How does "this" escape the constructor in Java? - java

I've heard about this happening in non thread-safe code due to improperly constructed objects but I really don't have the concept down, even after reading about in in Goetz's book. I'd like to solidify my understanding of this code smell as I maybe doing it and not even realize it. Please provide code in your explanation to make it stick, thanks.

Example : in a constructor, you create an event listener inner class (it has an implicit reference to the current object), and register it to a list of listener.
=> So your object can be used by another thread, even though it did not finish executing its constructor.
public class A {
private boolean isIt;
private String yesItIs;
public A() {
EventListener el = new EventListener() { ....};
StaticListeners.register(el);
isIt = true;
yesItIs = "yesItIs";
}
}
An additional problem that could happen later : the object A could be fully created, made available to all threads, use by another thread ... except that that thread could see the A instance as created, yesItIs with it "yesItIs" value, but not isIt! Believe it or not, this could happen ! What happen is:
=> synchronization is only half about blocking thread, the other half is about inter-thread visibility.
The reason for that Java choice is performance : inter-thread visibility would kill performance if all data would be shared with all threads, so only synchronized data is guaranteed to be shared...

Really simple example:
public class Test
{
private static Test lastCreatedInstance;
public Test()
{
lastCreatedInstance = this;
}
}

This is the reason why double-checked locking doesn't work. The naive code
if(obj == null)
{
synchronized(something)
{
if (obj == null) obj = BuildObject(...);
}
}
// do something with obj
is not safe because the assignment to the local variable can occur before the rest of the construction (constructor or factory method). Thus thread 1 can be in the BuildObject step, when thread 2 enters the same block, detects a non-null obj, and then proceeds to operate on an incomplete object (thread 1 having been scheduled out in mid-call).

public class MyClass{
String name;
public MyClass(String s)
{
if(s==null)
{
throw new IllegalArgumentException();
}
OtherClass.method(this);
name= s;
}
public getName(){ return name; }
}
In the above code, OtherClass.method() is passed an instance of MyClass which is at that point incompletely constructed, i.e. not yet fulfilling the contract that the name property is non-null.

Steve Gilham is correct in his assesment of why double checked locking is broken. If thread A enters that method and obj is null, that thread will begin to create an instance of the object and assign it obj. Thread B can possibly enter while thread A is still instantiating that object (but not completing) and will then view the object as not null but that object's field may not have been initialized. A partially constructed object.
However, the same type of problem can arrise if you allow the keyword this to escape the constructor. Say your constructor creates an instance of an object which forks a thread, and that object accepts your type of object. Now your object may have not be fully initialized, that is some of your fields may be null. A reference to your object by the one you have created in your constructor can now reference you as a non null object but get null field values.
A bit more explanation:
Your constructor can initialize every field in your class, but if you allow 'this' to escape before any of the other objects are created, they can be null (or default primative) when viewed by other threads if 1. They are not declared final or 2. They are not declared volatile

public class Test extends SomeUnknownClass{
public Test(){
this.addListner(new SomeEventListner(){
#Override
void act(){}
});
}
}
After this operation instanse of SomeEventListner will have a link to Test object, as a usual inner class.
More examples can be find here:
http://www.ibm.com/developerworks/java/library/j-jtp0618/index.html

Here's an example of how uninitialized this of OuterClass can be accessed from inside of inner class:
public class OuterClass {
public Integer num;
public OuterClass() {
Runnable runnable = new Runnable() { // might lead to this reference escape
#Override
public void run() {
// example of how uninitialized this of outer class
// can be accessed from inside of inner class
System.out.println(OuterClass.this.num); // will print null
}
};
new Thread(runnable).start();
new Thread().start(); // just some logic to keep JVM busy
new Thread().start(); // just some logic to keep JVM busy
this.num = 8;
System.out.println(this.num); // will print 8
}
public static void main(String[] args) {
new OuterClass();
}
}
Output:
null
8
Pay attention to OuterClass.this.num instruction in the code

