Develop Reference

Java private member: why does this compile? [duplicate]

How is the compiler not complaining when I write the following code?
public class MyClass
{
private int count;
public MyClass(int x){
this.count=x;
}
public void testPrivate(MyClass o){
System.out.println(o.count);
}
}
Even though it is an instance of the same class in which testPrivate is written, shouldn't it give a compilation error at System.out.println(o.count)? After all, I am trying to access a private variable directly.
The code even runs fine.

A private member is accessible from any method within the class in which it is declared, regardless of whether that method accesses its own (this) instance's private member or some other instance's private member.
This is stated in JLS 6.6.1:
...Otherwise, if the member or constructor is declared private, then access is permitted if and only if it occurs within the body of the top level class (§7.6) that encloses the declaration of the member or constructor.
This feature of Java allows you to write methods that accept an instance of the class as an argument (for example - clone(Object other), compareTo(Object other)) without relying on the class having non private getters for all the private properties that need to be accessed.

Private fields are private to the class as a whole, not just to the object.
Other classes do not know that MyClass has a field called count; however, A MyClass object knows that another MyClass object has the count field.

Accessors are not security! They are encapsulation, to keep others from having to know about the code.
Consider if someone wrote a Quantum Bogo Sort, but disappeared once he quashed the last bug -- understanding the code causes one to be either deleted from the universe or to go mad.
Despite this minor drawback, if properly encapsulated, this should become your prefered sorting algorithm, as all fields and methods except Sort should be private.
You don't know how it works, and you don't want to know how it works, but it works and that's enough. If on the other hand, everything is public, and you have to understand how it does what it does to use it correctly -- that's just too much bother, I'll stick with quicksort.

Though it is the instance of the same class in which testPrivate is
written, but shouldn't it through a compiler error at
System.out.println(o.count);
No. It will never throw a compilation error.
This is much similar to what a simple getter and setter does or a copy constructor does. Remember we can access private members using this.
public MyClass {
private String propertyOne;
private String propertyTwo;
// cannot access otherObject private members directly
// so we use getters
// But MyClass private members are accessible using this.
public MyClass(OtherClass otherObject) {
this.propertyOne = otherObject.getPropertyOne();
this.propertyTwo = otherObject.calculatePropertyTwo();
}
public void setPropertyOne(String propertyOne) {
this.propertyOne = propertyOne;
}
public String getPropertyOne() {
return this.propertyOne;
}
}
Your testPrivate method accepts an instance of MyClass. Since testPrivate is a method inside MyClass, it will have access to private properties.
public void testPrivate(MyClass o) {
this.propertyOne = o.propertOne;
}
Methods defined inside the class will always have access to it's private members, through this. and instance variable.
But if you define testPrivate outside of MyClass then, you won't have access to private members. There you will have to use a method or a setter or a getter.

Methods, Variables and Constructors that are declared private can only be accessed within the declared class itself. Check the official documentation

Inheritance a big confusion for me [duplicate]

This is an interview question.
Does subclasses inherit private
fields?
I answered "No", because we can't access them using the "normal OOP way". But the interviewer thinks that they are inherited, because we can access such fields indirectly or using reflection and they still exist in the object.
After I came back, I found the following quote in the javadoc:
Private Members in a Superclass
A
subclass does not inherit the private
members of its parent class.
Do you know any arguments for the interviewer's opinion?

Most of the confusion in the question/answers here surrounds the definition of Inheritance.
Obviously, as #DigitalRoss explains an OBJECT of a subclass must contain its superclass's private fields. As he states, having no access to a private member doesn't mean its not there.
However. This is different than the notion of inheritance for a class. As is the case in the java world, where there is a question of semantics the arbiter is the Java Language Specification (currently 3rd edition).
As the JLS states (https://docs.oracle.com/javase/specs/jls/se8/html/jls-8.html#jls-8.2):
Members of a class that are declared
private are not inherited by
subclasses of that class. Only members
of a class that are declared protected
or public are inherited by subclasses
declared in a package other than the
one in which the class is declared.
This addresses the exact question posed by the interviewer: "do subCLASSES inherit private fields". (emphasis added by me)
The answer is No. They do not. OBJECTS of subclasses contain private fields of their superclasses. The subclass itself has NO NOTION of private fields of its superclass.
Is it semantics of a pedantic nature? Yes. Is it a useful interview question? Probably not. But the JLS establishes the definition for the Java world, and it does so (in this case) unambiguously.
EDITED (removed a parallel quote from Bjarne Stroustrup which due to the differences between java and c++ probably only add to the confusion. I'll let my answer rest on the JLS :)

