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How is a child object constructed in Java?

In java, how is a child object constructed?
I just started inheritance and these few points are not very clear to me:
Does the child object depend only on the child class' constructor, or does it also depend on the parent's constructor? I need some details about that point.
Also, is super() always called by default in a child constructor?
Any other information regarding this topic is appreciated.

I don't think "A child object" is a good way to think about this.
You're making an object. Like all objects, it is an instance of some specific class, (After all, new SomeInterface() does not compile) and like (almost) all objects, it is made because some code someplace (doesn't have to be your code, of course) ran the java expression new SomeSpecificClass(args); somewhere.
We could say it is a 'child object' because SomeSpecificClass is a child class of some other class.
But that's rather useless. That means the only way to ever make a new 'non-child' object would be to write new Object(); - after all, all classes except java.lang.Object are a child class: If you write public class Foo {}, java will interpret that exactly the same as if you had written public class Foo extends java.lang.Object {}, after all.
So, barring useless* irrelevancies, all objects are child objects, and therefore as a term, 'child object', I'd not use that.
That also means that ALL object creation goes through this 'okay and in what order and how do the constructors work' song and dance routine.
How it works is probably most easily explained by desugaring it all. Javac (the compiler) injects things if you choose to omit them, because a lot of things that feel optional (such as a constructor, a super call, or an extend clause), at the class file / JVM level, aren't**.
Sugar #1 - extends clause
Already covered: if you have no extends clause on your class def, javac injects extends java.lang.Object for you.
Sugar #2 - no super call in constructor
A constructor must either call some specific super constructor on its very first line, or, it it must call some other constructor from the same class on its very first line (this(arg1, arg2);). If you don't, java will inject it for you:
public MyClass(String arg) { this.arg = arg; }
// is treated as:
public MyClass(String arg) {
super();
this.arg = arg;
}
Notably including a compiler error if your parent class has no zero-arg constructor available.
Sugar #3: No constructor
If you write a class that has no constructor, then java makes one for you:
public YourClass() {}
It will be public, it will have no args, and it will have no code on it. However, as per sugar #2 rule, this then gets expanded even further, to:
public YourClass() {super();}
Field inits and code blocks get rewritten to a single block.
The constructor isn't the only thing that runs when you make new objects. Imagine this code:
public class Example {
private final long now = System.currentTimeMillis();
}
This code works; you can compile it. You can make new instances of Example, and the now field will hold the time as it was when you invoked new Example(). So how does that work? That feels a lot like constructor code, no?
Well, this is how it works: Go through the source file top to bottom and find every non-static initializing code you can find:
public class Example {
int x = foo(); // all non-constant initial values count
{
x = 10;
// this bizarre constructor is legal java, and also
// counts as an initializer.
}
}
and then move all that over to the one and only initializer that classes get, in the order you saw them.
Ordering
So, via sugar rules we have reduced ALL classes to adhere to the following rules:
ALL classes have a parent class.
ALL classes have at least 1 constructor.
ALL constructors invoke either another constructor or a constructor from parent.
There is one 'initializer' code block.
Now the only question is, in what order are things executed?
The answer is crazy. Hold on to your hats.
This is the order:
First, set all fields to 0/false/null of the entire 'construct' (the construct involves every field from Child all the way down to Object, of course).
Start with the actual constructor invoked on Child. Run it directly, which means, start with the first line, which neccessarily is either a this() or a super() invocation.
Evaluate the entire line, notably, evaluate all expressions passed as arguments. Even if those are themselves invocations of other methods. But, javac will do some minor effort to try to prevent you from accessing your fields (because those are all uninitialized! I haven't mentioned initializers yet!!).
Yeah, really. This means this:
public class Example {
private final long x = System.currentTimeMillis();
public Example() {
super(x); // x will be .... 0
// how's that for 'final'?
}
}
This will either end up invoking the first line of some other constructor of yours (which is itself also either a this() or a super() call). Either we never get out of this forest and a stack overflow error aborts our attempt to create this object (because we have a loop of constructors that endlessly invoke each other), or, at some point, we run into a super() call, which means we now go to our parent class and repeat this entire song and dance routine once more.
We keep going, all the way to java.lang.Object, which by way of hardcoding, has no this() or super() call at all and is the only one that does.
Then, we stop first. Now the job is to run the rest of the code in the constructor of j.l.Object, but first, we run Object's initializer.
Then, object's constructor runs all the rest of the code in it.
Then, Parent's initializer is run. And then the rest of the parent constructor that was used. and if parent has been shifting sideways (this() invokes in its constructors), those are all run in reverse order as normal in method invocations.
We finally end up at Child; its initializer runs, then the constructor(s) run in order, and finally we're done.
Show me!
class Parent {
/* some utility methods so we can run this stuff */
static int print(String in) {
System.out.println("#" + in);
return 0;
// we use this to observe the flow.
// as this is a static method it has no bearing on constructor calls.
}
public static void main(String[] args) {
new Child(1, 2);
}
/* actual relevant code follows */
Parent(int arg) {
print("Parent-ctr");
print("the result of getNow: " + getNow());
}
int y = print("Parent-init");
long getNow() { return 10; }
}
class Child extends Parent {
Child(int a, int b) {
this(print("Child-ctr1-firstline"));
print("Child-ctr1-secondline");
}
int x = print("Child-init");
Child(int a) {
super(print("Child-ctr2-firstline"));
print("Child-ctr2-secondline");
}
final long now = System.currentTimeMillis();
#Override long getNow() { return now; }
}
and now for the great puzzler. Apply the above rules and try to figure out what this will print.
#Child-ctr1-firstline
#Child-ctr2-firstline
#Parent-init
#Parent-ctr
#the result of getNow: 0
#Child-init
#Child-ctr2-secondline
#Child-ctr1-secondline
Constructor execution ordering is effectively: the first line goes first, and the rest goes last.
a final field was 0, even though it seems like it should never be 0.
You always end up running your parent's constructor.
--
*) You can use them for locks or sentinel pointer values. Let's say 'mostly useless'.
**) You can hack a class file so that it describes a class without a parent class (not even j.l.Object); that's how java.lang.Object's class file works. But you can't make javac make this, you'd have to hack it together, and such a thing would be quite crazy and has no real useful purpose.

