为什么volatile在Java 5+不同步缓存副本的变量与主内存？ [英] Why volatile in java 5+ doesn't synchronize cached copies of variables with main memory?
问题描述
根据：
http：// www.ibm.com/developerworks/library/j-jtp03304/
在新的内存模型下，当线程A写入到易失性变量V，并且线程B从V读取，在V被写入时对A可见的任何变量值现在保证对B可见
在互联网上的许多地方，以下代码不应该打印错误：
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; //使jvm缓存的值为b
while（a == 0）{
}
if（b == 0） {
System.out.println（error）;
}
}
} .start（）;
}
b = 1;
a = 1;
}
}
b
更新：
对于任何感兴趣的人，这个bug已经解决并修复了Java 7u6 build b14。您可以在这里查看错误报告/修正
原始回答
当考虑内存可见性/顺序时，您需要考虑其发生之前的关系。 b！= 0 的重要前提条件是 a == 1 。如果 a！= 1 ，那么b可以是0或1.
c> a == 1 那么该线程保证看到 b == 1 。
Post Java 5，在OP示例中，一旦 while（a == 0）保证为1
我运行的模拟多个次，没有看到你的输出。
什么操作系统，Java版本& CPU
我在Windows 7，Java 1.6_24（尝试_31）
编辑2：
感谢OP和Walter Laan - 对我来说，只有当我从64位Java切换到32位Java时，
$ b $ b
tt ，或者 b 的静态作用似乎有重大影响证明这个删除 int tt = b; ，它应该总是工作。
c $ c> b 到 tt 将存储本地的字段，然后将用于if coniditonal因此，如果 b == 0 是真的，它可能意味着本地存储到 tt 为0（在这一点上它是一个比赛，将1分配给本地 tt ）这似乎只对32位Java 1.6& 7与客户端设置。
我比较了两个输出组件，而直接的区别在这里。 （请记住这些是片段）
p $ p>
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;隐式异常：分派到0x021dd7c2
0x021dd775：nop; * getstatic out
; - test $ 1 :: run @ 16（line 18）
0x021dd776：cmp（％ecx），％eax;隐式异常：分派到0x021dd7dc
0x021dd778：mov $ 0x39239500，％edx; * invokevirtual println
$ b
这未打印错误 code> 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;隐式异常：分派到0x0226d7f7
0x0226d795：nop; * getstatic out
; - test $ 1 :: run @ 16（line 18）
0x0226d796：cmp（％ecx），％eax;隐式异常：分派到0x0226d811
0x0226d798：mov $ 0x39239500，％edx; * invokevirtual println
在这个例子中，第一个条目是从打印错误的运行，而第二个是从一个没有。
似乎工作运行加载和分配 b 在测试之前等于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
; - 测试$ 1 :: run @ 13（行17）
缓存版本％edx
0x021dd75e：cmp $ 0x0， edx
0x021dd761：jne 0x021dd788; * ifne
; - 测试$ 1 :: run @ 13（行17）
对于有更多汇编经验的人重量：）
编辑4
并发开发人员得到了一个手，我做了测试和没有
int tt = b; 分配一些。我发现，当我将最大值从100增加到1000时，当包括 int tt = b 时，似乎有100％的错误率，
According to:
http://www.ibm.com/developerworks/library/j-jtp03304/
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
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 should be 1 for all the threads when a is 1.
However I sometimes get "error" printed. How is this possible?
Update:
For anyone interested this bug has been addressed and fixed for Java 7u6 build b14. You can see the bug report/fixes here
Original Answer
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.
Once a thread sees a == 1 then that thread is guaranteed to see b == 1.
Post Java 5, in the OP example, once the while(a == 0) breaks out b is guaranteed to be 1
Edit:
I ran the simulation many number of times and didn't see your output.
What OS, Java version & CPU are you testing under?
I am on Windows 7, Java 1.6_24 (trying with _31)
Edit 2:
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.
Edit 3:
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.
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.
I compared the two output assembly and the immediate difference was here. (Keep in mind these are snippets).
This printed "error"
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
And
This did not print "error"
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
In this example the first entry is from a run that printed "error" while the second was from one which didnt.
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)
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)
For those who have more experience with assembler please weigh in :)
Edit 4
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.
这篇关于为什么volatile在Java 5+不同步缓存副本的变量与主内存？的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！
