Question is about Java. There are three ways of obtaining classes of a type (note that code is just a collection of snippets for demonstration):
Class c1 = Class.forName("com.mypkg.MyClass"); //assumes initialize = true
// - OR -
Class c1 = Class.forName("com.mypkg.MyClass", true/false,
this.getClass().getClassLoader());
Class c2 = com.mypkg.MyClass.class;
// - OR -
import com.mypkg.MyClass;
Class c2 = MyClass.class;
MyClass mc = new MyClass();
Class c3 = mc.getClass();
My question is about initialization. For method 1, I can control whether class is initialized or not using the boolean argument. For method 3, since an object is created, obviously the class is initialized (right?).
But what about method 2? Does calling .class initialize the class if it has not already been initialized?And how can one programmatically check whether a class is already initialized or not?
Summary answer:
Please see accepted solution and its comments for full answer. This is just a summary.
Calling .class does not initialize the class if it has not already been initialized. You can check whether a class is getting initialized by using a static block in the class definition that prints a message.
Reason for original question:
According to javadoc of Class.forName, "A call to forName("X") causes the class named X to be initialized.". Also there are two Class.forName methods including one that accepts a boolean parameter called initialize. According to javadoc, "The class is initialized only if the initialize parameter is true and if it has not been initialized earlier."
How about testing it yourself, to see what your VM does?
Just use this class with all three methods (in independent runs, as it will only be initialized once at most!)
class Example {
static {
System.out.println("Class was initialized!");
}
public static int bananas = 0;
}
for details when a class is supposed to be initialized, see http://docs.oracle.com/javase/specs/#12.4.1
Roughly speaking, a class is supposed to be initialized when either
the first instance is created
a static method is invoked
a non-final static field is used
some more complex corner cases
So essentially, when you first use anything of the actual class, not just reference it.
But e.g. Example.bananas = 1; should also trigger class initialization, without having an instance.
On the bytecode level, the reference to the class is loaded by the one of the ldc instructions. The VM specification does not mention that the classes are initialized, so the safe thing to say is probably that the VM does not guarantee that such a reference initializes the class, but do mind that the VM specification does not require a VM to do initialization as lazily as possible either.
The class may legally be initialized at any time between it being first referenced and initialization being strictly required.
Related
This question already has answers here:
How come invoking a (static) method on a null reference doesn't throw NullPointerException?
(5 answers)
Closed 8 years ago.
Recently I was going through a page on javarevisited and found a block of code which asked the readers to determine what would be the output for it...
Though I got the output I am not satisfied with the result (WHICH CAME OUT TO BE "Hello") since I don't know how a static member is accessed from a null reference. What's happening in the background?
public class StaticDEMO {
private static String GREET = "Hello";
public static void main(String[] args) {
StaticDEMO demo = null;
System.out.println(demo.GREET);
// TODO code application logic here
}
}
This works because the JVM knows you are trying to access a static member on a specific class. Because you had to declare demo as a specific class (in this case a StaticDEMO), it knows to use that to find GREET.
To be clear, you don't run into this often (I actually had to type this code in to verify it, I can't say I've ever seen this). Mainly, it's good practice to always refer to static fields by their class, not an object instance (which may be null, as we can see!).
Meaning, prefer this:
System.out.println(StaticDEMO.GREET);
EDIT
I found a reference to this in the Java Specification: Chapter 15, Section 11: Field Access Expressions.
Example 15.11.1-2. Receiver Variable Is Irrelevant For static Field Access
The following program demonstrates that a null reference may be used to access a class (static) variable without causing an exception
[example not shown here for brevity]
Any method which is being decorated as static in Java means, that method is a class level memeber. That means, you do not need an object to accss static members. The static method/variable is maintained by the Class itself, and not by any instance of the class. Here in your example, the compiler already knows that the member is static member and it doesn't need any instance to access that static method.
Static members are stored with the Class, not with any specific instance of it. So, it doesn't matter that the instance is null - the member from the Class is still be accessible.
The JVM simply ignores null, because GREET is a class field and demo is irrelevant reference to refer Class field.
Static method not needed object reference to call it so you can call it even reference to the object is null.
Given the following classes:
public abstract class Super {
protected static Object staticVar;
protected static void staticMethod() {
System.out.println( staticVar );
}
}
public class Sub extends Super {
static {
staticVar = new Object();
}
// Declaring a method with the same signature here,
// thus hiding Super.staticMethod(), avoids staticVar being null
/*
public static void staticMethod() {
Super.staticMethod();
}
*/
}
public class UserClass {
public static void main( String[] args ) {
new UserClass().method();
}
void method() {
Sub.staticMethod(); // prints "null"
}
}
I'm not targeting at answers like "Because it's specified like this in the JLS.". I know it is, since JLS, 12.4.1 When Initialization Occurs reads just:
A class or interface type T will be initialized immediately before the first occurrence of any one of the following:
...