Related

ExecutorService taking a runnable which is still under construction

I would like to know if this piece of code is correct or not. Will this not lead to issues as I am submitting the runnable object to the executor service while constructing the object itself?
public class A implements Runnable {
public A() {
Executors.newSingleThreadExecutor().execute(this);
// some other initializations
}
}
Will this lead to any issues as we are trying to submit the object to the executor even before creating it completely? If the run() method is called even before all the initializing is done (if at all it's possible), will the variables still be null which were not yet initialized?
Please do not ask me to come up with the complete code, as I have been asking this as a general question which requires clarification.
Yes, there may be issues. The Executor might read a field that you set in the constructor, even before the corresponding code in the constructor was executed. In general you should not expose this from inside a constructor. Java provides useful guarantees for objects after their constructor finished, but in order to benefit from those you have to wait for the result of new X(...) before using it.
Will this lead to any issues as we are trying to submit the object to
the executor even before creating it completely?
For one thing, you can get final variable that are still changing value - that is quite bad per the semantics of final. It can lead to very hard-to-trace concurrency bugs in multi-threaded code.
This code will usually print a few zeros and even the occasional 4, even though the final field a is only ever assigned the value 4 and should never been seen having any other value than 4.
public class A implements Runnable {
private static ExecutorService threads = Executors.newSingleThreadExecutor();
final int a;
public A() {
threads.execute(this);
Thread.yield();
a = 4;
}
#Override
public void run() {
if (a != 4) {
System.out.println(a);
}
}
public static void main(String[] args) {
for (int i = 0; i < 50_000; i++) {
new A();
}
threads.shutdown();
}
}
If the run() method is called even before all the initializing is done
(if at all it's possible), will the variables still be null which were
not yet initialized?
Yes, the ones not yet initialized will be null for reference variables, or the default value (0, false, '\0', 0d, 0f, etc.) for the primitive types. It is even possible according to the specifications with long and double fields to see only 32 of the 64 bits initialized (although on 64 bit architectures it is unlikely that you will ever observe this)
There will almost certainly be issues. What you have is called a "this escape" where you pass a this reference in a ctor to an external method. It's super bad, the object is incompletely constructed at that point and anything could happen.
What you should probably do instead is make the constructor private and use a factory method to get a new instance and execute it, if that's the goal.
public class A implements Runnable {
public A getNew() {
A a = new A();
Executors.newSingleThreadExecutor().execute(a);
return a;
}
private A() {
// some other initializations
}
}

Why can method reference use non-final variables?

