Java Error Messages and Solutions

attempting to use incompatible return type

2011-05-20T00:30:00.002+06:00

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);
}
}

is not abstract and does not override abstract method

2011-05-19T23:59:00.008+06:00

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.

attempting to assign weaker access privileges

2011-05-19T23:11:00.002+06:00

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

modifier private not allowed here

2011-05-19T22:43:00.004+06:00

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
{
//...
}

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();
}

public interface TestInterface
{
void method(); // interface method declared as package private
}

No Persistence provider for EntityManager named

2011-05-17T22:29:00.006+06:00

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

is not a known entity type

2011-05-17T22:03:00.004+06:00

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.

Color parameter outside of expected range: Red Green Blue

2011-05-09T22:59:00.012+06:00

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

modifier static not allowed here

2011-05-09T22:35:00.002+06:00

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();
}

illegal combination of modifiers: abstract and static

2011-05-09T21:27:00.002+06:00

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
}

interface methods cannot have body

2011-05-09T21:06:00.002+06:00

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();
}

interface expected here

2011-05-09T20:49:00.002+06:00

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

no interface expected here

2011-05-09T20:22:00.006+06:00

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

invalid method declaration; return type required

2011-05-05T21:48:00.004+06:00

The Java error "invalid method declaration; return type required" can occur for any of the following reasons:

1. You are trying to define a constructor (of course without any return type; not even void) but the constructor name is not given same as the class name.

See the wrong code below:

public class Employee
{
private int id;
private String name;
public employee(int id,String name) // constructor name is not same as the class name.
{
this.id = id;
this.name = name;
}
}

So check the spelling of the constructor name and also note that Java is a case sensitive language.

2. If you are really defining a method (not a constructor) you have forgotten to declare the return type as the wrong code given below:

public class Test
{
public aMethod() // no return type given
{
System.out.println("This is a method");
}
}

The correct code is:

public class Test
{
public void aMethod() // return type void is given now
{
System.out.println("This is a method");
}
}

recursive constructor invocation

2011-05-03T21:17:00.006+06:00

If a constructor calls itself, then the error message "recursive constructor invocation" is shown. It may happen when we overload constructors and call the wrong constructor (itself) accidentally.

See the wrong code below:

public class Employee
{
private int id;
private String name;
public Employee(int id,String name)
{
this.id = id;
this.name = name;
}
public Employee(int id)
{
this(id);
/* this(id) calls the constructor having one parameter of int type. Thus this is calling itself.*/
}
public Employee()
{
}
}

In the above code, there are three constructors:
i) Employee(int id,String name)
ii) Employee(int id)
iii) Employee()

Within the second constructor, we have written this(id) which means it calls a constructor with one int parameter and see that it is the constructor itself. It means we have not invoked the right constructor. If we want to invoke the first constructor Employee(int id,String name) within the second constructor we should write this(id,null). We are providing null for the String name as no value is available for the name variable in the second constructor but now it calls the first constructor Employee(int id,String name) as we have provided two parameters now.

See the corrected code below:

public class Employee
{
private int id;
private String name;
public Employee(int id,String name)
{
this.id = id;
this.name = name;
}
public Employee(int id)
{
this(id,null); // this(id,null) calls another constructor having two parameters
}
public Employee()
{
}
}

Summary:

If you see the error message "recursive constructor invocation", check that you have provided correct no and type of parameters while calling one constructor in another one.

java.sql.SQLException: Access denied for user

2011-04-15T16:31:00.004+06:00

The full error message may be :

java.sql.SQLException: Access denied for user 'root'@'localhost' (using password : YES)

or

java.sql.SQLException: Access denied for user 'root'@'localhost' (using password: NO)

When we try to connect to a database through JDBC with wrong username or password - we see this error message.

See the sample code below:

Connection connection = DriverManager.getConnection("jdbc:mysql://localhost:3306/DatabaseName","root","mypassword");

In the above code, root is the username and mypassword is the password. If the username root or the password is not correct we see this error message "java.sql.SQLException: Access denied for user 'root'@'localhost' (using password : YES)".

If the password is not provided (blank password) as the code below but the user has password we see the error message "java.sql.SQLException: Access denied for user 'root'@'localhost' (using password: NO)"

Connection connection = DriverManager.getConnection("jdbc:mysql://localhost:3306/DatabaseName","root","");

If you are using JPA with eclipselink/toplink or other, correct the username and password in the persistence.xml file of your project.

persistence.xml may look like below:

<?xml version="1.0" encoding="UTF-8"?>
<persistence version="1.0" 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_1_0.xsd">
<persistence-unit name="MyPersistenceUnitPU" transaction-type="RESOURCE_LOCAL">
<provider>oracle.toplink.essentials.PersistenceProvider</provider>
<properties>
<property name="toplink.jdbc.user" value="root"/>
<property name="toplink.jdbc.password" value="mypassword"/>
<property name="toplink.jdbc.url" value="jdbc:mysql://localhost:3306/DatabaseName"/>
<property name="toplink.jdbc.driver" value="com.mysql.jdbc.Driver"/>
</properties>
</persistence-unit>
</persistence>

com.mysql.jdbc.exceptions.jdbc4.MySQLSyntaxErrorException: Unknown database

2011-04-15T16:16:00.002+06:00

Trying to connect to a MySQL server from Java? But seeing the error "Unknown database"?

