java - 无法理解 Java 规范中的 volatile 示例

Question

我大致了解了 volatile 在 Java 中的含义。但读书 Java SE Specification 8.3.1.4我无法理解该特定示例下方的文字。

class Test {
    static volatile int i = 0, j = 0;
    static void one() { i++; j++; }
    static void two() {
        System.out.println("i=" + i + " j=" + j);
    }
}

This allows method one and method two to be executed concurrently, but guarantees that accesses to the shared values for i and j occur exactly as many times, and in exactly the same order, as they appear to occur during execution of the program text by each thread. Therefore, the shared value for j is never greater than that for i, because each update to i must be reflected in the shared value for i before the update to j occurs. It is possible, however, that any given invocation of method two might observe a value for j that is much greater than the value observed for i, because method one might be executed many times between the moment when method two fetches the value of i and the moment when method two fetches the value of j.

怎么样

j never greater than i

，但同时

any given invocation of method two might observe a value for j that is much greater than the value observed for i

??

看起来很矛盾。

运行示例程序后，我得到的j 大于i。为什么要使用 volatile 呢？它在没有 volatile 的情况下给出几乎相同的结果(而且 i 可以大于 j，规范中的先前示例之一)。为什么这个示例在这里作为 synchronized 的替代方法？

Answer 1

在任何时候，j 都不大于i。

这与方法二观察到的不同，因为它在不同的时间访问变量 i 和 j。 i 先被访问，j 稍微晚一点被访问。

这不是同步版本的直接替代方案，因为行为不同。与不使用 volatile 的一个区别是，如果不使用 volatile，则始终可以打印 0 值。增量永远不需要可见。

该示例演示了可变访问的顺序。一个需要这个的例子可能是这样的:

volatile boolean flag = false;
volatile int value;

// Thread 1
if(!flag) {
    value = ...;
    flag = true;
}

// Thread 2
if(flag) {
    System.out.println(value);
    flag = false;
}

线程 2 读取线程 1 设置的值而不是旧值。