Saturday, September 20, 2014

Published 9/20/2014 by with 0 comment

Illegal modifier for parameter ... only final is permitted

A local variable, i.e., method level variable or a parameter cannot be declared as static or public/private/protected.

So local variables must be declared without these modifiers.

For example, the following code is wrong:

    void methodA(static int param) {

        System.out.println(param);
    }

    void methodB(private int param) {

        System.out.println(param);
    }

    void methodC() {

        private static int methodVariable = 10;
        System.out.println(methodVariable);

    }


To resolve the error, remove static and the access modifiers from the local variable declarations. Only the class level variables (declared outside of any method) can be declared with static and access modifiers.
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Wednesday, June 11, 2014

Published 6/11/2014 by with 1 comment

This method requires a body instead of a semicolon

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Published 6/11/2014 by with 6 comments

Font ' net/sf/jasperreports/fonts/pictonic/pictonic.ttf ...' is not available to the JVM

The full error message is:

Exception in thread "AWT-EventQueue-0" java.lang.ExceptionInInitializerError
       at net.sf.jasperreports.engine.fill.JRBaseFiller.(JRBaseFiller.java:184)
Caused by: net.sf.jasperreports.engine.util.JRFontNotFoundException: Font '
                     net/sf/jasperreports/fonts/pictonic/pictonic.ttf
                     net/sf/jasperreports/fonts/pictonic/pictonic.svg
                     net/sf/jasperreports/fonts/pictonic/pictonic.eot
                     net/sf/jasperreports/fonts/pictonic/pictonic.woff
              ' is not available to the JVM. See the Javadoc for more details.


Obviously there might have several reasons for this error. A Google Search or stackoverflow search will bring many results.

For me, in one instance, the problem was with the jasperreports version as I had more that one jasperreports jar in my classpath.

My reports were compiled with jasperreports-5.6.0 and in my classpath jasperreports-4.1.1.jar was also included and accidentally jasperreports-4.1.1.jar was "up" in the Build class path order.

Actually, my eclipse project had jasperreports-5.6.0 as a library jar and this project had another project as "Required projects in the build path" that contained the old version of jasperreports.

I moved the jasperreports-5.6.0 to "Up in the build classpath order" than the old version, and the problem got resolved.

Again, later I removed the project dependency and got the problem resolved as well.

So check, whether you have more than one jasperreports library  anyhow; if so, remove the unnecessary duplicate jar. 
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Thursday, December 12, 2013

Published 12/12/2013 by with 0 comment

The constructor ... is not visible

When you invoke the private constructor of a class in another class, you’ll get the error message “The constructor ... is not visible”. And when the private constructor of a super class is invoked from the subclass implicitly or explicitly, you’ll get the following:

Implicit super constructor ... is not visible for default constructor. Must define an explicit constructor.
Implicit super constructor ... is not visible. Must explicitly invoke another constructor.

See an example below:

public class SuperClass
{

       private SuperClass()
       {
              System.out.println("Super class private constructor");
       }
      
       public SuperClass(int n)
       {
              System.out.println("Super class protected constructor");
       }
}

In the above code, the class has two constructors. Access modifier for one is private and another is public. So if you create an instance of this class using the private constructor in another class, you’ll encounter the error.

You cannot write SuperClass  superClassObject = new SuperClass();
You should call other constructors that you can access as per access modifier rule or the other way of instantiating the class (if available in that class). Sometimes such class has some getInstance() like methods.

Again, when you inherit a class that has private default constructor, you have to write your own constructor and invoke a constructor (that is accessible) within your constructor. In such case, if you do not write any constructor in the subclass you’ll get “Implicit super constructor ... is not visible for default constructor. Must define an explicit constructor.”

This happens because when you do not define any constructor, a default constructor is automatically created by the Java compiler and the default constructor invokes the default constructor of the super class. This is how the constructor is invoked implicitly.

See the code below:
//Code1
public class Subclass extends SuperClass
{
      
}
Or
//Code2
public class Subclass extends SuperClass
{

       public Subclass()
       {
             
       }
}

I have not defined any constructor in code1. In code2, I have defined a constructor but have not invoked any constructor explicitly. So the super class default constructor will implicitly be called.

In the code below I have defined a constructor. But still I called the super class default constructor explicitly through “super()”. I’ll encounter the error.

public class Subclass extends SuperClass
{

       public Subclass()
       {
              super();
       }
}

Now the solutions are:
1.      
      We cannot call the private constructor of class from the subclass or other class. You should look the other constructors or the other way of instantiating class.

