This question already has answers here:
Are fields initialized before constructor code is run in Java?
(5 answers)
Closed 2 years ago.
I'm trying to understand why the output of the following Java code is as like this:
class A {
public A() {
System.out.print("A()");
}
public static void main(String[] args) {
}
}
class B {
public B() {
System.out.print("B()");
}
public static void main(String[] args) {}
}
class C extends A {
B b = new B();
}
public class E05SimpleInheritance {
public static void main(String args[]) {
new C();
}
}
Output:
"A()B()"
I would imagine that when the main method of the E05SimpleInheritance public class is called the following things should happen
Non-public class C is loaded and its fields are initialized(before calling the default constructor of class C)
Since its member 'b' is an object of class B, class B is loaded in memory
Since we construct an object of Class B its constructor is called which should print B()
Default constructor of C is called which automatically calls the constructor of the superclass A which print A()
So the final output should be B()A() which is obviously wrong so I do not really understand how the code flows in this case. Can you perhaps show me why is A()B() printed instead of B()A()
Your mistake is in step 1:
Non-public class C is loaded and its fields are initialized(before calling the default constructor of class C)
This is not what’s happening. In reality, non-static fields are initialised at the beginning of the constructor, not before it. And the base class constructor is implicitly (or explicitly) invoked even before that.
In other words, javac generates code for C which is equivalent to the following:
class C extends A {
B b;
C() {
super();
b = new B();
}
}
Related
This question already has answers here:
Java order of Initialization and Instantiation
(2 answers)
Closed 3 years ago.
Exactly when the Class members are initialized ?
In below code:
public class A{
public B b = new B();
private C c = new C(123);
public A (arg a){
// Do something
}
public someMethod(){
Log.i(TAG, " "+ b.getD());
}
}
main(){
A a = new A ("xyz"); }
When would be the Object of Class B & C created? And is it guaranteed to be created?
Should a professional app use such code or not ?
You can analyze the question this way:
class Scratch {
public static void main(String[] args) {
A a = new A ("xyz");
}
}
class A{
public B b = new B();
private C c = new C(123);
public A (String a){
System.out.println("new A()");
}
}
class B {
public B() {
System.out.println("new B()");
}
}
class C {
public C(int i) {
System.out.println("new C()");
}
}
which executes giving the following output:
new B()
new C()
new A()
which matches wiith the answer of #Jakir Hossain.
So: inline fields initializers are executed before code in constructor, following the same order which they are declared in.
Instances of classes B and C are created on A's instance creation, before A's constructor is executed. This ordering (and fields' initialization) are guaranteed.
When an object is created, the following things are done in the following order:
The object is allocated on the heap with the correct object type,
and all instance fields are "default initialized" to zero, false or
null.
The expressions in the super(...) or this(...) are evaluated and the
constructor for the next class up the chain is called. (This
recurses up the chain constructor, so that the Object constructor
gets executed first.)
The instance variable initializers and any instance initializer
blocks are executed in order.
The body of the constructor is executed.
The constructor returns.
Hope it helps you.
I wrote down this mini-program:
A class:
public class A
{
public A()
{
System.out.println(getS());
}
public String getS() { return s;}
}
B class:
public class B extends A
{
private String s = "hello2";
public String getS() { return s;}
}
main:
public static void main(String[] args)
{
B b = new B();
}
and it printed:
null
Why is that?
I know that the String that printed is B's string, but why it didn't initialized before?
According to this answer - the variable initialized before the constructor..
EDIT -
I edited the code so the unrelated code won't confuse
Here is what's going on: when you construct B, the first thing its constructor needs to do is constructing A. This is done before B's own field s is initialized.
A constructs its own s, and then calls getS. However, it does not get its own getS, because B provides an override for it. Recall that B.s has not been initialized yet. That is why you see null printed.
Follow-up reading: What's wrong with overridable method calls in constructors?
What is happening:
You create a B instance, this will call the super() so the constructor of A.