In inheritance, the construction of a child object depends on at least one parent constructor.
Calling the super () method is not mandatory. By default, Java will call the parent constructor without argument except if you precise a custom constructor.
Here an example
Mother
public class Mother {
int a;
public Mother() {
System.out.println("Mother without argument");
a = 1;
}
public Mother(int a) {
System.out.println("Mother with argument");
this.a = a;
}
}
child
public class Child extends Mother {
public Child() {
System.out.println("Child without argument");
}
public Child(int a) {
super(a);
System.out.println("Child with argument");
}
}
If you do this :
Child c1 = new Child();
you will get :
Mother without argument
Child without argument
If you do this :
Child c1 = new Child(a);
You will get :
Mother with argument
Child with argument
But if you change the second child constructor to and remove the super(arg) the parent constructor without argument will be called :
public Child(int a) {
// super(a);
System.out.println("Child with argument");
}
You will get :
Mother without argument
Child with argument
May be this course for beginners can help you Coursera java inheritance

In Java, is there a difference between initializing an object in constructor or in declaration? [duplicate]

Is there any advantage for either approach?
Example 1:
class A {
B b = new B();
}
Example 2:
class A {
B b;
A() {
b = new B();
}
}

There is no difference - the instance variable initialization is actually put in the constructor(s) by the compiler.
The first variant is more readable.
You can't have exception handling with the first variant.
There is additionally the initialization block, which is as well put in the constructor(s) by the compiler:
{
a = new A();
}
Check Sun's explanation and advice
From this tutorial:
Field declarations, however, are not part of any method, so they cannot be executed as statements are. Instead, the Java compiler generates instance-field initialization code automatically and puts it in the constructor or constructors for the class. The initialization code is inserted into a constructor in the order it appears in the source code, which means that a field initializer can use the initial values of fields declared before it.
Additionally, you might want to lazily initialize your field. In cases when initializing a field is an expensive operation, you may initialize it as soon as it is needed:
ExpensiveObject o;
public ExpensiveObject getExpensiveObject() {
if (o == null) {
o = new ExpensiveObject();
}
return o;
}
And ultimately (as pointed out by Bill), for the sake of dependency management, it is better to avoid using the new operator anywhere within your class. Instead, using Dependency Injection is preferable - i.e. letting someone else (another class/framework) instantiate and inject the dependencies in your class.