2.       If the default constructor of a super class is private, you have to write your own constructor in the subclass and explicitly call visible constructors.

For the above scenario, correct coding would be:

public class Subclass extends SuperClass
{

       public Subclass()
       {
              super(10);
       }

}
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Tuesday, November 26, 2013

Published 11/26/2013 by with 0 comment

java.lang.UnsupportedOperationException at java.util.AbstractList.remove

When you create a java.util.List object from an array using Arrays.asList method and after that if you try to clear or remove the elements from the list you'll get this run time error. The reason is that Arrays.asList method returns a fixed-size list backed by the specified array.

See in the example code below:

int[] myArray = {5,10,15,20};

List intList = Arrays.asList(myArray);

intList.clear();

Or

int[] myArray = {5,10,15,20};

List intList = Arrays.asList(myArray);

for(int i=0;i
{
intList.remove(i);
}

You'll get error from intList.clear() or intList.remove(i) because you cannot remove from a fixed-size array. And also you cannot add any new element in this list.

If you require to copy array elements in a list object and later need to add/delete elements from the list, you can do it in the following way.

  int[] myArray = {5,10,15,20};
List intList = new ArrayList();
System.out.println(intList); // print empty list
//copy the array elements into the list
for(int i=0;i
{
intList.add(myArray[i]);
}
System.out.println(intList); // print list after copying array elements
//You can now add
intList.add(25);
System.out.println(intList); // print list after adding new element
//You can remove
intList.remove(0);
System.out.println(intList); // print list after removing an element
//You can remove all the elements
intList.clear();
System.out.println(intList); // print list after removing all the elements

Below is the output of this sample code:

[]
[5, 10, 15, 20]
[5, 10, 15, 20, 25]
[10, 15, 20, 25]
[]

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Published 11/26/2013 by with 5 comments

Resource leak: 'scanner' is never closed

If you do not close the java.util.Scanner object after reading data, you can get an warning saying "Resource leak: 'scanner' is never closed".

In the code below see that the Scanner object is closed in a finally block after reading all inputs using the Scanner object.

int number1, number2;

Scanner scanner = new Scanner(System.in);

try
{
System.out.println("Ener the first number");
number1 = scanner.nextInt();
System.out.println("Ener the second number");
number2 = scanner.nextInt();
}
finally
{
scanner.close();
}

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Published 11/26/2013 by with 0 comment

Dimensions expected after this token / '[' expected

Java error "Dimensions expected after this token" or " '[' expected " can occur while creating array elements or allocating memory using the keyword new. If the square brackets [] are not used properly, you can get this error message.

See the code below:

int myArray[]=new int{10,20,30,40};

Here this is a syntax error. You have to put square brackets [] after new int. To initialize an array the correct code would be:

int myArray[]=new int[]{10,20,30,40};

or just

int myArray[]={10,20,30,40};
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Published 11/26/2013 by with 0 comment

Cannot invoke length() on the array type

The error message "Cannot invoke length() on the array type" or "cannot find symbol ... method length()" is related with the use of array.

Java array has a built in property length which is a variable, not method. If you put parenthesis () after the array property length, you'll get this error.

See the example below:

int[] number = {5,10,15};

for(int i=0; i<number.length(); i++)
{
    System.out.println(number[i]);
}

Here number.length() is wrong. To resolve the error, remove the parenthesis ().

The correct code:

for(int i=0; i<number.length; i++)
{
    System.out.println(number[i]);
}

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Friday, May 20, 2011

Published 5/20/2011 by with 2 comments

attempting to use incompatible return type

We may encounter this error message when we override a superclass method in the subclass with a wrong return type in the subclass method. To understand this, let's see the following two classes.

public class SuperClass
{
public double sum(double a,double b)
{
return (a+b);
}
}

public class Subclass extends SuperClass
{
public long sum(double a,double b)
{
return Math.round(a+b);
}
}

In the above example code, the super class method sum has double return type and in the subclass we have overridden this method where return type is long which is not compatible with double.
What should we do now to remove the error? We should either declare the return type of the subclass as double or change the parameter type.

