In Java, I understand that volatile keyword provides visibility to variables. The question is, if a variable is a reference to a mutable object, does volatile also provide visibility to the members inside that object?
In the example below, does it work correctly if multiple threads are accessing volatile Mutable m and changing the value?
example
class Mutable {
private int value;
public int get()
{
return a;
}
public int set(int value)
{
this.value = value;
}
}
class Test {
public volatile Mutable m;
}
This is sort of a side note explanation on some of the details of volatile. Writing this here because it is too much for an comment. I want to give some examples which show how volatile affects visibility, and how that changed in jdk 1.5.
Given the following example code:
public class MyClass
{
private int _n;
private volatile int _volN;
public void setN(int i) {
_n = i;
}
public void setVolN(int i) {
_volN = i;
}
public int getN() {
return _n;
}
public int getVolN() {
return _volN;
}
public static void main() {
final MyClass mc = new MyClass();
Thread t1 = new Thread() {
public void run() {
mc.setN(5);
mc.setVolN(5);
}
};
Thread t2 = new Thread() {
public void run() {
int volN = mc.getVolN();
int n = mc.getN();
System.out.println("Read: " + volN + ", " + n);
}
};
t1.start();
t2.start();
}
}
The behavior of this test code is well defined in jdk1.5+, but is not well defined pre-jdk1.5.
In the pre-jdk1.5 world, there was no defined relationship between volatile accesses and non-volatile accesses. therefore, the output of this program could be:
In the jdk1.5+ world, the semantics of volatile were changed so that volatile accesses affect non-volatile accesses in exactly the same way as synchronization. therefore, only certain outputs are possible in the jdk1.5+ world:
Output 3. is not possible because the reading of "5" from the volatile _volN establishes a synchronization point between the 2 threads, which means all actions from t1 taken before the assignment to _volN must be visible to t2.
Further reading: