如何理解JSR-133发生 - 之前太弱图6/7(How to understand JSR-133 Happens-Before is too Weak Figure6/7)

 在JSR-133中 ，有两个例子说明大概正确同步的程序。 
 
 第一个例子由图6给出，其中： 
 
x == y == 0
 
 然后，线程1通过以下方式修改状态： 
 
r1 = x; 
if (r1 != 0) 
y = 1; 
 
 线程2通过以下方式修改状态： 
 
r2 = y;
if (r2 != 0)
x = 1;
 
 作者声明这个程序是正确同步的，但是，这个程序执行的结果是r1 == r2 == 1 。 后者的价值甚至是可能的？ 
 
 此外，图7表示为初始状态： 
 
x == y == 0
 
 随着线程1修改： 
 
r1 = x; 
y = r1; 
 
 和线程2执行： 
 
r2 = y;
x = r2;
 
 这怎么会导致r1 == r2 == 42 ？ 

In the JSR-133, there are two example that state presumably correctly synchronized programs.
 
The first example is given by figure 6 where:
 
x == y == 0
 
Then, thread 1 modifies the state by:
 
r1 = x; 
if (r1 != 0) 
y = 1; 
 
and thread 2 modifies the state by:
 
r2 = y;
if (r2 != 0)
x = 1;
 
The author states that this program is correctly synchronized，However, there is an execution of this program that results in r1 == r2 == 1. How would this latter value even be possible?
 
Furthermore, figure 7 states as an initial state:
 
x == y == 0
 
With thread 1 modifying:
 
r1 = x; 
y = r1; 
 
and thread 2 executing:
 
r2 = y;
x = r2;
 
How could this result in r1 == r2 == 42?

原文：https://stackoverflow.com/questions/34013636