See the correct code of the subclass below:

public class Subclass extends SuperClass
{
public double sum(double a,double b)
{
return Math.round(a+b);
}
public long sum(long a,long b)
{
return (a+b);
}
}


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Thursday, May 19, 2011

Published 5/19/2011 by with 5 comments

is not abstract and does not override abstract method

See the code below:


import javax.swing.JFrame;
import javax.swing.JButton;
import java.awt.Container;
import java.awt.FlowLayout;
import java.awt.event.ActionEvent;
import java.awt.event.ActionListener;

public class MyFrame extends JFrame implements ActionListener{
private JButton exitButton = new JButton("Exit");
public MyFrame()
{
Container container = getContentPane();
container.setLayout(new FlowLayout());
container.add(exitButton);
  exitButton.addActionListener(this);
setSize(300,200);
}
}

If this code is compiled, we will see the error "MyFrame is not abstract and does not override abstract method actionPerformed(java.awt.event.ActionEvent) in java.awt.event.ActionListener"

Why? When we implement an interface(ActionListener in this example) or extend an abstract class in a concrete class, we must override the abstract method(s). In the above example we have implemented ActionListener but have not overridden the abstract method actionPerformed.

The correct code of the above is:

import javax.swing.JFrame;
import javax.swing.JButton;
import java.awt.Container;
import java.awt.FlowLayout;
import java.awt.event.ActionEvent;
import java.awt.event.ActionListener;

public class MyFrame extends JFrame implements ActionListener
{
private JButton exitButton = new JButton("Exit");
public MyFrame()
{
Container container = getContentPane();
container.setLayout(new FlowLayout());
container.add(exitButton);
exitButton.addActionListener(this);
setSize(300,200);
}
public void actionPerformed(ActionEvent ae)
{
System.exit(0);
}
}


Sometimes it happens that we override the abstract method with wrong name spelling or wrong signature, then also we see the same error message. Note that Java is a case sensitive language; most often we write the method name with wrong case (example, actionperformed/ActionPerformed instead of actionPerformed) and see this error message.

For implementing an interface or extending an abstract class, we must override all of the abstract methods (if more than one), though we do not need all. For example, if we implement KeyListener we must override each of the three abstract methods keyPressed, keyTyped and keyReleased though we may not need all of them. In that case, method body can be kept blank. Instead of interface, we can use adapter class (if available) to get rid of overriding all abstract methods.

In summary, override the abstract method with correct name spelling and signature to remove the error.
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Published 5/19/2011 by with 2 comments

attempting to assign weaker access privileges

Java error "attempting to assign weaker access privileges" can happen when we override a superclass method in the subclass if the access modifier is not correctly chosen.

See the sample code below:

public class SuperClass
{
public void display()
{
System.out.println("this is super class");
}
}


public class Subclass extends SuperClass
{
private void display()
{
System.out.println("this is subclass");
}
}

If the above Subclass is compiled the error message will be:
display() in Subclass cannot override display() in SuperClass; attempting to assign weaker access privileges; was public

Let's understand the error message:
The method display has public access in the SuperClass and private access in the Subclass, that means when overriding this method, a weaker access privilege is going to be set (private is weaker access privilege than public).

So we need to know the allowed access modifier in the subclass for method overriding. The table below shows this:


Access modifier in super class method
Allowed access modifier for overriding
private
private, package, protected, public
package
package, protected, public
protected
protected, public
public
public


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Published 5/19/2011 by with 4 comments

modifier private not allowed here

Java error "modifier private not allowed here" can occur for any of the following reasons:

1. An outer class declared as private like below:
private class OuterClass
{
    //...
}
An outer class only be public or package private not private or protected. So you can write an outer class as below:
public class OuterClass
{
   //...
}
or
class OuterClass
{
   //...
}
This is true for an interface also. An interface can only be public or package not private or protected.


2. An interface method declared as private.

public interface TestInterface
{
   private void method(); // wrong as the interface method is private
}

An interface can contain only public or package private method.

The correct code of the above is:

public interface TestInterface
{
   public void method();
}
or
public interface TestInterface
{
   void method(); // interface method declared as package private
}
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Tuesday, May 17, 2011

Published 5/17/2011 by with 0 comment

No Persistence provider for EntityManager named

Are you using JPA and getting the error "Exception in thread "main" javax.persistence.PersistenceException: No Persistence provider for EntityManager named ..."

