java - 一个线程中对非 volatile 成员变量的赋值是否保证在另一个线程中可见？

Question

考虑下面的 Java 示例。请注意，两个类成员变量都未声明为 volatile。如果我正确理解内存模型和“先于发生”规则，Java 实现可以优化 run() 方法，使其永远运行，即使另一个线程调用 stopNow() 方法。发生这种情况是因为 run() 方法中没有任何内容强制线程多次读取 stop 的值。那是对的吗？如果不是，为什么不呢？

class Example implements Runnable {
    boolean stop = false;
    int value = 0;

    public void stopNow() {
       stop = true;
    }

    public int getValue() {
        return value;
    }

    @Override
    public void run() {
        // Loop until stop is set to true.
        while (!stop) {
            ++value;
        }
        return;
    }
}

Answer 1

可以被另一个线程修改，但这不是保证。这也不是线程安全的。要保证从另一个线程看到一个变量，您需要执行以下任何操作

Changes to fields made by one thread are guaranteed to be visible to other threads only under the following conditions:

• A writing thread releases a synchronization lock and a reading thread subsequently acquires that same synchronization lock.
• If a field is declared as volatile, any value written to it is flushed and made visible by the writer thread before the writer
  thread performs any further memory operation (i.e., for the purposes
  at hand it is flushed immediately). Reader threads must reload the
  values of volatile fields upon each access.
• The first time a thread accesses a field of an object, it sees either the initial value of the field or a value since written by some other thread.
• As a thread terminates, all written variables are flushed to main memory. For example, if one thread synchronizes on the termination of another thread using Thread.join, then it is guaranteed to see the
  effects made by that thread (see §4.3.2).

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