java8的lambda表达式的thread

    public static  void  aa(String[] args){
        System.out.println("Thread.currentThread().getName() = " + Thread.currentThread().getName());
        try {
            Thread.sleep(5000);
        } catch (InterruptedException e) {
            e.printStackTrace();
        }
        System.out.println("args.toString()1111 = " + args.toString());
    }
    public static void main(String[] args) {
        Thread b= new Thread(new Runnable() {
            @Override
            public void run() {
                aa(args);
            }
        }, "zzxx");
        b.start();
        Thread c= new Thread(()->{
                aa(args);
        }, "zzxxxx");
        c.start();
    }

结果：
Thread.currentThread().getName() = zzxx
Thread.currentThread().getName() = zzxxxx
args.toString()1111 = [Ljava.lang.String;@4d6c1ee7
args.toString()1111 = [Ljava.lang.String;@4d6c1ee7