Creating an object of the Object class in Java - 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

Related

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.

Make an object read only in Java [duplicate]

If an object reference is passed to a method, is it possible to make the object "Read Only" to the method?
Not strictly speaking. That is, a reference that can mutate an object can not be turned into a reference that can not mutate an object. Also, there is not way to express that a type is immutable or mutable, other than using conventions.
The only feature that ensure some form of immutability would be final fields - once written they can not be modified.
That said, there are ways to design classes so that unwanted mutation are prevented. Here are some techniques:
Defensive Copying. Pass a copy of the object, so that if it is mutated it doesn't break your internal invariants.
Use access modifiers and/or interface to expose only read-only methods. You can use access modifieres (public/private/protected), possibly combined with interface, so that only certain methods are visible to the other object. If the methods that are exposed are read-only by nature, you are safe.
Make your object immutable by default. Any operation on the object returns actually a copy of the object.
Also, note that the API in the SDK have sometimes methods that return an immutable version of an object, e.g. Collections.unmodifiableList. An attempt to mutate an immutable list will throw an exception. This does not enforce immutability statically (at compile-time with the static type system), but is is a cheap and effective way to enforce it dynamically (at run-time).
There has been many research proposals of Java extension to better control of aliasing, and accessibility. For instance, addition of a readonly keyword. None of them is as far as I know planned for inclusion in future version of Java. You can have a look at these pointers if you're interested:
Why We Should Not Add ''Read-Only'' to Java (yet) -- it lists and compare most of the proposals
The Checker Framework: Custom pluggable types for Java -- a non intrusive way to extend the type system, notably with immutable types.
The Checker Framework is very interesting. In the Checker Framework, look at Generic Universe Types checker, IGJ immutability checker, and Javari immutability checker. The framework works using annotations, so it is not intrusive.
No, not without decorating, compositing, cloning, etc.
There's no general mechanism for that. You'll need to write special-case code to achieve it, like writing an immutable wrapper (see Collections.unmodifiableList).
You could achieve a similar thing in most cases by cloning the Object as the first statement of the method, such as this...
public void readOnlyMethod(Object test){
test = test.clone();
// other code here
}
So if you called readOnlyMethod() and pass in any Object, a clone of the Object will be taken. The clone uses the same name as the parameter of the method, so there's no risk of accidentally changing the original Object.
No. But you could try to clone the object before passing it, so any changes made by the method won't affect the original object.
making it implement a interface which has only read only methods (no setter methods) this gives a copy of an object (road-only copy) and returning the read only instance of interface instead of returning the instance of an object itself
You could define all parameters of the objects as final but that makes the object read only to everyone.
I believe your real question is about avoiding escape references.
As pointed out in some answers to extract an Interface from class and expose only get methods. It will prevent modification by accident but it is again not a foolproof solution to avoid above problem.
Consider below example:
Customer.java:
public class Customer implements CustomerReadOnly {
private String name;
private ArrayList<String> list;
public Customer(String name) {
this.name=name;
this.list = new ArrayList<>();
this.list.add("First");
this.list.add("Second");
}
#Override
public String getName() {
return name;
}
public void setName(String name) {
this.name = name;
}
#Override
public ArrayList<String> getList() {
return list;
}
public void setList(ArrayList<String> list) {
this.list = list;
}
}
CustomerReadOnly.java:
public interface CustomerReadOnly {
String getName();
ArrayList<String> getList();
}
Main.java:
public class Test {
public static void main(String[] args) {
CustomerReadOnly c1 = new Customer("John");
System.out.println("printing list of class before modification");
for(String s : c1.getList()) {
System.out.println(s);
}
ArrayList<String> list = c1.getList();
list.set(0, "Not first");
System.out.println("printing list created here");
for(String s : list) {
System.out.println(s);
}
System.out.println("printing list of class after modification");
for(String s : c1.getList()) {
System.out.println(s);
}
}
}
Ouput:
printing list of class before modification
First
Second
printing list created here
Not first
Second
printing list of class after modification
Not first
Second
So, as you can see extracting interface and exposing only get methods works only if you don't have any mutable member variable.
If you have a collection as a member variable whose reference you don't want to get escape from class, you can use Collections.unmodifiableList() as pointed out in ewernli's answer.
With this no external code can modify the underlying collection and your data is fully read only.
But again when it comes to custom objects for doing the same, I am aware of the Interface method only as well which can prevent modification by accident but not sure about the foolproof way to avoid reference escape.
Depending on where you want the rule enforced. If you are working collaboratively on a project, use final with a comment telling the next person they are not meant to modify this value. Otherwise wouldn't you simply write the method to not touch the object?
public static void main(String[] args) {
cantTouchThis("Cant touch this");
}
/**
*
* #param value - break it down
*/
public static void cantTouchThis(final String value) {
System.out.println("Value: " + value);
value = "Nah nah nah nah"; //Compile time error
}
So specifically to this method, the value will never be written to, and it is enforced at compile time making the solution extremely robust. Outside the scope of this method, the object remains unaltered without having to create any sort of wrapper.
private boolean isExecuteWriteQueue = false;
public boolean isWriting(){
final boolean b = isExecuteWriteQueue;
return b;
}
Expanding on ewernli's answer...
If you own the classes, you can use read-only interfaces so that methods using a read-only reference of the object can only get read-only copies of the children; while the main class returns the writable versions.
example
public interface ReadOnlyA {
public ReadOnlyA getA();
}
public class A implements ReadOnlyA {
#Override
public A getA() {
return this;
}
public static void main(String[] cheese) {
ReadOnlyA test= new A();
ReadOnlyA b1 = test.getA();
A b2 = test.getA(); //compile error
}
}
If you don't own the classes, you could extend the class, overriding the setters to throw an error or no-op, and use separate setters. This would effectively make the base class reference the read-only one, however this can easily lead to confusion and hard to understand bugs, so make sure it is well documented.

