spring创建多例对象的时候，默认对照xml中bean的id会是什么呢？

Question

@Service
@Scope("prototype")
public class DemoService implements BeanNameAware{

    private String beanName;

    public void outputResult() {
        System.out.println("获取bean："+this+",beanName:"+beanName);
    }
    @Override
    public void setBeanName(String name) {
        this.beanName = name;
    }
}

获取demoService的示例

    public void newBean() {
        ThreadPoolExecutor executor = (ThreadPoolExecutor) Executors.newCachedThreadPool();
        for (int i = 0; i < 5; i++) {
            executor.execute(new Runnable() {
                @Override
                public void run() {
                    DemoService demo = context.getBean(DemoService.class);
                    demo.outputResult();
                }
            });
        }
    }

每一次获取到的beanName都是demoService，觉得获取到的不是bean的id而是bean的name，所以在并发获取多个实例的时候每个bean的id应该是什么？不是不应该重复吗？

Answer 1

这是 scope="prototype"时获取的beanid

换成单例之后