I had some confusion about inner classes and lambda expression, and I tried to ask a question about that, but then another doubt arose, and It's probable better posting another question than commenting the previous one.
Straight to the point: I know (thank you Jon) that something like this won't compile
public class Main {
public static void main(String[] args) {
One one = new One();
F f = new F(){ //1
public void foo(){one.bar();} //compilation error
};
one = new One();
}
}
class One { void bar() {} }
interface F { void foo(); }
due to how Java manages closures, because one is not [effectively] final and so on.
But then, how come is this allowed?
public class Main {
public static void main(String[] args) {
One one = new One();
F f = one::bar; //2
one = new One();
}
}
class One { void bar() {} }
interface F { void foo(); }
Is not //2 equivalent to //1? Am I not, in the second case, facing the risks of "working with an out-of-date variable"?
I mean, in the latter case, after one = new One(); is executed f still have an out of date copy of one (i.e. references the old object). Isn't this the kind of ambiguity we're trying to avoid?
A method reference is not a lambda expression, although they can be used in the same way. I think that is what is causing the confusion. Below is a simplification of how Java works, it is not how it really works, but it is close enough.
Say we have a lambda expression:
Runnable f = () -> one.bar();
This is the equivalent of an anonymous class that implements Runnable:
Runnable f = new Runnable() {
public void run() {
one.bar();
}
}
Here the same rules apply as for an anonymous class (or method local class). This means that one needs to effectively final for it to work.
On the other hand the method handle:
Runnable f = one::bar;
Is more like:
Runnable f = new MethodHandle(one, one.getClass().getMethod("bar"));
With MethodHandle being:
public class MethodHandle implements Runnable {
private final Object object;
private final Method method;
public MethodHandle(Object object, java.lang.reflect.Method method) {
this.object = Object;
this.method = method;
}
#Override
public void run() {
method.invoke(object);
}
}
In this case, the object assigned to one is assigned as part of the method handle created, so one itself doesn't need to be effectively final for this to work.
Your second example is simply not a lambda expression. It's a method reference. In this particular case, it chooses a method from a particular object, which is currently referenced by the variable one. But the reference is to the object, not to the variable one.
This is the same as the classical Java case:
One one = new One();
One two = one;
one = new One();
two.bar();
So what if one changed? two references the object that one used to be, and can access its method.
Your first example, on the other hand, is an anonymous class, which is a classical Java structure that can refer to local variables around it. The code refers to the actual variable one, not the object to which it refers. This is restricted for the reasons that Jon mentioned in the answer you referred to. Note that the change in Java 8 is merely that the variable has to be effectively final. That is, it still can't be changed after initialization. The compiler simply became sophisticated enough to determine which cases will not be confusing even when the final modifier is not explicitly used.
The consensus appears to be that this is because when you do it using an anonymous class, one refers to a variable, whereas when you do it using a method reference, the value of one is captured when the method handle is created. In fact, I think that in both cases one is a value rather than a variable. Let's consider anonymous classes, lambda expressions and method references in a bit more detail.
Anonymous classes
Consider the following example:
static Supplier<String> getStringSupplier() {
final Object o = new Object();
return new Supplier<String>() {
#Override
public String get() {
return o.toString();
}
};
}
public static void main(String[] args) {
Supplier<String> supplier = getStringSupplier();
System.out.println(supplier.get()); // Use o after the getStringSupplier method returned.
}
In this example, we are calling toString on o after the method getStringSupplier has returned, so when it appears in the get method, o cannot refer to a local variable of the getStringSupplier method. In fact it is essentially equivalent to this:
static Supplier<String> getStringSupplier() {
final Object o = new Object();
return new StringSupplier(o);
}
private static class StringSupplier implements Supplier<String> {
private final Object o;
StringSupplier(Object o) {
this.o = o;
}
#Override
public String get() {
return o.toString();
}
}
Anonymous classes make it look as if you are using local variables, when in fact the values of these variables are captured.
In contrast to this, if a method of an anonymous class references the fields of the enclosing instance, the values of these fields are not captured, and the instance of the anonymous class does not hold references to them; instead the anonymous class holds a reference to the enclosing instance and can access its fields (either directly or via synthetic accessors, depending on the visibility). One advantage is that an extra reference to just one object, rather than several, is required.
Lambda expressions
Lambda expressions also close over values, not variables. The reason given by Brian Goetz here is that
idioms like this:
int sum = 0;
list.forEach(e -> { sum += e.size(); }); // ERROR
are fundamentally serial; it is quite difficult to write lambda bodies
like this that do not have race conditions. Unless we are willing to
enforce -- preferably at compile time -- that such a function cannot
escape its capturing thread, this feature may well cause more trouble
than it solves.
Method references
The fact that method references capture the value of the variable when the method handle is created is easy to check.
For example, the following code prints "a" twice:
String s = "a";
Supplier<String> supplier = s::toString;
System.out.println(supplier.get());
s = "b";
System.out.println(supplier.get());
Summary
So in summary, lambda expressions and method references close over values, not variables. Anonymous classes also close over values in the case of local variables. In the case of fields, the situation is more complicated, but the behaviour is essentially the same as capturing the values because the fields must be effectively final.
In view of this, the question is, why do the rules that apply to anonymous classes and lambda expressions not apply to method references, i.e. why are you allowed to write o::toString when o is not effectively final? I do not know the answer to that, but it does seem to me to be an inconsistency. I guess it's because you can't do as much harm with a method reference; examples like the one quoted above for lambda expressions do not apply.
No. In your first example you define the implementation of F inline and try to access the instance variable one.
In the second example you basically define your lambda expression to be the call of bar() on the object one.
Now this might be a bit confusing. The benefit of this notation is that you can define a method (most of the time it is a static method or in a static context) once and then reference the same method from various lambda expressions:
msg -> System.out::println(msg);

Can someone give an example of the situation described here about thread-safe variables?