You might have used an entity manager as the following code:


public void persist(Object object) {
        EntityManagerFactory emf = Persistence.createEntityManagerFactory("TestPU");
        EntityManager em = emf.createEntityManager();
        em.getTransaction().begin();
        try {
            em.persist(object);
            em.getTransaction().commit();
        } catch (Exception e) {
            e.printStackTrace();
            em.getTransaction().rollback();
        } finally {
            em.close();
        }
    }


And you might have a persistence.xml file as below:

<?xml version="1.0" encoding="UTF-8"?>
<persistence version="2.0" xmlns="http://java.sun.com/xml/ns/persistence" xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xsi:schemaLocation="http://java.sun.com/xml/ns/persistence http://java.sun.com/xml/ns/persistence/persistence_2_0.xsd">
<persistence-unit name="PersonPU" transaction-type="RESOURCE_LOCAL">
<provider>org.eclipse.persistence.jpa.PersistenceProvider</provider>
<class>person.entity.Person</class>
<properties>
<property name="javax.persistence.jdbc.url" value="jdbc:mysql://localhost:3306/persondb"/>
<property name="javax.persistence.jdbc.password" value="mypassword"/>
<property name="javax.persistence.jdbc.driver" value="com.mysql.jdbc.Driver"/>
<property name="javax.persistence.jdbc.user" value="root"/>
</properties>
</persistence-unit>
</persistence>


Now see that, this error may happen for any of the following reasons:



1. In the code EntityManagerFactory emf = Persistence.createEntityManagerFactory("TestPU"); TestPU is passed in the method which must be a valid persistence unit name. In the persistence.xml file the name is defined as persistence-unit name="PersonPU". These two names must be same. So to remove the error the correct code will be :

EntityManagerFactory emf = Persistence.createEntityManagerFactory("PersonPU");

2. If the persistence unit name is correctly used but still the error exists, then confirm that the required jar file containing the JPA implementation classes are available in the classpath. Required jar files for the above code are eclipselink-2.0.0.jar and eclipselink-javax.persistence-2.0.jar

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Published 5/17/2011 by with 1 comment

is not a known entity type

To understand the error, let us assume that we have the following classes:

1) An entity class
package person.entity;
import java.io.Serializable;
import java.util.Date;
import javax.persistence.Basic;
import javax.persistence.Column;
import javax.persistence.Entity;
import javax.persistence.GeneratedValue;
import javax.persistence.GenerationType;
import javax.persistence.Id;
import javax.persistence.NamedQueries;
import javax.persistence.NamedQuery;
import javax.persistence.Table;
import javax.persistence.Temporal;
import javax.persistence.TemporalType;
import javax.xml.bind.annotation.XmlRootElement;
/**
 *
 * @author Shamsuddin
 */
@Entity
@Table(name = "person")
@XmlRootElement
@NamedQueries({
    @NamedQuery(name = "Person.findAll", query = "SELECT p FROM Person p"),
    @NamedQuery(name = "Person.findById", query = "SELECT p FROM Person p WHERE p.id = :id"),
    @NamedQuery(name = "Person.findByFirstName", query = "SELECT p FROM Person p WHERE p.firstName = :firstName"),
    @NamedQuery(name = "Person.findByLastName", query = "SELECT p FROM Person p WHERE p.lastName = :lastName"),
    @NamedQuery(name = "Person.findByDateOfBirth", query = "SELECT p FROM Person p WHERE p.dateOfBirth = :dateOfBirth"),
    @NamedQuery(name = "Person.findByEmail", query = "SELECT p FROM Person p WHERE p.email = :email")})
public class Person implements Serializable {
    private static final long serialVersionUID = 1L;
    @Id
    @GeneratedValue(strategy = GenerationType.IDENTITY)
    @Basic(optional = false)
    @Column(name = "id")
    private Integer id;
    @Column(name = "first_name")
    private String firstName;
    @Column(name = "last_name")
    private String lastName;
    @Column(name = "date_of_birth")
    @Temporal(TemporalType.TIMESTAMP)
    private Date dateOfBirth;
    @Column(name = "email")
    private String email;
    public Person() {
    }
    public Person(Integer id) {
        this.id = id;
    }
    public Integer getId() {
        return id;
    }
    public void setId(Integer id) {
        this.id = id;
    }
    public String getFirstName() {
        return firstName;
    }
    public void setFirstName(String firstName) {
        this.firstName = firstName;
    }
    public String getLastName() {
        return lastName;
    }
    public void setLastName(String lastName) {
        this.lastName = lastName;
    }
    public Date getDateOfBirth() {
        return dateOfBirth;
    }
    public void setDateOfBirth(Date dateOfBirth) {
        this.dateOfBirth = dateOfBirth;
    }
    public String getEmail() {
        return email;
    }
    public void setEmail(String email) {
        this.email = email;
    }
    @Override
    public int hashCode() {
        int hash = 0;
        hash += (id != null ? id.hashCode() : 0);
        return hash;
    }
    @Override
    public boolean equals(Object object) {
        // TODO: Warning - this method won't work in the case the id fields are not set
        if (!(object instanceof Person)) {
            return false;
        }
        Person other = (Person) object;
        if ((this.id == null && other.id != null) || (this.id != null && !this.id.equals(other.id))) {
            return false;
        }
        return true;
    }
    @Override
    public String toString() {
        return "person.entity.Person[ id=" + id + " ]";
    }
 
