌不可靠‌，因为：

• Thread.activeCount() 返回的线程数包含主线程自身，判断逻辑不准确。
• Thread.yield() 仅让出当前线程的 CPU 时间片，但无法确保所有子线程执行完毕。

主线程可能‌提前输出结果‌，而子线程尚未完成所有累加操作，导致输出值远小于 1000, 其实个人答错这道题是因为忘记volatile 只能保证可见性，无法保证原子性，还有就是这个yield我从来没有用过。

class NumberDemo {
    //private AtomicInteger count = new AtomicInteger(0);
    private volatile int count = 0;

    public synchronized void add() {
        this.count++;
    }

    public int getCount() {
        return this.count;
    }
}

public class ThreadDemo {
    public static void main(String[] args) throws InterruptedException {
        NumberDemo d = new NumberDemo();
        Thread [] threads = new Thread[10];
        //CountDownLatch latch = new CountDownLatch(10);
        for (int i = 0; i < 10; i++) {
            threads[i] = new Thread(()-> {
                for (int j = 1; j <= 100; j++) {
                    d.add();
                }
            });
            threads[i].start();
        }
        for(Thread t: threads){
            t.join();
        }
        System.out.println("count is :" + d.getCount());

    }
}

2. 现成的run 问题

       Thread t  = new Thread(new Runnable() {
            @Override
            p