Java - how much space does anonymous object take?

supposing the following scenario:
void thisIsCalledManyTimes(){
// ....
someObject.executeIfNecessary( new Runnable(){
void run(){ //do sth here}
});
}
how much space would the anonymous object take? I understand that each object anonymous obj would have a pointer to the same implementation of run in its method lookup table.
The source code of Runnable doesn't specify any fields, and so the anonymous class won't take any more space than an Object, with two differences. An inner class has an implicit reference to the outer class instance, so you would want to factor this in. It will also take copies of final variables referenced from the outer class.
An anonymous class takes up much the same amount of space as a non-anonymous class. All objects keep a reference to the class they are an instance of.
The only thing different about an anonymous class is that it will contain a reference to
the outer class instance
a copy of any final variables used.
All instances of run will indeed share the same "pointer" to the method look-up table. However, the precise size of the object depends on the content of the run and the ... above it. The anonymous class the compiler generates provides space for storing the variables referenced by the run method, in addition to space for storing a reference to the this object of the enclosing class.
For example,
private String name;
void thisIsCalledManyTimes(){
final int value1 = 123;
final double value2 = 456.789.
someObject.executeIfNecessary( new Runnable(){
void run(){
System.out.println(name); // Referenced through the enclosing "this"
System.out.println(value1); // Variable will be inserted by the compiler
System.out.println(value2); // Variable will be inserted by the compiler
}
});
}
reserves the space in the object to store value1, value2, and this; each instance of the anonymous object will get a copy of these variables.
Use profiler to check size of your class' instance. It's rather small as there are no properties - it's just as large as instance of java.lang.Object. It depends on JVM version.
If you need to make this more efficent just do :
void thisIsCalledManyTimes(){
// ....
if (someObject.isItNecessary()) {
someObject.execute( new Runnable(){
void run(){ //do sth here}
});
}
}
However I won't bother about this too much as this.

How does "this" escape the constructor in 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

Categories

Resources