为什么java 5+中的volatile不能确保来自另一个线程的可见性？ [英] Why doesn't volatile in java 5+ ensure visibility from another thread?
问题描述
根据：
http：// www.ibm.com/developerworks/library/j-jtp03304/
在新内存模型下，当线程A写入时对于一个易失性变量V，并且线程B从V读取，在写入V时对A可见的任何变量值现在都保证对B可见
Under the new memory model, when thread A writes to a volatile variable V, and thread B reads from V, any variable values that were visible to A at the time that V was written are guaranteed now to be visible to B
互联网上的许多地方声明以下代码永远不会打印错误：
And many places on the internet state that the following code should never print "error":
public class Test {
volatile static private int a;
static private int b;
public static void main(String [] args) throws Exception {
for (int i = 0; i < 100; i++) {
new Thread() {
@Override
public void run() {
int tt = b; // makes the jvm cache the value of b
while (a==0) {
}
if (b == 0) {
System.out.println("error");
}
}
}.start();
}
b = 1;
a = 1;
}
}
b a 为1时，所有线程的$ c> 应为1。
但是我有时会打印错误。这怎么可能？
推荐答案
更新：
对于任何感兴趣的人，这个bug已经针对Java 7u6 build b14进行了解决和修复。你可以在这里看到错误报告/修复
For anyone interested this bug has been addressed and fixed for Java 7u6 build b14. You can see the bug report/fixes here
- Report
- Changeset
- Buglist
原始答案
在考虑内存可见性/订单时，您需要考虑其发生的情况关系。 b！= 0 的重要前提条件是 a == 1 。如果 a！= 1 ，则b可以是0或1.
When thinking in terms of memory visibility/order you would need to think about its happens-before relationship. The important pre condition for b != 0 is for a == 1. If a != 1 then b can be either 0 or 1.
一旦线程看到 a == 1 然后该线程保证看到 b == 1 。
Once a thread sees a == 1 then that thread is guaranteed to see b == 1.
在OP示例中发布Java 5，一旦而（a == 0）爆发b是保证是1
Post Java 5, in the OP example, once the while(a == 0) breaks out b is guaranteed to be 1
编辑：
我运行的模拟很多时间并没有看到你的输出。
I ran the simulation many number of times and didn't see your output.
什么操作系统，Java版本&您正在测试CPU吗？
What OS, Java version & CPU are you testing under?
我在Windows 7上，Java 1.6_24（尝试使用_31）
I am on Windows 7, Java 1.6_24 (trying with _31)
编辑2：
对OP和Walter Laan的称赞 - 对我来说，这只发生在我从64位Java切换到32位Java时，打开（但可能不会被排除在外）一个64位的Windows 7。
Kudos to the OP and Walter Laan - For me it only happened when I switched from 64 bit Java to 32 bit Java, on (but may not be excluded to) a 64 bit windows 7.
编辑3：
对 tt 的分配，或者更确切地说， b 的静态选择似乎会产生重大影响（对证明这删除了 int tt = b; 并且它应该始终有效。
The assignment to tt, or rather the staticget of b seems to have a significant impact (to prove this remove the int tt = b; and it should always work.
看起来加载 b 进入 tt 将在本地存储该字段，然后将在if coniditonal中使用该字段（对该值的引用不是 tt ）。所以，如果 b == 0 为真，则可能意味着本地商店为 tt 为0（此时它的竞争是将1分配给本地 tt ）。这似乎只适用于32位Java 1.6& 7与cl设置。
It appears the load of b into tt will store the field locally which will then be used in the if coniditonal (the reference to that value not tt). So if b == 0 is true it probably means that the local store to tt was 0 (at this point its a race to assign 1 to local tt). This seems only to be true for 32 Bit Java 1.6 & 7 with client set.
我比较了两个输出组件，这里的直接区别就在于此。 （请记住这些是片段）。
I compared the two output assembly and the immediate difference was here. (Keep in mind these are snippets).
此印刷的错误
0x021dd753: test %eax,0x180100 ; {poll}
0x021dd759: cmp $0x0,%ecx
0x021dd75c: je 0x021dd748 ;*ifeq
; - Test$1::run@7 (line 13)
0x021dd75e: cmp $0x0,%edx
0x021dd761: jne 0x021dd788 ;*ifne
; - Test$1::run@13 (line 17)
0x021dd767: nop
0x021dd768: jmp 0x021dd7b8 ; {no_reloc}
0x021dd76d: xchg %ax,%ax
0x021dd770: jmp 0x021dd7d2 ; implicit exception: dispatches to 0x021dd7c2
0x021dd775: nop ;*getstatic out
; - Test$1::run@16 (line 18)
0x021dd776: cmp (%ecx),%eax ; implicit exception: dispatches to 0x021dd7dc
0x021dd778: mov $0x39239500,%edx ;*invokevirtual println
和
这不会打印错误
0x0226d763: test %eax,0x180100 ; {poll}
0x0226d769: cmp $0x0,%edx
0x0226d76c: je 0x0226d758 ;*ifeq
; - Test$1::run@7 (line 13)
0x0226d76e: mov $0x341b77f8,%edx ; {oop('Test')}
0x0226d773: mov 0x154(%edx),%edx ;*getstatic b
; - Test::access$0@0 (line 3)
; - Test$1::run@10 (line 17)
0x0226d779: cmp $0x0,%edx
0x0226d77c: jne 0x0226d7a8 ;*ifne
; - Test$1::run@13 (line 17)
0x0226d782: nopw 0x0(%eax,%eax,1)
0x0226d788: jmp 0x0226d7ed ; {no_reloc}
0x0226d78d: xchg %ax,%ax
0x0226d790: jmp 0x0226d807 ; implicit exception: dispatches to 0x0226d7f7
0x0226d795: nop ;*getstatic out
; - Test$1::run@16 (line 18)
0x0226d796: cmp (%ecx),%eax ; implicit exception: dispatches to 0x0226d811
0x0226d798: mov $0x39239500,%edx ;*invokevirtual println
在这个例子中，第一个条目来自打印错误的运行，而第二个条目来自没有的错误。
In this example the first entry is from a run that printed "error" while the second was from one which didnt.
似乎加载并分配了工作运行 b 在测试之前正确等于0。
It seems that the working run loaded and assigned b correctly before testing it equal to 0.
0x0226d76e: mov $0x341b77f8,%edx ; {oop('Test')}
0x0226d773: mov 0x154(%edx),%edx ;*getstatic b
; - Test::access$0@0 (line 3)
; - Test$1::run@10 (line 17)
0x0226d779: cmp $0x0,%edx
0x0226d77c: jne 0x0226d7a8 ;*ifne
; - Test$1::run@13 (line 17)
在打印错误的运行中加载缓存版本％edx
While the run that printed "error" loaded the cached version of %edx
0x021dd75e: cmp $0x0,%edx
0x021dd761: jne 0x021dd788 ;*ifne
; - Test$1::run@13 (line 17)
对于那些对汇编程序有更多经验的人请权衡：）
For those who have more experience with assembler please weigh in :)
编辑4
应该是我的最后编辑，如并发开发者得到它，我做了测试有和没有
int tt = b; 分配更多。我发现当我将max从100增加到1000时，当包含 int tt = b 时，似乎有100％的错误率，当排除它时有0％的几率。
Should be my last edit, as the concurrency dev's get a hand on it, I did test with and without the
int tt = b; assignment some more. I found that when I increase the max from 100 to 1000 there seems to be a 100% error rate when int tt = b is included and a 0% chance when it is excluded.
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