Yes
It's important to realize that while there are two classes, there is only one object.
So, yes, of course it inherited the private fields. They are, presumably, essential for proper object functionality, and while an object of the parent class is not an object of the derived class, an instance of the derived class is mostly definitely an instance of the parent class. It could't very well be that without all of the fields.
No, you can't directly access them. Yes, they are inherited. They have to be.
It's a good question!
Update:
Err, "No"
Well, I guess we all learned something. Since the JLS originated the exact "not inherited" wording, it is correct to answer "no". Since the subclass can't access or modify the private fields, then, in other words, they are not inherited. But there really is just one object, it really does contain the private fields, and so if someone takes the JLS and tutorial wording the wrong way, it will be quite difficult to understand OOP, Java objects, and what is really happening.
Update to update:
The controversy here involves a fundamental ambiguity: what exactly is being discussed? The object? Or are we talking in some sense about the class itself? A lot of latitude is allowed when describing the class as opposed to the object. So the subclass does not inherit private fields, but an object that is an instance of the subclass certainly does contain the private fields.

No. Private fields are not inherited... and that's why Protected was invented. It is by design. I guess this justified the existence of protected modifier.
Now coming to the contexts. What you mean by inherited -- if it is there in the object created from derived class? yes, it is.
If you mean can it be useful to derived class. Well, no.
Now, when you come to functional programming the private field of super class is not inherited in a meaningful way for the subclass. For the subclass, a private field of super class is same as a private field of any other class.
Functionally, it's not inherited. But ideally, it is.
OK, just looked into Java tutorial they quote this:
Private Members in a Superclass
A subclass does not inherit the private members of its parent class. However, if the superclass has public or protected methods for accessing its private fields, these can also be used by the subclass.
refer: http://download.oracle.com/javase/tutorial/java/IandI/subclasses.html
I agree, that the field is there. But, subclass does not get any privilege on that private field. To a subclass, the private field is same as any private field of any other class.
I believe it's purely matter of point-of-view. You may mould the argument either side. It's better justify both way.

It depends on your definition of "inherit". Does the subclass still have the fields in memory? Definitely. Can it access them directly? No. It's just subtleties of the definition; the point is to understand what's really happening.

I will demonstrate the concept with code. Subclasses ACTUALLY inherit the private variables of super class. The only problem is that they are not accessible to the
child objects unless you provide public getters and setters for the private variables
in the super class.
Consider two class in package Dump. Child extends Parent.
If I remember correctly, a child object in memory consists of two regions. One is the parent part only and the other is the child part only. A child can access the private
section in the code of its parent only via a public method in the parent.
Think of it this way. Borat's father Boltok has a safe containing $100,000. He does not want to share his "private" variable safe. So, he does not provide a key for the safe. Borat inherits the safe. But, what good is it if he cannot even open it ? If only his
dad had provided the key.
Parent -
package Dump;
public class Parent {
private String reallyHidden;
private String notReallyHidden;
public String getNotReallyHidden() {
return notReallyHidden;
}
public void setNotReallyHidden(String notReallyHidden) {
this.notReallyHidden = notReallyHidden;
}
}//Parent
Child -
package Dump;
public class Child extends Parent {
private String childOnly;
public String getChildOnly() {
return childOnly;
}
public void setChildOnly(String childOnly) {
this.childOnly = childOnly;
}
public static void main(String [] args){
System.out.println("Testing...");
Child c1 = new Child();
c1.setChildOnly("childOnly");
c1.setNotReallyHidden("notReallyHidden");
//Attempting to access parent's reallyHidden
c1.reallyHidden;//Does not even compile
}//main
}//Child