I'm learning about Java multi-threading and came across a very good tutorial online. But I am not sure if I understand a part where the writer explains about thread-safe objects, variables and such. To quote him,
public void someMethod(){
LocalObject localObject = new LocalObject();
localObject.callMethod();
method2(localObject);
}
public void method2(LocalObject localObject){
localObject.setValue("value");
}
...the whole method someMethod() is thread safe. Even if the LocalObject instance is passed as parameter to other methods in the same class, or in other classes, the use of it is thread safe. The only exception is of course, if one of the methods called with the LocalObject as parameter, stores the LocalObject instance in a way that allows access to it from other threads.
I understand why the LocalObject instance is thread-safe. But I would like to see an example of the exception case (the last line in the above block quote). If someone could write a code snippet that fits what's written in the last line, that would be very helpful. Thank you!
public class SomeClass {
private LocalObject cachedLocalObject;
public void someMethod() {
LocalObject localObject = new LocalObject();
localObject.callMethod();
method2(localObject);
}
public void method2(LocalObject localObject) {
this.cachedLocalObject = localObject;
localObject.setValue("value");
}
public LocalObject getCachedLocalObject() { return cachedLocalObject; }
}
The combination of caching the object in method2() and then exposing it for external use in getCachedLocalObject() breaks threadsafety: some other thread can use getCachedLocalObject() to obtain and modify cachedLocalObject.
The contrast is demonstrated in the next part: "Object Members". As long as the object remains local to thread, it will be inherently thread-safe. But as soon as the reference is assigned to an object's field, any thread with a reference to the parent object can gain access to its fields, rendering them (potentially) not thread-safe.

java thread accessing outer object before it's created

Yes, this is an academic question, I know people will complain that I'm not posting any code
but I'm genuinely struck with this question, really don't know where to begin. I would really appreciate an explanation and maybe some code example.
If an object constructor starts a new thread that executes the method
run of an anonymous inner class object, it is possible that this new
thread can access its surrounding outer object before it has been
fully constructed and its fields fully initialized. How would you
prevent this from happening?
This is called "leaking this". Here you have the code
public class Test {
// this is guaranteed to be initialized after the constructor
private final int val;
public Test(int v) {
new Thread(new Runnable() {
#Override public void run() {
System.out.println("Val is " + val);
}
}).start();
this.val = v;
}
}
Guess what it will (may, since it's a thread) print. I used a final field to stress that the object is accessed before it has been fully initialized (final fields must be definitely assigned after the last line of every constructor)
How do you recover
You don't want to pass this around when you are in a constructor. This also mean you don't want to call non-final virtual methods in the very same class (non-static, non-private), and not using inner classes (anonymous classes are inner classes), that are implicitely linked to the enclosing instance, thus it's as they could access this.
Think about the single-threaded situation first:
Whenever you create an object via new, its constructor is called which (hopefully) initializes the fields of the new object before a reference to this object is returned. That is, from the point of view of the caller, this new is almost like an atomic operation:
Before calling new, there is no object. After returning from new, the object exists fully initialized.
So all is good.
The situation changes slightly when multiple threads come into play. But we have to read your quote carefully:
...has been fully constructed and its fields fully initialized.
The crucial point is fully. The subject line of your question says "before created", but what is meant here is not before the object has been created, but between object creation and initialization. In a multi-threaded situation, new can no longer be considered (pseudo-)atomic because of this (time flows from left to right):
Thread1 --> create object --> initialize object --> return from `new`
^
|
| (messing with the object)
Thread2 ------------------/
So how can Thread2 mess with the object? It would need a reference to that object but since new will only return the object after is both been created and initialized, this should be impossible, right?
Well, no - there is one way where it's still possible -- namely if Thread 2 is created inside the object's constructor. Then the situation would be like this:
Thread1 --> create object --> create Thread2 --> initialize object --> return from `new`
| ^
| |
| | (messing with the object)
\-----/
Since Thread2 is created after the object has been created (but before it has been fully initialized), there is already a reference to the object that Thread2 could get a hold of. One way is simply if the constructor of Thread2 explicitly takes a reference to the object as a parameter. Another way is by using a non-static inner class of the object for Thread2's run method.
I would change the title of the question, as threads are not accessing themselves, but the second one to the first one. I mean:
You have one thread, creating an object.
Inside the constructor for this object, you declare an anonymous inner class that implements Runnable.
In the same constructor of the first thread, you start a new thread to run your anonymous inner class.
Thus, you're having two threads. If you want to assure that the new thread doesn't do anything before the constructor is "fully ended", I would use some locks in the constructor. This way, the 2nd thread can be started but will wait until the first thread ends.
public class A {
int final number;
A() {
new Thread(
new Runnable() {
public void run() {
System.out.pritnln("Number: " + number);
}
}).start();
number = 2;
}
}
I do not fully agree with Pablos answer because it heavily depends on your initialization method.
public class ThreadQuestion {
public volatile int number = 0;
public static void main(String[] args) {
ThreadQuestion q = new ThreadQuestion();
}
public ThreadQuestion() {
Thread t = new Thread(new Runnable() {
#Override
public void run() {
System.out.println(number);
}
});
try {
Thread.sleep(500);
} catch(Exception e) {
e.printStackTrace();
}
number = 1;
t.start();
}
}
When you
place t.start() at the end, the correct data is printed.
place t.start() before the sleep command, it will print 0
remove the sleep command and place t.start() before the assignment it can print 1 (not determinable)
Play a mind game on 3.) you can say a "tiny" assignment of 1 simple data type will work as expected but if you create a database connection it will not achieve a reliable result.
Do not hesitate to raise any question.
So a situation like this?
public class MyClass {
private Object something;
public MyClass() {
new Thread() {
public void run() {
something = new Object();
}
}.start();
}
}
Depending on the actual code used, the behaviour could vary. This is why constructors should be carefully made so that they don't for example call non-private methods (a subclass could override it, allowing the superclass this to be accessed from a subclass before the superclass is fully initialized). Although this particular example deals with a single class and a thread, it's related to the reference leaking problem.

