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Tú Đình
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Tú ĐìnhBeginner
Asked: September 17, 20212021-09-17T04:52:16+00:00 2021-09-17T04:52:16+00:00In: Java

Xin hỏi về volatile trong xử lí đa luồng java

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Em chào a, e là Tú, e có biết nhóm Unity ạ. E đang đọc và thực hiện lại 1 số ví dụ của a trong trang này của a https://tvd12.com/volatile-and-atomic/ . Em đang bị đánh giá ở phần VolatileVí dụ khi em thử bỏ qua biến biến trong biến hoạt động thì luồng 2 của em vẫn hoạt động bình thường, e không hiểu tại sao, mong a chỉ dẫn

threadjava
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  1. monkey Enlightened
    2021-09-17T04:58:12+00:00Added an answer on September 17, 2021 at 4:58 am
    This answer was edited.

    ý em là code thế này:

    private boolean active;
    
    public void prepare() throws InterruptedException {
        new Thread(() -> {
            System.out.println("application preparing ...");
            sleep(3);
            active = true;
        })
            .start();
    }
    
    public void start() throws Exception {
        new Thread(() -> {
            while(!active);
            System.out.println("application started");
        })
            .start();
    }
    

    Nó vẫn in được application started đúng không em?

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    • Tú Đình Beginner
      2021-09-17T05:00:50+00:00Replied to answer on September 17, 2021 at 5:00 am

      vâng đúng rồi a,

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    • Tú Đình Beginner
      2021-09-17T10:53:06+00:00Replied to answer on September 17, 2021 at 10:53 am

      vâng đúng rồi a,

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  2. Tú Đình Beginner
    2021-09-17T05:00:26+00:00Added an answer on September 17, 2021 at 5:00 am
    package dev.ozudo.volatile_atomic;
    
    public class VolatileExample
    {
        private  boolean active;
    
        public  void prepare() throws  InterruptedException{
            new Thread(()->{
                System.out.println("application preparing");
                System.out.println("name = "+Thread.currentThread().getName());
                sleep(5);
                active = true;
            }).start();
        }
    
        public  void start() throws Exception{
            new Thread(()->{
                long time = System.currentTimeMillis();
                while (!active){
                   // System.out.println("application started");
                }
                if(active){
                    System.out.println("name = "+Thread.currentThread().getName());
                    System.out.println("stop app = "+(System.currentTimeMillis()-time));
                }
            }).start();
        }
        public  void start2() throws Exception{
            new Thread(()->{
                long time = System.currentTimeMillis();
                while (!active){
                    // System.out.println("application started");
                }
                if(active){
                    System.out.println("name = "+Thread.currentThread().getName());
                    System.out.println("stop app 2222 = "+(System.currentTimeMillis()-time));
                }
            }).start();
        }
    
        private  static  void  sleep(int seconds){
            try {
                Thread.sleep(seconds*1000);
            } catch (InterruptedException e){
                e.printStackTrace();
            }
        }
    
        public static void main(String[] args) throws  Exception {
            VolatileExample example  = new VolatileExample();
            example.prepare();
            example.start();
            example.start2();
            sleep(1);
        }
    }
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    • Tú Đình Beginner
      2021-09-17T05:01:33+00:00Replied to answer on September 17, 2021 at 5:01 am

      kết quả của em :
      application preparing
      name = Thread-0
      name = Thread-1
      name = Thread-2
      stop app = 5008
      stop app 2222 = 5008

      Process finished with exit code 0

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      • tvd12 Enlightened
        2021-09-17T06:19:28+00:00Replied to answer on September 17, 2021 at 6:19 am

        do em có thêm dòng System.out.println("name = "+Thread.currentThread().getName()); em ạ. Anh cũng chưa tìm hiểu sâu về bên trong xem thằng volatile nó làm gì, nhưng theo một số bạn bè của anh giải thích thì khi em dùng cái lệnh System.out thì nó làm cho cái thanh ghi nó bị đầy, và sau khi chạy xong thì nó nạp lại cái giá trị active đã thay đổi vào thanh ghi và thread có vòng while nó nhận được sự thay đổi đó. Tuy nhiên nó đúng với C, còn với Java cần kiểm chứng thêm em nhé.

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        • Tú Đình Beginner
          2021-09-18T03:57:32+00:00Replied to answer on September 18, 2021 at 3:57 am

          Vâng em cảm ơn a

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