挥发性和原子性在一起它有意义吗？

I noticed that all examples about atomic writes without volatile keyword. Surely it is right.

我注意到所有关于没有volatile关键字的原子写入的例子。当然是对的。

What will happen if I add the volatile modifier to my atomic link ?

如果我将volatile修饰符添加到原子链接会发生什么？

Is there difference between :

有没有区别：

public class VolatilAtomic {
    volatile AtomicInteger atomicInteger = new AtomicInteger();
}

and

和

 public class VolatilAtomic {
         AtomicInteger atomicInteger = new AtomicInteger();
    }

？

3 个解决方案

#1

There would be a difference if you reassign the variable: atomicInteger = new AtomicInteger() in a method, in which case marking the variable as volatile will guarantee that the assignment will be visible by other threads.

如果在方法中重新分配变量：atomicInteger = new AtomicInteger（）会有所不同，在这种情况下将变量标记为volatile将保证其他线程可以看到赋值。

But if you only use the AtomicInteger instance that is created with each instance of your class and never reassign it, then volatile is unnecessary.

但是，如果您只使用使用类的每个实例创建的AtomicInteger实例并且从不重新分配它，那么volatile是不必要的。

#2

Generally speaking, final would be more appropriate.

一般来说，最终会更合适。

The important thing about modifiers is that the alter the reference, not the object referenced. e.g.

修饰符的重要之处在于改变引用，而不是引用的对象。例如

final int[] a = { 0 };
a[0] = 5;    // ok, not final
a = b;       // not ok as it is final

similarly

同样

volatile int[] a = { 0 };
a[0] = 5;    // not a volatile write
a = b;       // volatile write.

#3

Both provide the same visibility, but only AtomicInteger provide atomicity.

两者都提供相同的可见性，但只有AtomicInteger提供原子性。

#1