T is a class and a static method declared by T is invoked.
...
I'm interested in whether there is a good reason why there is not a sentence like:
T is a subclass of S and a static method declared by S is invoked on T.
Be careful in your title, static fields and methods are NOT inherited. This means that when you comment staticMethod() in Sub , Sub.staticMethod() actually calls Super.staticMethod() then Sub static initializer is not executed.
However, the question is more interesting than I thought at the first sight : in my point of view, this shouldn't compile without a warning, just like when one calls a static method on an instance of the class.
EDIT: As #GeroldBroser pointed it, the first statement of this answer is wrong. Static methods are inherited as well but never overriden, simply hidden. I'm leaving the answer as is for history.
I think it has to do with this part of the jvm spec:
Each frame (§2.6) contains a reference to the run-time constant pool (§2.5.5) for the type of the current method to support dynamic linking of the method code. The class file code for a method refers to methods to be invoked and variables to be accessed via symbolic references. Dynamic linking translates these symbolic method references into concrete method references, loading classes as necessary to resolve as-yet-undefined symbols, and translates variable accesses into appropriate offsets in storage structures associated with the run-time location of these variables.
This late binding of the methods and variables makes changes in other classes that a method uses less likely to break this code.
In chapter 5 in the jvm spec they also mention:
A class or interface C may be initialized, among other things, as a result of:
The execution of any one of the Java Virtual Machine instructions new, getstatic, putstatic, or invokestatic that references C (§new, §getstatic, §putstatic, §invokestatic). These instructions reference a class or interface directly or indirectly through either a field reference or a method reference.
...
Upon execution of a getstatic, putstatic, or invokestatic instruction, the class or interface that declared the resolved field or method is initialized if it has not been initialized already.
It seems to me the first bit of documentation states that any symbolic reference is simply resolved and invoked without regard as to where it came from. This documentation about method resolution has the following to say about that:
[M]ethod resolution attempts to locate the referenced method in C and its superclasses:
If C declares exactly one method with the name specified by the method reference, and the declaration is a signature polymorphic method (§2.9), then method lookup succeeds. All the class names mentioned in the descriptor are resolved (§5.4.3.1).
The resolved method is the signature polymorphic method declaration. It is not necessary for C to declare a method with the descriptor specified by the method reference.
Otherwise, if C declares a method with the name and descriptor specified by the method reference, method lookup succeeds.
Otherwise, if C has a superclass, step 2 of method resolution is recursively invoked on the direct superclass of C.
So the fact that it's called from a subclass seems to simply be ignored. Why do it this way? In the documentation you provided they say:
The intent is that a class or interface type has a set of initializers that put it in a consistent state, and that this state is the first state that is observed by other classes.
In your example, you alter the state of Super when Sub is statically initialized. If initialization happened when you called Sub.staticMethod you would get different behavior for what the jvm considers the same method. This might be the inconsistency they were talking about avoiding.
Also, here's some of the decompiled class file code that executes staticMethod, showing use of invokestatic:
Constant pool:
...
#2 = Methodref #18.#19 // Sub.staticMethod:()V
...
Code:
stack=0, locals=1, args_size=1
0: invokestatic #2 // Method Sub.staticMethod:()V
3: return
The JLS is specifically allowing the JVM to avoid loading the Sub class, it's in the section quoted in the question:
A reference to a static field (§8.3.1.1) causes initialization of only the class or interface that actually declares it, even though it might be referred to through the name of a subclass, a subinterface, or a class that implements an interface.
The reason is to avoid having the JVM load classes unnecessarily. Initializing static variables is not an issue because they are not getting referenced anyway.
The reason is quite simple: for JVM not to do extra work prematurely (Java is lazy in its nature).
Whether you write Super.staticMethod() or Sub.staticMethod(), the same implementation is called. And this parent's implementation typically does not depend on subclasses. Static methods of Super are not supposed to access members of Sub, so what's the point in initializing Sub then?
Your example seems to be artificial and not well-designed.
Making subclass rewrite static fields of superclass does not sound like a good idea. In this case an outcome of Super's methods will depend on which class is touched first. This also makes hard to have multiple children of Super with their own behavior. To cut it short, static members are not for polymorphism - that's what OOP principles say.
According to this article, when you call static method or use static filed of a class, only that class will be initialized.