Here it will do the print using the getter getS(). This will use the getter of B since this is the type of this but in this getter, the String is not yet instanciate since it is still doing the super class construction, so it return null.
Note that the String s in A is hidden by the one in B
The order during an instance is :
the static (from the super then the class)
the super class declaration (statement then constructor)
the block statement
the constructor
As Seen with :
public class A{
static{System.out.println("sA");}
{System.out.println("A1");}
public Main() {
System.out.println("new A");
}
{System.out.println("A2");}
public static void main(String[] args) {
new A();
}
}
class B extends Main {
static{System.out.println("sB");}
{ System.out.println("B1"); }
public B() {
System.out.println("new B");
}
{ System.out.println("B2"); }
}
Output :
sA
sB
A1
A2
new A
B1
B2
new B
it prints null because you have polymorphism in java. You Overrided method getS(). So when you call it from A, you try to call getS() from class B. But you didn't create instance of class B yet, because you need to finish class A first. So String in class B haven't been initialized yet, because of it you get null.
I am facing a problem with the below program , it has two classes A and B and one method getObject() in Class A that has been overridden in class B, Class B extends Class A.
I am calling getObject() from the base class constructor which I believe it will call the overridden method in class B, and at runtime the error is null pointer exception, why the Object has not been initialized, even so it has?
class A {
Object object = new Object();
public A() {
System.out.println("A Class");
getObject();
}
public void getObject() {
System.out.println("Class A Version");
System.out.println(object.toString());
}
}
class B extends A {
Object object = new Object();
public B() {
System.out.println("B Class");
}
#Override
public void getObject() {
System.out.println("Class B Version");
System.out.println(object.toString());
}
}
public class Init {
public static void main(String[] args) {
new B();
}
}
output
A Class
Exception in thread "main" Class B Version
java.lang.NullPointerException
at net.mindview.util.B.getObject(Init.java:28)
at net.mindview.util.A.<init>(Init.java:8)
at net.mindview.util.B.<init>(Init.java:21)
at net.mindview.util.Init.main(Init.java:34)
According to this answer:
The correct order of initialisation is:
Static variable initialisers and static initialisation blocks, in textual order, if the class hasn't been previously initialised.
The super() call in the constructor, whether explicit or implicit.
Instance variable initialisers and instance initialisation blocks, in textual order.
Remaining body of constructor after super().
See sections §2.17.5-6 of the Java Virtual Machine Specification.
So basically, object in class B is not initialized yet during constructor of class A is being executed (point 2: implicit super() call).
Within B constructor you call A construct which is using B's getObject() which is using not initialized object from B.
Use debugger step by step and you will see the flow.
The NPE is coming from the class itself, not object. While the constructor is working, this is null.
B extends A, so there's an implicit super constructor call inserted. That prints your "A Class" line, but then the method call is equivalent to this.getObject();, and this is null.
I think the key point of this problem is that you should know that the constructor of super class (A in case) will be called first. let's assume antoher condition that A also extends C , in case, constructor of c will be called first.
so the order of call should be : 1 construtor of C, 2 constructor A and then constructor of B at the last.
For your case, because Constructor A was called first , at this point object is not initialized yet which cause the null pointer excetion
Execution orders are:
class A {
Object object = new Object(); // --------------- 2
public A() {
System.out.println("A Class"); // ---------- 3
getObject(); // ---------------------------- 4 (calls B.getObject())
}
public void getObject() {
System.out.println("Class A Version");
System.out.println(object.toString());
}
}
class B extends A {
Object object = new Object(); // --------------- 7 (not executed)
public B() {
System.out.println("B Class"); // ---------- 8 (not executed)
}
#Override
public void getObject() {
System.out.println("Class B Version"); // -- 5
System.out.println(object.toString()); // -- 6 (exception!)
}
}
public class Init {
public static void main(String[] args) {
new B(); // ------------------------------- 1
}
}
You should never call overridable method in constructors as it introduces ambiguity in terms of which method to call. In you case your super class A calls B's getObject(). At this Point B is yet not instantiated so the call fails and gives you NPE.
ot sure if a similar question do exists or have been answered earlier. I did a search but couldn't find anything related.
