有效的重新排序 - 在新的JMM下

问题描述

我只想知道下面的重新排序在新的JMM模式下是否有效。

 原始代码：
 instanceVar1 = value; //正常读操作，没有volatile 
 synchronized（this）{
 instanceVar2 = value2; //正常读操作，没有volatile 
} 
 instanceVar3 = value3; //正常读操作，没有volatile 
  

上述代码可以重新排序为以下执行。 p>

 案例1：
 
 synchronized（this）{
 instanceVar2 = value2; //正常读操作，没有volatile 
 instanceVar1 = value; //正常读操作，没有volatile 
} 
 instanceVar3 = value3; //正常读操作，无变化
  

另一种情况：

 情况2：
 
 synchronized（this）{
 instanceVar3 = value3; //正常读操作，没有volatile 
 instanceVar2 = value2; //正常读操作，没有volatile 
 instanceVar1 = value; //正常读操作，没有volatile 
} 
  

另一种情况：

 情况3：
 
 instanceVar3 = value3; //正常读操作，没有volatile 
 synchronized（this）{
 instanceVar2 = value2; //正常读操作，没有volatile 
 instanceVar1 = value; //正常读操作，没有volatile 
} 
  

另一种情况：

 案例4：
 
 instanceVar3 = value3; //正常读操作，没有volatile 
 synchronized（this）{
 instanceVar2 = value2; //正常读操作，无挥发
} 
 instanceVar1 = value; //正常读操作，无挥发
  

$ b b

上述4种情况都是在新JMM模型下原始代码的有效重新排序。
根据我对
的理解，我已经给出了所有上述的重新排序。

解决方案

考虑如何正常加载/存储与监视器重新排序输入和退出：

案例1使用监视器输入重新排序正常加载/存储，这是有效的重新排序。 p>

案例2使用监视器输入重新排序正常加载/存储，后跟正常加载/存储的监视器退出，这是有效的重新排序。

查看类似的示例：。

案例3和4重新排序监视器输入，后跟正常加载/存储，不是有效。

I just wanted to know whether the below reorderings are valid one or not under new JMM model

Original Code: 
     instanceVar1 = value ;//  normal read operation, no volatile
     synchronized(this) {
       instanceVar2 = value2; //normal read operation, no volatile   
     }
     instanceVar3 = value3;  //normal read operation, no volatile

The above code can be reordered into the following executions.

Case 1:

     synchronized(this) {
       instanceVar2 = value2; //normal read operation, no volatile   
       instanceVar1 = value ;//  normal read operation, no volatile
     }
     instanceVar3 = value3;  //normal read operation, no volatile

Another case :

Case 2:

  synchronized(this) {
       instanceVar3 = value3;  //normal read operation, no volatile
       instanceVar2 = value2; //normal read operation, no volatile   
       instanceVar1 = value ;//  normal read operation, no volatile
     }

Another Case :

Case 3: 

    instanceVar3 = value3;  //normal read operation, no volatile
    synchronized(this) {
       instanceVar2 = value2; //normal read operation, no volatile   
       instanceVar1 = value ;//  normal read operation, no volatile
     }

Another Case :

Case 4: 

    instanceVar3 = value3;  //normal read operation, no volatile
    synchronized(this) {
       instanceVar2 = value2; //normal read operation, no volatile   
     }
    instanceVar1 = value ;//  normal read operation, no volatile

Do all the above 4 cases are valid reordering of original Code under new JMM Model.I have given all the above reorderings based on my understanding ofhttp://gee.cs.oswego.edu/dl/jmm/cookbook.html

解决方案

Consider how the normal load/stores are reordered with the monitor enter and exits:

Case 1 reorders a normal load/store with a monitor enter which is a valid reordering.

Case 2 reorders a normal load/store with a monitor enter, and a monitor exit followed by a normal load/store, which are valid reorderings.

See a similar example: Roach Motels and Java Memory Model. This shows that accesses can be moved into a synchronized block but not back out again.

Case 3 and 4 reorders a monitor enter followed by a normal load/store which is not valid.

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