Another option would be to use Dependency Injection.
class A{
B b;
A(B b) {
this.b = b;
}
}
This removes the responsibility of creating the B object from the constructor of A. This will make your code more testable and easier to maintain in the long run. The idea is to reduce the coupling between the two classes A and B. A benefit that this gives you is that you can now pass any object that extends B (or implements B if it is an interface) to A's constructor and it will work. One disadvantage is that you give up encapsulation of the B object, so it is exposed to the caller of the A constructor. You'll have to consider if the benefits are worth this trade-off, but in many cases they are.

I got burned in an interesting way today:
class MyClass extends FooClass {
String a = null;
public MyClass() {
super(); // Superclass calls init();
}
#Override
protected void init() {
super.init();
if (something)
a = getStringYadaYada();
}
}
See the mistake? It turns out that the a = null initializer gets called after the superclass constructor is called. Since the superclass constructor calls init(), the initialization of a is followed by the a = null initialization.

my personal "rule" (hardly ever broken) is to:
declare all variables at the start of
a block
make all variables final unless they
cannot be
declare one variable per line
never initialize a variable where
declared
only initialize something in a
constructor when it needs data from
the constructor to do the
initialization
So I would have code like:
public class X
{
public static final int USED_AS_A_CASE_LABEL = 1; // only exception - the compiler makes me
private static final int A;
private final int b;
private int c;
static
{
A = 42;
}
{
b = 7;
}
public X(final int val)
{
c = val;
}
public void foo(final boolean f)
{
final int d;
final int e;
d = 7;
// I will eat my own eyes before using ?: - personal taste.
if(f)
{
e = 1;
}
else
{
e = 2;
}
}
}
This way I am always 100% certain where to look for variables declarations (at the start of a block), and their assignments (as soon as it makes sense after the declaration). This winds up potentially being more efficient as well since you never initialize a variable with a value that is not used (for example declare and init vars and then throw an exception before half of those vars needed to have a value). You also do not wind up doing pointless initialization (like int i = 0; and then later on, before "i" is used, do i = 5;.
I value consistency very much, so following this "rule" is something I do all the time, and it makes it much easier to work with the code since you don't have to hunt around to find things.
Your mileage may vary.

Example 2 is less flexible. If you add another constructor, you need to remember to instantiate the field in that constructor as well. Just instantiate the field directly, or introduce lazy loading somewhere in a getter.
If instantiation requires more than just a simple new, use an initializer block. This will be run regardless of the constructor used. E.g.
public class A {
private Properties properties;
{
try {
properties = new Properties();
properties.load(Thread.currentThread().getContextClassLoader().getResourceAsStream("file.properties"));
} catch (IOException e) {
throw new ConfigurationException("Failed to load properties file.", e); // It's a subclass of RuntimeException.
}
}
// ...
}

Using either dependency injection or lazy initialization is always preferable, as already explained thoroughly in other answers.
When you don't want or can't use those patterns, and for primitive data types, there are three compelling reasons that I can think of why it's preferable to initialize the class attributes outside the constructor:
avoided repetition = if you have more than one constructor, or when you will need to add more, you won't have to repeat the initialization over and over in all the constructors bodies;
improved readability = you can easily tell with a glance which variables will have to be initialized from outside the class;
reduced lines of code = for every initialization done at the declaration there will be a line less in the constructor.

I take it is almost just a matter of taste, as long as initialization is simple and doesn't need any logic.
The constructor approach is a bit more fragile if you don't use an initializer block, because if you later on add a second constructor and forget to initialize b there, you'll get a null b only when using that last constructor.
See http://java.sun.com/docs/books/tutorial/java/javaOO/initial.html for more details about initialization in Java (and for explanations on initalizer blocks and other not well known initialization features).

I've not seen the following in the replies:
A possible advantage of having the initialisation at the time of declaration might be with nowadays IDE's where you can very easily jump to the declaration of a variable (mostly
Ctrl-<hover_over_the_variable>-<left_mouse_click>) from anywhere in your code. You then immediately see the value of that variable. Otherwise, you have to "search" for the place where the initialisation is done (mostly: constructor).
This advantage is of course secondary to all other logical reasonings, but for some people that "feature" might be more important.