Here is the example screen shot.
for some reason jvm think that static block is no good, and its not executed
I believe, it is because you are not using any methods for subclass, so jvm sees no reason to "init" the class itself, the method call is statically bound to parent at compile time - there is late binding for static methods
http://ideone.com/pUyVj4
static {
System.out.println("init");
staticVar = new Object();
}
Add some other method, and call it before the sub
Sub.someOtherMethod();
new UsersClass().method();
or do explicit Class.forName("Sub");
Class.forName("Sub");
new UsersClass().method();
When static block is executed Static Initializers
A static initializer declared in a class is executed when the class is initialized
when you call Sub.staticMethod(); that means class in not initialized.Your are just refernce
When a class is initialized
When a Class is initialized in Java After class loading, initialization of class takes place which means initializing all static members of class. A Class is initialized in Java when :
1) an Instance of class is created using either new() keyword or using reflection using class.forName(), which may throw ClassNotFoundException in Java.
2) an static method of Class is invoked.
3) an static field of Class is assigned.
4) an static field of class is used which is not a constant variable.
5) if Class is a top level class and an assert statement lexically nested within class is executed.
When a class is loaded and initialized in JVM - Java
that's why your getting null(default value of instance variable).
public class Sub extends Super {
static {
staticVar = new Object();
}
public static void staticMethod() {
Super.staticMethod();
}
}
in this case class is initialize and you get hashcode of new object().If you do not override staticMethod() means your referring super class method
and Sub class is not initialized.
Just having a play around with Java reflection and I think I'm getting the hang of it for the most part. I understand from this question/answer that, for the most part, I'm limited to static variables. If I have an instance of the class though, I can access non-static variables, which does make sense, I get that much.
Say I have the following two classes:
public class A
{
private static int _staticInt;
public static void main(String[] args)
{
B instanceOfB = new B();
}
}
public class B
{
private int _nonStaticInt;
public Game() {}
}
I understand how to access _staticInt, that's not an issue.
My understanding is that I can get the Field for _nonStaticInt in the same way (i.e. Field f = B.class.getDeclaredField("_nonStaticInt");). From other research (javadocs, trails, etc) I have gathered that I need an instance of B in order to get the value of _nonStaticInt.
So my question; Since main is static, is it possible to access instanceOfB in order to access the value of _nonStaticInt? I don't think it is possible, but I thought it's always best to consult people that are more knowledgable than myself before giving up on the idea.
Since main is static, is it possible to access instanceOfB in order to access the value of _nonStaticInt?
"No." Local variables (being in a static method or not) cannot be accessed with the Java Reflection API. Reflection only works at the type level, not the byte-code level2.
The stated understanding of the linked question is correct; reflection access of a non-static (instance) field logically requires an instance. That is, the issue then isn't about reflecting on the B type, the issue is about obtaining the B instance (which is assigned to a local variable) to reflect upon.
To do this the B instance has to be "bled" somehow - e.g. assigned to a static field or passed as an argument to a method/constructor from main1 - so that it can be used with reflection later as the object who's instance members are to be accessed.
The cleanest approach would probably be to pass the B instance down through the appropriate context (or "DI"), perhaps with the aide of IoC .. and maybe changing the type to avoid the use of reflection entirely.
1 Another possible way to "bleed" the B instance is to attach a debugger and inspect/use the local variable within the main methods executing frame - but this sounds like trying to swat a fly with a club.
2 Even tooling like BCEL/ASM wouldn't immediately help during the execution of the main method. Rather it would be used to deconstruct the method, add in the required hooks/code to "bleed" or use the instance created, and then construct a modified method to execute.
Yes, you can get the value of _nonStaticInt in that same way:
B instanceOfB = new B();
Field f = B.class.getDeclaredField("_nonStaticInt");
// Because the variable is private you need this:
f.setAccessible(true);
Object content = f.get(instanceOfB);
System.out.println(content);
The value will be 0, that is the default value for an int.
This question already has answers here:
Initialize class fields in constructor or at declaration?
(16 answers)
Closed 10 years ago.
What is the difference between the following two, and which is more preferable??
public class foo {
int i = 2;
}
public class foo {
int i;
foo() {
i = 2;
}
}
In your example, there is no difference in behavioural semantics. In Java, all instance field initializers (and instance blocks) are executed after superclass initialization, and before the body of the constructor; see JLS 12.5.
The difference lies in code readability and (in other examples) avoiding repetitious coding and fragility1. These need to be assessed on a case-by-case basis.
It is also worth noting that there are some cases where you have to initialize in the constructor; i.e. when the initialization depends on a constructor parameter.