}

2. A java class that encapsulates an EntityManager

package person.manager;
import javax.persistence.EntityManager;
import javax.persistence.EntityManagerFactory;
import javax.persistence.Persistence;
/**
 *
 * @author Shamsuddin
 */
public class MyEntityManager {
   EntityManagerFactory emf = Persistence.createEntityManagerFactory("PersonPU");
   EntityManager em = emf.createEntityManager();
    public void persist(Object object) {
        em.getTransaction().begin();
        try {
            em.persist(object);
            em.getTransaction().commit();
        } catch (Exception e) {
            e.printStackTrace();
            em.getTransaction().rollback();
        } finally {
            em.close();
        }
    }
 
}

3. A main class

package person;
import person.entity.Person;
import person.manager.MyEntityManager;
/**
 *
 * @author Shamsuddin
 */
public class PersonMain {
    public static void main(String args[])
    {
        MyEntityManager entityManager = new MyEntityManager();
     
        Person person = new Person();
        person.setFirstName("Shamsuddin");
        person.setLastName("Ahammad");
     
        entityManager.persist(person);
     
     
    }
}


If we run the PersonMain class, in a situation, following error may be encountered:



java.lang.IllegalArgumentException: Object: person.entity.Person[ id=null ] is not a known entity type. The compiler indicates the error in the line entityManager.persist(person); meaning person is not an instance of an entity class.



Now Let's see:

1. See that Person is an entity class which has an annotation @Entity that means we have declared this class as an entity class. If your entity class does not have @Entity annotation, this may be the cause of this error. In that case, add annotation @Entity in the entity class. In our case, though the entity class is annotated properly, other reasons may cause this error(see below).

2. As we are using Java Persistence API (JPA), we have a persistence.xml file containing the following xml code:
<?xml version="1.0" encoding="UTF-8"?>
<persistence version="2.0" xmlns="http://java.sun.com/xml/ns/persistence" xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xsi:schemaLocation="http://java.sun.com/xml/ns/persistence http://java.sun.com/xml/ns/persistence/persistence_2_0.xsd">
  <persistence-unit name="PersonPU" transaction-type="RESOURCE_LOCAL">
    <provider>org.eclipse.persistence.jpa.PersistenceProvider</provider>
    <properties>
      <property name="javax.persistence.jdbc.url" value="jdbc:mysql://localhost:3306/persondb"/>
      <property name="javax.persistence.jdbc.password" value="mypassword"/>
      <property name="javax.persistence.jdbc.driver" value="com.mysql.jdbc.Driver"/>
      <property name="javax.persistence.jdbc.user" value="root"/>
    </properties>
  </persistence-unit>
</persistence>

We have error in this persistence.xml file. What's that? The persistence.xml must declare the entity class by writing <class>person.entity.Person</class>

So the correct persistence.xml will be:

<?xml version="1.0" encoding="UTF-8"?>
<persistence version="2.0" xmlns="http://java.sun.com/xml/ns/persistence" xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xsi:schemaLocation="http://java.sun.com/xml/ns/persistence http://java.sun.com/xml/ns/persistence/persistence_2_0.xsd">
  <persistence-unit name="PersonPU" transaction-type="RESOURCE_LOCAL">
    <provider>org.eclipse.persistence.jpa.PersistenceProvider</provider>
    <class>person.entity.Person</class>    <properties>
      <property name="javax.persistence.jdbc.url" value="jdbc:mysql://localhost:3306/persondb"/>
      <property name="javax.persistence.jdbc.password" value="mypassword"/>
      <property name="javax.persistence.jdbc.driver" value="com.mysql.jdbc.Driver"/>
      <property name="javax.persistence.jdbc.user" value="root"/>
    </properties>
  </persistence-unit>
</persistence>

If you use netbeans you can add the entity class declaration in the persistence.xml file very easily using the design view.
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Monday, May 9, 2011

Published 5/09/2011 by with 0 comment

Color parameter outside of expected range: Red Green Blue

Java error "Color parameter outside of expected range: Red Green Blue" can be encountered when we use the java.awt.Color class with wrong arguments for RGB (Red Green Blue).