Creating an object of the Object class in Java

In Java, every class implicitly extends the Object class. So, does this mean we can create an object of the Object class ?
public static void main(String[] args) {
Object ob=new Object();
// code here ....
}
When I tried it, it compiled and ran successfully. In that case, can someone please explain when do we generally create an object of the Object class ?
You could instantiate an instance of an Object if you want to do a synchronization lock.
public void SomeClass {
private Object lock = new Object();
private SomeState state;
public void mutateSomeSharedState() {
synchronized(lock) {
//mutate some state
}
}
public SomeState readState() {
synchronized(lock) {
//access state
}
}
}
It might be necessary to do this when this is already used to lock some other state of the same object, or if you want to have your lock be private (ie, no one else can utilize it). Even if it isn't necessary, some people prefer to do things that way. This is merely an example of when someone might do it.
Normally we don't create an object of the Object class directly. Usually, we create instances of direct/indirect subclasses of Object.
A scenario where we create an instance of Object is to create an object to synchronize threads.
Eg:
Object lock = new Object();
//...
synchronize( lock ) {
//...
//...
}
However the Object class is used a lot to describe parameters of methods and of instance variables that may assume values of different classes (polymorphism).
Eg:
void example(Object arg) {
// ...
System.out.println( "example=" + arg.toString() );
}
Object foo = returnObject();
Sometimes the use of Generics may be better than using Object to describe parameters and variables.
For the most part I believe Object is no longer used explicitly.
Since Java's debut of Generics, casting to the Object class is almost non-existent.
Since java.lang.Object is the super most class, it can be substituted with any instance we create. This concept is very useful when you not aware of the type( eg: A method which conditionally returns different types , Collections with multiple types)
Also commonly used when you want to instantiate class from String,or execute a method using reflections.
However, direct usage of Object is getting redundant due to Generics.
Cheers
Satheesh

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