No. They don't inherit it.
The fact some other class may use it indirectly says nothing about inheritance, but about encapsulation.
For instance:
class Some {
private int count;
public void increment() {
count++;
}
public String toString() {
return Integer.toString( count );
}
}
class UseIt {
void useIt() {
Some s = new Some();
s.increment();
s.increment();
s.increment();
int v = Integer.parseInt( s.toString() );
// hey, can you say you inherit it?
}
}
You can also get the value of count inside UseIt via reflection. It doesn't means, you inherit it.
UPDATE
Even though the value is there, it is not inherited by the subclass.
For instance a subclass defined as:
class SomeOther extends Some {
private int count = 1000;
#Override
public void increment() {
super.increment();
count *= 10000;
}
}
class UseIt {
public static void main( String ... args ) {
s = new SomeOther();
s.increment();
s.increment();
s.increment();
v = Integer.parseInt( s.toString() );
// what is the value of v?
}
}
This is exactly the same situation as the first example. The attribute count is hidden and not inherited by the subclass at all. Still, as DigitalRoss points out, the value is there, but not by means on inheritance.
Put it this way. If your father is wealthy and gives you a credit card, you can still buy thing with his money, but doesn't mean you have inherited all that money, does it?
Other update
It is very interesting though, to know why the attribute is there.
I frankly don't have the exact term to describe it, but it's the JVM and the way it works that loads also the "not inherited" parent definition.
We could actually change the parent and the subclass will still work.
For instance:
//A.java
class A {
private int i;
public String toString() { return ""+ i; }
}
// B.java
class B extends A {}
// Main.java
class Main {
public static void main( String [] args ) {
System.out.println( new B().toString() );
}
}
// Compile all the files
javac A.java B.java Main.java
// Run Main
java Main
// Outout is 0 as expected as B is using the A 'toString' definition
0
// Change A.java
class A {
public String toString() {
return "Nothing here";
}
}
// Recompile ONLY A.java
javac A.java
java Main
// B wasn't modified and yet it shows a different behaviour, this is not due to
// inheritance but the way Java loads the class
Output: Nothing here
I guess the exact term could be found here: The JavaTM Virtual Machine Specification

Well, my answer to interviewer's question is - Private members are not inherited in sub-classes but they are accessible to subclass or subclass's object only via public getter or setter methods or any such appropriate methods of original class. The normal practice is to keep the members private and access them using getter and setter methods which are public. So whats the point in only inheriting getter and setter methods when the private member they deal with are not available to the object? Here 'inherited' simply means it is available directly in the sub-class to play around by newly introduced methods in sub-class.
Save the below file as ParentClass.java and try it yourself ->
public class ParentClass {
private int x;
public int getX() {
return x;
}
public void setX(int x) {
this.x = x;
}
}
class SubClass extends ParentClass {
private int y;
public int getY() {
return y;
}
public void setY(int y) {
this.y = y;
}
public void setXofParent(int x) {
setX(x);
}
}
class Main {
public static void main(String[] args) {
SubClass s = new SubClass();
s.setX(10);
s.setY(12);
System.out.println("X is :"+s.getX());
System.out.println("Y is :"+s.getY());
s.setXofParent(13);
System.out.println("Now X is :"+s.getX());
}
}
Output:
X is :10
Y is :12
Now X is :13
If we try to use private variable x of ParentClass in SubClass's method then it is not directly accessible for any modifications (means not inherited). But x can be modified in SubClass via setX() method of original class as done in setXofParent() method OR it can be modified using ChildClass object using setX() method or setXofParent() method which ultimately calls setX(). So here setX() and getX() are kind of gates to the private member x of a ParentClass.
Another simple example is Clock superclass has hours and mins as private members and appropriate getter and setter methods as public. Then comes DigitalClock as a sub-class of Clock. Here if the DigitalClock's object doesn't contain hours and mins members then things are screwed up.

Ok, this is a very interesting problem I researched a lot and came to a conclusion that private members of a superclass are indeed available (but not accessible) in the subclass's objects. To prove this, here is a sample code with a parent class and a child class and I am writing child class object to a txt file and reading a private member named 'bhavesh' in the file, hence proving it is indeed available in the child class but not accessible due to the access modifier.
import java.io.Serializable;
public class ParentClass implements Serializable {
public ParentClass() {
}
public int a=32131,b,c;
private int bhavesh=5555,rr,weq,refw;
}
import java.io.*;
import java.io.Serializable;
public class ChildClass extends ParentClass{
public ChildClass() {
super();
}
public static void main(String[] args) {
ChildClass childObj = new ChildClass();
ObjectOutputStream oos;
try {
oos = new ObjectOutputStream(new FileOutputStream("C:\\MyData1.txt"));
oos.writeObject(childObj); //Writing child class object and not parent class object
System.out.println("Writing complete !");
} catch (IOException e) {
}
}
}
Open MyData1.txt and search for the private member named 'bhavesh'. Please let me know what you guys think.