Both of the methods are acceptable. Note that in the latter case b=new B() may not get initialized if there is another constructor present. Think of initializer code outside constructor as a common constructor and the code is executed.

I think Example 2 is preferable. I think the best practice is to declare outside the constructor and initialize in the constructor.

The second is an example of lazy initialization. First one is more simple initialization, they are essentially same.

There is one more subtle reason to initialize outside the constructor that no one has mentioned before (very specific I must say). If you are using UML tools to generate class diagrams from the code (reverse engineering), most of the tools I believe will note the initialization of Example 1 and will transfer it to a diagram (if you prefer it to show the initial values, like I do). They will not take these initial values from Example 2. Again, this is a very specific reason - if you are working with UML tools, but once I learned that, I am trying to take all my default values outside of constructor unless, as was mentioned before, there is an issue of possible exception throwing or complicated logic.

The second option is preferable as allows to use different logic in ctors for class instantiation and use ctors chaining. E.g.
class A {
int b;
// secondary ctor
A(String b) {
this(Integer.valueOf(b));
}
// primary ctor
A(int b) {
this.b = b;
}
}
So the second options is more flexible.

It's quite different actually:
The declaration happens before construction. So say if one has initialized the variable (b in this case) at both the places, the constructor's initialization will replace the one done at the class level.
So declare variables at the class level, initialize them in the constructor.

class MyClass extends FooClass {
String a = null;
public MyClass() {
super(); // Superclass calls init();
}
#Override
protected void init() {
super.init();
if (something)
a = getStringYadaYada();
}
}
Regarding the above,
String a = null;
null init could be avoided since anyway it's the default.
However, if you were to need another default value,
then, because of the uncontrolled initialization order,
I would fix as follow:
class MyClass extends FooClass
{
String a;
{
if( a==null ) a="my custom default value";
}
...

How can i call a variable from another method

I need to know how to call a variable from one method to another
Can anyone help me?
public static void number(){
number = 1;
}
public static void callNumber(){
/*How can I call number to this method???
*/
}

Actually, "call a variable from an other method" is not very explicit, since a variable in a method is either global (used in the method but naturally available in the entire program), or a local variable of the method.
And in this last situation it is impossible to get this value.
Then either you declare your variable externally and it is trivial, or you specifiy a type value to your method "number()":
public static int number() {
int number = ...;
return number;
}
and you call it:
public static void callNumber() {
int numberReturned = number();
// other things...
}
Note: your code number = 1; specifies that your variable is global...
The trick is to set "number" available either by the return of the method, or by specifying this variable global.
I don't know if I've answered your question, if not try to be more explicit.

Between static methods, variables can be shared by making them global,
or by sending them as parameters(noas described by #Gaétan Séchaud).
However, if those two methods has a continuos connection between them, and they handle some variables needed to be shared, it smells like a class is needed.

how to declare a helper method in java, to be used locally being able to reference variables

I have a class with several methods. Now I would like to define a helper method that should be only visible to method A, like good old "sub-functions" .
public class MyClass {
public methodA() {
int visibleVariable=10;
int result;
//here somehow declare the helperMethod which can access the visibleVariable and just
//adds the passed in parameter
result = helperMethod(1);
result = helperMethod(2);
}
}
The helperMethod is only used by MethodA and should access MethodA's declared variables - avoiding passing in explicitly many parameters which are already declared within methodA.
Is that possible?
EDIT:
The helper mehod is just used to avoid repeating some 20 lines of code which differ in only 1 place. And this 1 place could easily be parameterized while all the other variables in methodA remain unchanged in these 2 cases

Well you could declare a local class and put the method in there:
public class Test {
public static void main(String[] args) {
final int x = 10;
class Local {
int addToX(int value) {
return x + value;
}
}
Local local = new Local();
int result1 = local.addToX(1);
int result2 = local.addToX(2);
System.out.println(result1);
System.out.println(result2);
}
}
But that would be a very unusual code. Usually this suggests that you need to take a step back and look at your design again. Do you actually have a different type that you should be creating?
(If another type (or interface) already provided the right signature, you could use an anonymous inner class instead. That wouldn't be much better...)

