示例代码：
class Sample{
private int v;
public void setV(){
Lock a=new Lock();
a.lock();
try{
v=1;
}finally{
a.unlock();
}
}
public int getV(){
return v;
}
}
如果我有一个线程不断调用getV,我只是在另一个线程中执行setV一次,那读取线程是否保证在写入后立即看到新值？还是需要使“V”变化或atomicreference？
如果答案是否定的,那么我应该把它改成：
class Sample{
private int v;
private Lock a=new Lock();
public void setV(){
a.lock();
try{
v=1;
}finally{
a.unlock();
}
}
public int getV(){
a.lock();
try{
int r=v;
}finally{
a.unlock();
}
return r;
}
}
解决方法
从
documentation：
All Lock implementations must enforce the same memory synchronization semantics as provided by the built-in monitor lock:
- A successful lock operation acts like a successful monitorEnter action
- A successful unlock operation acts like a successful monitorExit action
如果您在两个线程中使用Lock(即读取和写入),读取线程将看到新值,因为monitorEnter会刷新缓存.否则,您需要声明变量volatile以强制在读取线程中从内存读取.