It would seem that a subclass does inherit the private fields in that these very fields are utilized in the inner workings of the subclass (philosophically speaking). A subclass, in its constructor, calls the superclass constructor. The superclass private fields are obviously inherited by the subclass calling the superclass constructor if the superclass constructor has initialized these fields in its constructor. That's just an example. But of course without accessor methods the subclass cannot access the superclass private fields (it's like not being able to pop the back panel of an iPhone to take the battery out to reset the phone... but the battery is still there).
PS
One of the many definitions of inheritance that I have come across:
"Inheritance -- a programming technique that allows a derived class to extend the functionality of a base class, inheriting all of its STATE (emphasis is mine) and behaviour."
The private fields, even if not accessible by the subclass, are the inherited state of the superclass.

For example,
class Person {
private String name;
public String getName () {
return this.name;
}
Person(String name) {
this.name = name;
}
}
public class Student extends Person {
Student(String name) {
super(name);
}
public String getStudentName() {
return this.getName(); // works
// "return this.name;" doesn't work, and the error is "The field Person.name is not visible"
}
}
public class Main {
public static void main(String[] args) {
Student s = new Student("Bill");
String name = s.getName(); // works
// "String name = s.name;" doesn't work, and the error is "The field Person.name is not visible"
System.out.println(name);
}
}

Padding bits/Alignment and the inclusion of Object Class in the VTABLE is not considered. So the object of the subclass does have a place for the private members of the Super class. However, it cannot be accessed from the subclass's objects...

I would have to answer that private fields in Java are inherited. Allow me to demonstrate:
public class Foo {
private int x; // This is the private field.
public Foo() {
x = 0; // Sets int x to 0.
}
//The following methods are declared "final" so that they can't be overridden.
public final void update() { x++; } // Increments x by 1.
public final int getX() { return x; } // Returns the x value.
}
public class Bar extends Foo {
public Bar() {
super(); // Because this extends a class with a constructor, it is required to run before anything else.
update(); //Runs the inherited update() method twice
update();
System.out.println(getX()); // Prints the inherited "x" int.
}
}
If you run in a program Bar bar = new Bar();, then you will always see the number "2" in the output box. Because the integer "x" is encapsulated with the methods update() and getX(), then it can be proven that the integer is inherited.
The confusion is that because you can't directly access the integer "x", then people argue that it isn't inherited. However, every non-static thing in a class, be it field or method, is inherited.

No, private fields are not inherited. The only reason is that subclass can not access them directly.

I believe, answer is totally dependent on the question, which has been asked. I mean, if question is
Can we directly access the private field of the super-class from
their sub-class ?
Then answer is No, if we go through the access specifier details, it is mentioned, private members are accessible only within the class itself.
But, if question is
Can we access the private field of the super-class from
their sub-class ?
Which means, it doesn't matters, what you will do to access the private member. In that case, we can make public method in the super-class and you can access the private member. So, in this case you are creating one interface/bridge to access the private member.
Other OOPs language like C++, have the friend function concept, by which we can access the private member of other class.

We can simply state that when a superclass is inherited, then the private members of superclass actually become private members of the subclass and cannot be further inherited or are inacessible to the objects of subclass.

A private class member or constructor is accessible only within the body of the top level class (§7.6) that encloses the declaration of the member or constructor. It is not inherited by subclasses. https://docs.oracle.com/javase/specs/jls/se7/html/jls-6.html#jls-6.6

A subclass does not inherit the private members of its parent class. However, if the superclass has public or protected methods for accessing its private fields, these can also be used by the subclass
reference:
https://docs.oracle.com/javase/tutorial/java/IandI/subclasses.html#:~:text=Private%20Members%20in%20a%20Superclass,be%20used%20by%20the%20subclass.

I can try to help you.
When a subclass(named B, for example) extends a superclass (named A, for example), it automatically inherits fields (such as attributes and/or methods) from its superclass.
Now, B in its Memory Layout has the space for every field in class A even the private ones. The fact is that Java doesn't allow the subclass B to use the private fields because they are private.