Please see following example. There's a class A with noargument contructor. Class B extends Class A but no default contructor exists. The output is B-> get c, and I think java uses the runtime polymorphism logic to invoke the B object method.
Here would like to understand if its correct and how Java works here (Maybe what's the underlying logic)
Thanks
public class A {
private String cStr="";
private String dStr="";
A()
{
getcStr();
}
public String getcStr() {
System.out.println("A-> get c");
return cStr;
}
public String getdStr() {
System.out.println("A-> get d");
getcStr();
return dStr;
}
}
public class B extends A{
private String cStr;
public String getcStr() {
System.out.println("B-> get c");
return cStr;
}
}
public static void main (String[] args)
{
A a= new B();
a.getdStr();
}
EDIT: the output for the a.getdStr() is A-> get d B-> get c
If a class doesn't have a default constructor, in your case B, then java will implicitly insert a no argument constructor something like this
public B() {
super();
}
To answer your question, Yes, due to run time polymorphism, Java invoke's B's getcStr() method. Because B's constructor is calling A's constructor via super()
This is a famous anti-pattern. And its anti-pattern because you are trying to invoke a overriden method on B before B is fully constructed
Imagine your B class was like this
public class B extends A{
private String cStr = "abc";
public String getcStr() {
System.out.println(cStr);
return cStr;
}
}
Now the output of your program would have been null instead of "abc" because B is not fully constructed yet (that means cStr is not initialized to "abc")
there is no default constructor for B so it'll call A() constructor.....
Now new B(); creates the object of B so java will first look into class B that is any procedure like getcstr() exists if procedure found there it will invoke it otherwise check in the super class for that procedure......
In Python, class methods can be inherited. e.g.
>>> class A:
... #classmethod
... def main(cls):
... return cls()
...
>>> class B(A): pass
...
>>> b=B.main()
>>> b
<__main__.B instance at 0x00A6FA58>
How would you do the equivalent in Java? I currently have:
public class A{
public void show(){
System.out.println("A");
}
public void run(){
show();
}
public static void main( String[] arg ) {
new A().run();
}
}
public class B extends A{
#Override
public void show(){
System.out.println("B");
}
}
I'd like to call B.main() and have it print "B", but clearly it will print "A" instead, since "new A()" is hardcoded.
How would you change "new A()" so that it's parameterized to use the class it's in when called, and not the hard-coded class A?
Static methods in java are not classmethods they are staticmethods. In general it is not possible to know which class reference the static method was called from.
Your class B does not have a main method and static methods are not inherited.
The only way I can see this happening is to find whatever is calling A.main( String[] arg ) and change it to call B.main instead.
B.main:
public static void main( String[] arg ) {
new B().run();
}
How is your program started? Is there a batch file, shortcut, etc? Something you can change? Where does A.main get called?
I think this isn't possible. Here's why:
In Java, the implementation of a method is determined by the instance's run-time type. So, to execute B.show(), you need to have an instance of B. The only way I could see to do this, if the method that constructs the instance is supposed to be inherited, is to use Class.newInstance() to construct an instance of a type that's not known at runtime.
The problem with that is that within a static method, you have no reference to the containing class, so you don't know whose newInstance method to call.
Why do you want to do this, though? There may be some better way to achieve whatever it is you want to achieve.
In your example I wouldn't put your main method inside of A. This is setup as the entry point into the system (you can't be in B if you are specifically entering into A).
In the example below I created class A, B, and C. Class C instantiates A and B and runs them. Notice that in C I created an A, a B, and another A that I instantiate as a B. My output is:
A
B
B
Hopefully this makes sense.
public class A {
public void show(){
System.out.println("A");
}
public void run(){
show();
}
}
public class B extends A {
#Override
public void show(){
System.out.println("B");
}
}
public class C {
public static void main(String[] args) {
A a = new A();
B b = new B();
A anothera = new B();
a.show();
b.show();
anothera.show();
}
}