Java线程中的内存顺序和可见性

Question

class S {
    public int x = 100; 
}

class T {
    public void do(S s){
       new Thread( () -> { 
            System.out.println(s.x);
        };).start();
    }
}


class M {
    public static void main(String[] args){
       T t = new T();
       S s = new S();
       s.x = 101;
       t.do(s);
    }
}

Hello,

Is T::do guaranteed to see always sx == 101? Why it is or why it is not?

Thanks in advance for your help.

Answer 1

When you start a thread, this introduces a memory barrier where anything which happened before you started the thread will be visible. Note: starting a thread takes a very long time in computer terms.

BTW System.out.println is a synchronized method which add full read/write memory barriers although in this case, that won't matter/