As others have pointed out exerpt from JLS:
https://docs.oracle.com/javase/specs/jls/se8/html/jls-8.html#jls-8.2:
Members of a class that are declared private are not inherited by
subclasses of that class. Only members of a class that are declared
protected or public are inherited by subclasses declared in a package
other than the one in which the class is declared.
The answer is NO without a doubt, without any iffs and buts. That revolves around the definition of inheritance. By definition inheritance is for classes, not for objects. Objects are created using class definitions. Inheritance is just another block to add to definition of a class. So, does a class inherits any private member of super class? NO

A subclass does not inherit the private members of its parent class. However, if the superclass has public or protected methods for accessing its private fields, these can also be used by the subclass.

Private members (state and behavior) are inherited. They (can) affect the behavior and size of the object which is instantiated by the class. Not to mention that they are very well visible to the subclasses via all the encaptulation-breaking mechanisms that are available, or can be assumed by their implementers.
Although inheritance has a "defacto" definition, it definitely has no link to "visibility" aspects, which get assumed by the "no" answers.
So, there is no need to be diplomatic. JLS is just wrong at this point.
Any assumption that they are not "inherited" is unsafe and dangerous.
So among two defacto (partially) conflicting definitions (which I will not repeat), the only one that should be followed is the one that is safer (or safe).

Access a private variable from the superclass (JAVA)

Ok so I have studied java all semester and thought I had a clear understanding about inheritance and super/sub classes. Today we were given as assignment for making a superclass called enemy, with sub classes of different types of enemies. I did everything fine and all of my subclasses are working, but when I went back to read the guidelines we must follow, I found this sentence:
"All member variables of the super class must be private. Any access to a variable must be done through protected methods in the subclasses."
From what I have learned, this makes no sense to me. If a variable is private within the superclass, doesn't that disallow access even from a subclass? The last part that talks about protected methods in the subclasses also doesn't make any sense to me. How does this help and/or allow any access whatsoever to the super class?
From what I've learned about inheritance, Below is what i thought was true:
Access Levels
Modifier Class Package Subclass World
public Y Y Y Y
protected Y Y Y N
no modifier Y Y N N
private Y N N N
If I'm understanding something wrong here please do explain! I don't want to confront the instructor about giving us faulty instructions, if I'm the one not understanding it correctly!

The part
Any access to an a variable must be done through protected methods in the sub classes.
... just means that the subclasses have to call protected methods that are defined in the superclass. Since these methods are protected they can be accessed by the subclasses.
First you would define a base class like this:
public class Base {
private int x; // field is private
protected int getX() { // define getter
return x;
}
protected void setX(int x) { // define setter
this.x = x;
}
}
Then you would use it in your child class like this:
class Child extends Base{
void foo() {
int x = getX(); // we can access the method since it is protected.
setX(42); // this works too.
}
}

Probably the sentence is not worded correctly, from what I understand it makes perfect sense to me :
1. the superclass has private fields and protected methods to access them
2. the subclasses access the fields by using that methods.

Those are just the constraints of the assignment, not Java itself. You could give the superclass protected data members, and then access those directly from the subclass. However, the professor likely wishes to teach you about how a superclass can protect its data members from direct access by a subclass, and protected methods would be a way to do that.

When you inheritance other class, you cannot access your private attributes directly. So, if you have a class named "A" and other called "B", and make B extends A, B cannot access private attributes of A.
Think this like a protection. This way, you can write some attributes in class "A" that you dont want others classes access it through inheritance.
The "B" class can access only public, protected and default attributes directly in "A" class. But if you want to access a private attribute in "A" class for any reasons, you can write a method in "A" to return this attribute.
public class A{
private int foo;
public int getFoo(){
return this.foo;
}
}
public class B extends A{
public void doSomething(){
getFoo(); //return the private foo attribute of superclass
}
}

You are right in your thinking that, literally speaking, you can't access a superclass's private field. The text of your assignment uses wording which is not 100% strict, but is 100% customary in Java parlance: the so-called "getter" methods, also called "accessor" methods, are seen as "accessing the field", even though, strictly speaking, they merely return the current value of the field—which is definitely not the same as giving access to the field itself. You just need to get used to this (and many more) conventions in the Java jargon.

package com.action.product;
public class ProductAction{
public static String strCategoryName;
public String getStrCategoryName() {
return strCategoryName;
}
public void setStrCategoryName(String strCategoryName) {
this.strCategoryName = strCategoryName;
}
}
-----------------------------------------------------------------------
package com.DAO.product;
public class ProductDAO extends ProductAction {
#SuppressWarnings("static-access")
public String addnewProductValidation(){
System.out.println("======through_name======="+super.strCategoryName);
System.out.println("======through_getters======="+super.getStrCategoryName());
}
}
Can initialize variables as static. No need to create object

