乱序执行

• 指令重排序
• 为什么
  • CPU快, 指令1阻塞时，指令2先执行
• 判断
  • 代码不影响单线程最终一致性
• 例子，多线程

  • static int x = 0, y = 0;
    static int a = 0, b = 0;
    void main(){
        for (long i = 0; i < Long.MAX_VALUE; i++>) {
            x=0;
            y=0;
            a=0;
            b=0;
            CountDownLatch latch = new CountDownLatch(2);
     
            Thread one = new Thread(() -> {
                public void run() {
                    a = 1;
                    x = b;
                    latch.countDown();
                }
            });
            
            Thread other = new Thread(() -> {
                public void run() {
                    b = 1;
                    y = a;
                    latch.countDown();
                }
            });
            one.start();
            other.start();
            latch.await();
            if (x == 0 && y == 0) {
                break;
            }
     
        }
    }

• 例子，类初始化指令换顺序

  • class T {
        int m = 8;
    }
    T t = new T();
    汇编码
        0 new #2 <T>
            # 变量半初始化状态为0
        3 dup
        4 invokespecial #3 <T.<init>>
        7 astore_1
            # 4,7乱序执行, 先建立了关联再初始化, 变量中间状态值为0。线程访问时中间状态逸出
        8 return
    class C {
        private int num = 0;
        public C() {
            new Thread(() -> System.out.println(this.num)).start();
        }
        void main() {
            new C();
            System.in.read();
        }
    }