1 - The repetitiousness and fragility issues are flip-sides of the same thing. If you have multiple constructors, the "initialize in constructor" approach tends to lead to repetition. And if you add extra fields, you might to add the initialization to all relevant constructors; i.e. fragility.
If you have two or more constructors and intialization value differs in each of them, then you should use constructor initialization as there is no way to do the same with member initialization...
however if you have just one constructor...you can use member initialization for better code clarity..
In particular this case there is no difference in these two variants. First variant is more preferable, because initializations of fields inside constructor, as usual, use external values from constructor arguments.
First of all I think the second example should look like this:
public class foo{
int i;
foo(){
i = 0;
}
}
Otherwise i is just a local variable in the C'tor scope.
Second, the first example shows initialization which is called before the class C'tor is invoked. this is good if you want this to happen no matter what C'tor is used.
It also enables you to declare i as readonly.
In your first example, i is an instance variable of class foo (better name would be Foo). It's initialised at class loading.
In your second example, i is also an instance varaible but in this case initialised in the foo() constructor.
There is no real difference here, and especially with primitives.
However, in a multi-threaded environment, if you do intend to initialise your ivars in your constructor, and those ivars are non-primitive, you need to avoid the risk of exposing a partially constructed object. The reason for this is that constructors aren't synchronised and can't have the synchronised keyword applied but then two threads can't be constructing the same object.
So, to avoid this, you should never expose this in your constructor. One way of doing so is to call non-final methods. Doing so, say calling an abstract method, allows some unknown code to do something with your unfinished object. Obviously, this can't be done if you initialise in your declaration.
p.s. I thought there was something on this in Effective Java but couldn't find anything.
As per standard book constructor is a special type of function which is used to initialize objects.As constructor is defined as a function and inside class function can have only two type either static or non static.My doubt is what constructor is ?
1.)As constructor is called without object so it must be static
Test test =new Test();//Test() is being called without object
so must be static
My doubt is if constructor is static method then how can we frequently used this inside
constructor
Test(){
System.out.println(this);
}
Does the output Test#12aw212 mean constructors are non-static?
Your second example hits the spot. this reference is available in the constructor, which means constructor is executed against some object - the one that is currently being created.
In principle when you create a new object (by using new operator), JVM will allocate some memory for it and then call a constructor on that newly created object. Also JVM makes sure that no other method is called before the constructor (that's what makes it special).
Actually, on machine level, constructor is a function with one special, implicit this parameter. This special parameter (passed by the runtime) makes the difference between object and static methods. In other words:
foo.bar(42);
is translated to:
bar(foo, 42);
where first parameter is named this. On the other hand static methods are called as-is:
Foo.bar(42);
translates to:
bar(42);
Foo here is just a namespace existing barely in the source code.
Constructors are non-static. Every method first parameter is implicit this (except static) and constructor is one of that.
Constructors are NOT static functions. When you do Test test =new Test(); a new Test object is created and then the constructor is called on that object (I mean this points to the newly created object).
The new keyword here is the trick. You're correct in noting that in general, if you're calling it without an object, a method is static. However in this special case (i.e., preceded by the new keyword) the compiler knows to call the constructor.
The new operator returns a reference to the object it created.
new Test(); // creates an instance.
The System.out.println(this); is called after the new operator has instantiated the object
Not static. Read about constructors http://www.javaworld.com/jw-10-2000/jw-1013-constructors.html.
Neither.
Methods can be divided into 2 types: static/non-static methods, aka class/instance methods.
But constructors are not methods.
When we talk about static class then it comes to our mind that methods are called with class name,But in case of constructor ,Constructor is initialized when object is created So this proves to be non-static.
Constructors are neither static (as called using class name) or non-static as executed while creating an object.
Static:
Temp t= new Temp();
The new operator creates memory in the heap area and passes it to the constructor as Temp(this) implicitly. It then initializes a non-static instance variable defined in a class called this to the local parameter variable this.
Below example is just for understanding the concept, if someone tries to compile it, it will give the compile-time error.
class Temp{
int a;
Temp this; //inserted by compiler.
Temp(Temp this){ //passed by compiler
this.this=this; // initialise this instance variable here.
this.a=10;//when we write only a=10; and all the non-static member access by this implicitly.
return this; // so that we can't return any value from constructor.
}
}
Constructor is static because:
It is helping to create object.
It is called without object.
Constructor is used to initialize the object and has the behavior of non-static methods,as non-static methods belong to objects so as constructor also and its invoked by the JVM to initialize the objects with the reference of object,created by new operator