Private Access Specifier usage in Java Inheritance

We can access the Super Class methods which consists of operations on private data members and print the results.But why can't I print the private data members of Super Class with the SubClass object calling them in my main function? Someone please explain me.
Here is the example below.
class SuperClass1
{
private int a;
private int b;
SuperClass1(int p,int q)
{
a=p;
b=q;
}
SuperClass1(SuperClass1 obj)
{
a=obj.a;
b=obj.b;
}
SuperClass1()
{
a=-1;
b=-1;
}
int Vol()
{
return a*b;
}
}
class SubClass1 extends SuperClass1
{
int c;
SubClass1(int p,int q,int r)
{
super(p,q);
c=r;
}
SubClass1(SubClass1 obj)
{
super(obj);
c=obj.c;
}
SubClass1()
{
super();
c=-1;
}
}
public class Super
{
public static void main(String[] args)
{
SubClass1 obj1=new SubClass1();
//System.out.println("The values of obj1 are:"+obj1.a+""+obj1.b+""+obj1.c);
int vol=obj1.Vol();
System.out.println("The volume is :"+vol);
}
}

security and encapsulation
The superclass is letting its subclasses use only the public and protected methods/fields.
This allows the designer of the superclass to change the implementation of these methods if he sees it better, without breaking the subclass's correctness.
A text book example is a complex number class.
The programmer using this class only needs its functionality, he doesn't care if the implementation is with imaginary and real fields or with radius and theta fields [two distinct ways to represent complex number].
It allows the designer of the ComplexNumber class more freedom if he wants to change the class in later versions, and it also allows the user less worries: he doesn't need to take care for all the details, some are being taken care of for him.
Bonus: note you can break this behavior and access private fields and methods by using reflection - but when you do so - all bets are off, and you do it on your own responsibility.

Your question isn't very clear without an example, but I suspect that the "methods which consist of operations on private data members" aren't private. It doesn't matter that they work by accessing private data - they're not private themselves. It would be pretty pointless having access modifiers if public methods could only access other public members etc.
The whole point of encapsulation is that only the class itself should care about implementation details such as the fields in question, but can expose a contract in terms of its public (and protected) API. Code outside the class shouldn't care about the private implementation details.

JLS says:
Members of a class that are declared private are not inherited by
subclasses of that class. Only members of a class that are declared
protected or public are inherited by subclasses declared in a package
other than the one in which the class is declared.
So, to answer you question. No, private members are not accessible by subclasses.

Private members are not inherited; only the protected and public members are.
If possible, you can do one of the following:
Make the private properties of the superclass protected
Make public getters (and setters if needed) for the private properties

How to create an "abstract field"?

I know abstract fields do not exist in java. I also read this question but the solutions proposed won't solve my problem. Maybe there is no solution, but it's worth asking :)
Problem
I have an abstract class that does an operation in the constructor depending on the value of one of its fields.
The problem is that the value of this field will change depending on the subclass.
How can I do so that the operation is done on the value of the field redefined by the subclass ?
If I just "override" the field in the subclass the operation is done on the value of the field in the abstract class.
I'm open to any solution that would ensure that the operation will be done during the instantiation of the subclass (ie putting the operation in a method called by each subclass in the constructor is not a valid solution, because someone might extend the abstract class and forget to call the method).
Also, I don't want to give the value of the field as an argument of the constructor.
Is there any solution to do that, or should I just change my design ?
Edit:
My subclasses are actually some tools used by my main program, so the constructor has to be public and take exactly the arguments with which they will be called:
tools[0]=new Hand(this);
tools[1]=new Pencil(this);
tools[2]=new AddObject(this);
(the subclasses are Hand, Pencil and AddObject that all extend the abstract class Tool)
That's why I don't want to change the constructor.
The solution I'm about to use is to slightly change the above code to:
tools[0]=new Hand(this);
tools[0].init();
tools[1]=new Pencil(this);
tools[1].init();
tools[2]=new AddObject(this);
tools[2].init();
and use an abstract getter to acces the field.