Given the variables you declare at the top of your method can be marked as final (meaning they don't change after being initialized) You can define your helper method inside a helper class like below. All the variables at the top could be passed via the constructor.
public class HelperClass() {
private final int value1;
private final int value2;
public HelperClass(int value1, int value2) {
this.value1 = value1;
this.value2 = value2;
}
public int helperMethod(int valuex) {
int result = -1;
// do calculation
return result;
}
}
you can create an instance of HelperClass and use it inside the method

It is not possible. It is also not good design. Violating the rules of variable scope is a sure-fire way to make your code buggy, unreadable and unreliable. If you really have so many related variables, consider putting them into their own class and giving a method to that class.
If what you mean is more akin to a lambda expression, then no, this is not possible in Java at this time (but hopefully in Java 8).

No, it is not possible.
I would advise you create a private method in your class that does the work. As you are author of the code, you are in control of which other methods access the private method. Moreover, private methods will not be accessible from the outside.
In my experience, methods should not declare a load of variables. If they do, there is a good chance that your design is flawed. Think about constants and if you couldn't declare some of those as private final variables in your class. Alternatively, thinking OO, you could be missing an object to carry those variables and offer you some functionality related to the processing of those variables.

methodA() is not a method, it's missing a return type.
You can't access variables declared in a method from another method directly.
You either has to pass them as arguments or declare methodA in its own class together with the helpermethods.
This is probably the best way to do it:
public class MyClass {
public void methodA() {
int visibleVariable=10;
int result;
result = helperMethod(1, visibleVariable);
result = helperMethod(2, visibleVariable);
}
public int helperMethod(int index, int visibleVariable) {
// do something with visibleVariable
return 0;
}
}

Final keyword in method signatures [duplicate]

This question already has answers here:
Closed 10 years ago.
Possible Duplicate:
Final arguments in interface methods - what’s the point?
While trying to experiment a few things, I've ran into a problem that it's described in this page.
interface B {
public int something(final int a);
}
abstract class C {
public int other(final int b);
}
class A extends C implements B {
public int something(int a) {
return a++;
}
public int other(int b) {
return b++
}
}
Why is such feature possible? I don't know why it's possible to to make a final parameter into a non-final one by just overriding the method. Why is the final keyword ignored in a method signature? And how do I obligate sub-classes to use in their methods final variables?

Java passes arguments to a method by value.
Therefore, no changes to a parameter can propagate back to the caller. It follows that whether or not the parameter is declared final makes absolutely no difference to the caller. As such, it is part of the implementation of the method rather than part of its interface.
What's your motivation for wanting to "obligate sub-classes to use in their methods final variables"?

final for a parameter only means that the value must not be changed within the method body. This is not a part of the method signature, and is not relevant to subclasses.
It should be invalid to have final parameters in interface or abstract methods, because it's meaningless.

Final variables are the only ones that can be used in closures. So if you want to do something like this:
void myMethod(int val) {
MyClass cls = new MyClass() {
#override
void doAction() {
callMethod(val); // use the val argument in the anonymous class - closure!
}
};
useClass(cls);
}
This won't compile, as the compiler requires val to be final. So changing the method signature to
void myMethod(final int val)
will solve the problem. Local final variable will do just as well:
void myMethod(int val) {
final int val0;
// now use val0 in the anonymous class

Java's final is not C++ const; there is no such thing as const-correctness in Java.
In Java, one achieves const-ness using immutable classes. It turns out to be quite effective because unlike C++, one cannot simply mess with memory. (You can use Field.setAccessible(true), and then use Reflection. But even that corruption-vector can be prevented by running the JVM with an appropriately configured security manager.)

The final keyword for arguments is not part of the method signature, and is only important for the body of the method, because Java passes all arguments by value (a copy of the value is always made for the method call).
I only use the final keyword (for arguments) if the compiler forces me to make it final, because the argument is used inside an anonymous class defined in the method.

In Java parameters are passed by value. Whether a parameter is final or not only affects that method, not the caller. I don't see why a class needs to obligate the subtypes.

Note that final parameters have one main purpose: you can't assign new values to them.
Also note that parameters are always passed by value, thus the caller won't see any assignments to the parameter inside the method.
If you really want to force parameters to be final (in order to prevent bugs that might be introduced when reassigning a parameter accidentially), employ a code anaylzer such as checkstyle.