几道基础的线程题

第一题

　　子线程循环十次，每次从1加到10，最后在主线程中相加得到结果?

　

public class CallableTest {
    public static void main(String[] args) {
        ThreadDemo demo = new ThreadDemo();
        
        FutureTask<Integer> result = new FutureTask<Integer>(demo);
        
        int sum = 0;
        for(int i = 0;i < 10;i++){
            Thread thread = new Thread(result);
            thread.start();
            try {
                Integer num = result.get();
                sum += num;
            } catch (InterruptedException e) {
                e.printStackTrace();
            } catch (ExecutionException e) {
                e.printStackTrace();
            }
        }
        
        System.out.println(sum);
    }
    
}

class ThreadDemo implements Callable<Integer> {
    @Override
    public Integer call() throws Exception {
        int sum = 0;
        for(int i = 1;i <= 10;i++){
            sum += i;
        }
        return sum;
    }
}

 第二题

　　设计 4 个线程，其中两个线程每次对 j 增加 1，另外两个线程对 j 每次减少
　　1。写出程序。

public class AddSubThread {
    public static void main(String[] args) {
        Operater operater = new Operater();
        ThreadDemoAdd add = new ThreadDemoAdd(operater);
        ThreadDemoSub sub = new ThreadDemoSub(operater);
        
        while(true) {
            for(int i = 0;i < 2;i++){
                new Thread(add).start();
            }
            
            for(int i = 0;i < 2;i++){
                new Thread(sub).start();
            }
        }
    }
}

class ThreadDemoAdd implements Runnable{
    private Operater operater;
    
    public ThreadDemoAdd(Operater operater){
        this.operater = operater;
    }
    
    @Override
    public void run() {
        operater.add();
    }
    
}

class ThreadDemoSub implements Runnable {
　　private Operater operater;
    
    public ThreadDemoSub(Operater operater){
        this.operater = operater;
    }

    @Override
    public void run() {
        operater.sub();
    }
    
}

class Operater {
    private int i = 0;
    
    public synchronized void add(){
        i++;
        System.out.println("i :" + i);
    }
    
    public synchronized void sub(){
        i--;
        System.out.println("i :" + i);
    }
}