How about abstract getter/setter for field?
abstract class AbstractSuper {
public AbstractSuper() {
if (getFldName().equals("abc")) {
//....
}
}
abstract public void setFldName();
abstract public String getFldName();
}
class Sub extends AbstractSuper {
#Override
public void setFldName() {
///....
}
#Override
public String getFldName() {
return "def";
}
}

Also, I don't want to give the value
of the field as an argument of the
constructor.
Why not? It's the perfect solution. Make the constructor protected and offer no default constructor, and subclass implementers are forced to supply a value in their constructors - which can be public and pass a constant value to the superclass, making the parameter invisible to users of the subclasses.
public abstract class Tool{
protected int id;
protected Main main;
protected Tool(int id, Main main)
{
this.id = id;
this.main = main;
}
}
public class Pencil{
public static final int PENCIL_ID = 2;
public Pencil(Main main)
{
super(PENCIL_ID, main);
}
}

How about using the Template pattern?
public abstract class Template {
private String field;
public void Template() {
field = init();
}
abstract String init();
}
In this way, you force all subclasses to implement the init() method, which, since it being called by the constructor, will assign the field for you.

You can't do this in the constructor since the super class is going to be initialized before anything in the subclass. So accessing values that are specific to your subclass will fail in your super constructor.
Consider using a factory method to create your object. For instance:
private MyClass() { super() }
private void init() {
// do something with the field
}
public static MyClass create() {
MyClass result = new MyClass();
result.init();
return result;
}
You have an issue in this particular sample where MyClass can't be subclassed, but you could make the constructor protected. Make sure your base class has a public / protected constructor also for this code. It's just meant to illustrate you probably need two step initialization for what you want to do.
Another potential solution you could use is using a Factory class that creates all variants of this abstract class and you could pass the field into the constructor. Your Factory would be the only one that knows about the field and users of the Factory could be oblivious to it.
EDIT: Even without the factory, you could make your abstract base class require the field in the the constructor so all subclasses have to pass in a value to it when instantiated.

Also, I don't want to give the value of the field as an argument of the constructor.
Is there any solution to do that, or should I just change my design ?
Yes, I think you should change your design so that the subclass passes the value to the constructor. Since the subclass portion of your object isn't initialized until after the superclass constructor has returned, there's really no other clean way of doing it. Sure, this'd work:
class Super {
protected abstract int abstractField();
protected Super() { System.out.println("Abstract field: " + abstractField); }
}
class Sub {
protected int abstractField(){ return 1337; }
}
... since the implementation of abstractField() doesn't operate on object state. However, you can't guarantee that subclasses won't think it's a great idea to be a little more dynamic, and let abstractField() returns a non-constant value:
class Sub2 {
private int value = 5;
protected int abstractField(){ return value; }
public void setValue(int v){ value = v; }
}
class Sub3 {
private final int value;
public Sub3(int v){ value = v; }
protected int abstractField(){ return value; }
}
This does not do what you'd expect it to, since the initializers and constructors of subclasses run after those of the superclass. Both new Sub2() and new Sub3(42) would print Abstract field: 0 since the value fields haven't been initialized when abstractField() is called.
Passing the value to the constructor also has the added benefit that the field you store the value in can be final.

If the value is determined by the type of subclass, why do you need a field at all? You can have a simple abstract method which is implemented to return a different value for each subclass.

I think you need a factory (aka "virtual constructor") that can act on that parameter.
If it's hard to do in a given language, you're probably thinking about it incorrectly.

If I understand you correctly: You want the abstract class's constructor to do something depending on a field in the abstract class but which is set (hopefully) by the subclass?
If I got this wrong you can stop reading ...
But if I got it right then you are trying to do something that is impossible. The fields of a class are instantiated in lexical order (and so if you declare fields "below", or "after", the constructor then those will not be instantiated before the constructor is called). Additionally, the JVM runs through the entire superclass before doing anything with the subclass (which is why the "super()" call in a subclass's constructor needs to be the first instruction in the constructor ... because this is merely "advice" to the JVM on how to run the superclass's constructor).
So a subclass starts to instantiate only after the superclass has been fully instantiated (and the superclass's is constructor has returned).
And this is why you can't have abstract fields: An abstract field would not exist in the abstract class (but only in the subclass) and so is seriously(!) "off limits" to the super (abstract) class ... because the JVM can't bind anything references to the field (cause it doesn't exist).
Hope this helps.