问题
As is known guarant that if we have some object reference and this reference has final field - we will see all reachable fields from final field(at least when constructor was finished)
example 1:
class Foo{
private final Map map;
Foo(){
map = new HashMap();
map.put(1,"object");
}
public void bar(){
System.out.println(map.get(1));
}
}
As I undertand at this case we have guarantee that bar() method always output object because:
1. I listed full code of class Foo and map is final;
2. If some thread will see reference of Foo and this reference != null, then we have guarantees that reachable from final map reference value will be actual.
also I think that
Example 2:
class Foo {
private final Map map;
private Map nonFinalMap;
Foo() {
nonFinalMap = new HashMap();
nonFinalMap.put(2, "ololo");
map = new HashMap();
map.put(1, "object");
}
public void bar() {
System.out.println(map.get(1));
}
public void bar2() {
System.out.println(nonFinalMap.get(2));
}
}
here we have same gurantees about bar() method but bar2 can throw NullPointerException despite nonFinalMap assignment occurs before map assignment.
I want to know how about volatile:
Example 3:
class Foo{
private volatile Map map;
Foo(){
map = new HashMap();
map.put(1,"object");
}
public void bar(){
System.out.println(map.get(1));
}
}
As I understand bar() method cannot throw NullPoinerException but it can print null; (I am fully not sure about this aspect)
Example 4:
class Foo {
private volatile Map map;
private Map nonVolatileMap;
Foo() {
nonVolatileMap= new HashMap();
nonVolatileMap.put(2, "ololo");
map = new HashMap();
map.put(1, "object");
}
public void bar() {
System.out.println(map.get(1));
}
public void bar2() {
System.out.println(nonFinalMap.get(2));
}
}
I think here we have same gurantees about bar() method also bar2() cannot throw NullPointerException because nonVolatileMap assignment written higher volatile map assignment but it can output null
Added after Elliott Frisch comment
Publication through race example:
public class Main {
private static Foo foo;
public static void main(String[] args) {
new Thread(new Runnable() {
@Override
public void run() {
foo = new Foo();
}
}).start();
new Thread(new Runnable() {
@Override
public void run() {
while (foo == null) ; // empty loop
foo.bar();
}
}).start();
}
}
Please proove or correct my comments to code snippets.
回答1:
In the realm of current Java Memory Model, volatile does not equal final. In other words, you cannot replace final with volatile, and think the safe construction guarantees are the same. Notably, this can theoretically happen:
public class M {
volatile int x;
M(int v) { this.x = v; }
int x() { return x; }
}
// thread 1
m = new M(42);
// thread 2
M lm;
while ((lm = m) == null); // wait for it
print(lm.x()); // allowed to print "0"
So, writing the volatile field in constructor is not as safe.
Intuition: there is a race on m in the example above. That race is not eliminated by making the field M.x volatile, only making the m itself volatile would help. In other words, volatile modifier in that example is at the wrong place to be useful. In safe publication, you have to have "writes -> volatile write -> volatile read that observes volatile write -> reads (now observing writes prior the volatile write)", and instead you have "volatile write -> write -> read -> volatile read (that does not observe the volatile write)".
Trivia 1: This property means we can optimize volatile-s much more aggressively in constructors. This corroborates the intuition that unobserved volatile store (and indeed it is not observed until constructor with non-escaping this finishes) can be relaxed.
Trivia 2: This also means you cannot safely initialize volatile variables. Replace M with AtomicInteger in the example above, and you have a peculiar real-life behavior! Call new AtomicInteger(42) in one thread, publish the instance unsafely, and do get() in another thread -- are you guaranteed to observe 42? JMM, as stated, says "nope". Newer revisions of Java Memory Model strive to guarantee safe construction for all initializations, to capture this case. And many non-x86 ports where that matters have already strengthened this to be safe.
Trivia 3: Doug Lea: "This final vs volatile issue has led to some twisty constructions in java.util.concurrent to allow 0 as the base/default value in cases where it would not naturally be. This rule sucks and should be changed."
That said, the example can be made more cunning:
public class C {
int v;
C(int v) { this.x = v; }
int x() { return x; }
}
public class M {
volatile C c;
M(int v) { this.c = new C(v); }
int x() {
while (c == null); // wait!
return c.x();
}
}
// thread 1
m = new M(42);
// thread 2
M lm;
while ((lm = m) == null); // wait for it
print(lm.x()); // always prints "42"
If there is a transitive read through volatile field after volatile read observed the value written by volatile write in constructor, the usual safe publication rules kick in.
来源：https://stackoverflow.com/questions/42012658/how-deep-volatile-publication-guarantees