See the wrong code below:
import javax.swing.JFrame;
import java.awt.Container;
import java.awt.Color;
public class MyFrame extends JFrame
{
public MyFrame()
{
Container container = getContentPane();
container.setBackground(new Color(-5,256,300)); // wrong value for RGB
setSize(800,600);
}
public static void main(String args[])
{
MyFrame myFrame = new MyFrame();
myFrame.setVisible(true);
}
}



See that, we have used -5, 256, 300 for Red, Green and Blue values respectively in the java.awt.Color class constructor. These arguments are illegal as the supported values are in the range 0 to 255.

So a correct code is:

container.setBackground(new Color(5,156,200)); // RGB values are in the range 0 to 255

If the constructor with float arguments is used than the correct RGB values are in the range 0.0 to 1.0

For more details http://download.oracle.com/javase/1.5.0/docs/api/java/awt/Color.html
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Published 5/09/2011 by with 1 comment

modifier static not allowed here


Java error "modifier static not allowed here" can occur for any of the following reasons:

1. Constructor is declared as static(!)

A constructor cannot be static. In the wrong code below:

public class Person
{
private String firstName;
private String lastName;
public static Person()
{
//...
}
}

Here the constructor Person is declared as static. As the constructor must be non-static, the correct code is:

public class Person
{
private String firstName;
private String lastName;
public Person()
{
//...
}
}



2. Interface contains static method(!)

All methods of any interface is abstract and hence they must be non static. See the code below:

public interface MyInterface
{
public static void myMetod();
}

The above code is wrong as the interface method is declared as static. Corrected code is:

public interface MyInterface
{
public void myMetod();
}
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Published 5/09/2011 by with 1 comment

illegal combination of modifiers: abstract and static

An abstract method cannot be static because the abstract method must be overridden i.e, body of the abstract method must be defined before invocation.

See the wrong code below:

public abstract class Employee
{
public static abstract double earnings(); // error
}

Here the abstract method earnings() is declared as static. An abstract method must be non-static. So the correct code is:


public abstract class Employee
{
public abstract double earnings(); // no error now
}


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Published 5/09/2011 by with 1 comment

interface methods cannot have body

Java error "interface methods cannot have body" occurs when body of the method within an interface is defined.

See the code below:

public interface MyInterface
{
public void myMetod()
{
}
}

Here the method myMethod has body starting at the symbol { and ending at }. An interface can have abstract methods (method without any body). The class that implements the interface defines the body of the interface methods.

So to remove the error, the method signature/header will end by ; (semi colon) and there will be no body section for the method. Below is the correct code:



public interface MyInterface
{
public void myMetod();
}


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Published 5/09/2011 by with 0 comment

interface expected here

The code below leads to an error "interface expected here"

public interface MyInterface extends Thread
{
}

Here an interface is created (MyInterface) and it inherits(extends) the Thread class which is not possible at all. Why? Because an interface can inherit only another interface, not any class. As Thread is a class, it cannot be inherited by any interface.

The code below is correct as the interface inherits another interface Runnable



public interface MyInterface extends Runnable
{
}


A similar type of error is posted at  no interface expected here

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Published 5/09/2011 by with 0 comment

no interface expected here

Any Java code as below will lead an error "no interface expected here"

import java.io.Serializable;
public class Person extends Serializable
{
private String firstName;
private String lastName;
//...
}

See that, here Person is a class (public class Person) and it inherits an interface (extends Serializable). Note that a class can inherit another class only, not any interface. A class can implement one or more interface only. Serializable is an interface, not a class.

Actually the correct code of the above will be:


import java.io.Serializable;
public class Person implements Serializable
{
private String firstName;
private String lastName;
//...
}

Here it is notable that, only an interface can inherit another interface.

A similar type of error is posted at interface expected here
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