The reason is that the database does not exist in the MySQL server. You might have accidentally misspelled the database name in the JDBC connection url.

See the code below:

Connection connection = DriverManager.getConnection("jdbc:mysql://localhost:3306/sales","username","password");

Here sales is the database name. If you misspelled this or the database sales does not exist in the server, you see this error message.

com.mysql.jdbc.exceptions.jdbc4.CommunicationsException: Communications link failure

2011-04-15T15:59:00.003+06:00

This error message may include the following text also:

"The last packet sent successfully to the server was 0 milliseconds ago. The driver has not received any packets from the server."

When we try to connect to a MySQL server "Communications link failure" error may occur for any of the following errors:

i) MySQL server is not running. Check for this and run the sever if it is stopped.

ii) Host name is wrong: If the server is running in the same computer, the correct host name is localhost otherwise it is a valid IP address or computer name.

iii) Wrong port no given : By default MySQL server runs in port 3306. If it is different, check that correct port no is given.

Sample code:

public static Connection getConnection()
{
Connection connection = null;
try
{
Class.forName("com.mysql.jdbc.Driver");
connection = DriverManager.getConnection("jdbc:mysql://localhost:3306/database","username","password");
System.out.println("Connection successful");
}
catch(ClassNotFoundException cnfe)
{
System.err.println("MySQL driver not found");
cnfe.printStackTrace();
}
catch(SQLException sqle)
{
System.err.println("Database/Connection error");
sqle.printStackTrace();
}

return connection;
}

ClassNotFoundException: com.mysql.jdbc.Driver

2011-04-15T15:41:00.002+06:00

This error occurs when the required JDBC driver file is not in the CLASSPATH. Also, if the driver name is not correct ClassNotFoundException occurs. To solve this problem, at first check the driver name; if the driver name is correct, do one of the followings as per requirement.

i) Add an entry for the driver file (mysql-connector-java-5.1.10.jar, for example) in the CLASSPATH variable.

or

ii) If you are not familiar with the CLASSPATH variable, place the mysql-connector-java-5.1.10.jar file in \jre\lib\ext folder in the JDK install directory (for example C:\Program Files\Java\jdk1.6.0_20\jre\lib\ext)

or

iii) If you are doing this in NetBeans or other IDEs, add MySQL JDBC Driver Library in the project.

call to super must be first statement in constructor

2011-04-10T21:09:00.002+06:00

The error message is very clear actually:

i) We can call super in the constructor of a subclass
ii) If we call super in the constructor of a subclass, it must be the first statement in that constructor; i.e, before writing any other statement we call to super must be made.

It is mentionable that, super(parameters) calls the constructor of the super class.

See the code below:

The super class...

public class Employee
{
private int id;
private String name;
public Employee()
{
}
public Employee(int id,String name)
{
this.id = id;
this.name = name;
}
}

The subclass...

public class SalariedEmployee extends Employee
{
private double salary;
public SalariedEmployee()
{
}
public SalariedEmployee(int id,String name,double salary)
{
this.salary = salary; //this is the first statement now
super(id,name); //error : call to super must be first statement in constructor
}
}

The correct subclass will be...

public class SalariedEmployee extends Employee
{
private double salary;
public SalariedEmployee()
{
}
public SalariedEmployee(int id,String name,double salary)
{
super(id,name); // now call to super is the first statement in the constructor
this.salary = salary; // it is now the second statement
}
}

java.lang.ArrayIndexOutOfBoundsException

2009-10-30T12:27:00.011+07:00

It's a very common runtime error message or exception we face when we use array wrongly. Very specifically this exception is thrown to indicate that an array has been accessed with an illegal index (location / position / subscript). The index is either negative or greater than or equal to the size of the array.

See the following code:

int data[]={5,7,18,21,27,15,3};

//show the array elements
for(int i=0;i<=7;i++)
{
System.out.println(data[i]);
}

Here the array size is 7 as the array is initialized by assigning 7 elements {5,7,18,21,27,15,3}. So the valid array index is 0 to 6. In the loop condition, we have given i&lt=7 and i is used as the array index inside the loop - so when i is 7 it tries to access the array element 7 but 7 is not a valid index. This causes the exception. Correct loop condition is i<7 or i<=6. The recommended approach is using the array built in variable length instead of the static size. Correct the code as below:

int data[]={5,7,18,21,27,15,3};

//show the array elements
for(int i=0;i<data.length;i++)//data is the array name
{
System.out.println(data[i]);
}

array dimension missing

2009-10-30T12:06:00.005+07:00

'array dimension missing' is shown when the array size is not specified while allocating memory for the array or creating an array.

See below:

int data[ ]=new int[ ];

See that, blank [ ] is given in new int[ ]. We must give an int as the size of the array. Memory will be allocated for the array according to the given int within the [ ] as below:

int data[]=new int[5];

illegal initializer

2009-10-30T11:45:00.006+07:00

You may see this error for writing wrong expression to initialize an array. See the code below:

int data={42,15,27,20,19};

If you compile this, you will see the following error message:

illegal initializer for int

int data={42,15,27,20,19};

Here the array is not declared properly as we have not put [ ] for declaring the array. The correct code is:

int data[ ]={42,15,27,20,19};

int[ ] data={42,15,27,20,19};

incompatible types

2009-10-30T11:09:00.007+07:00

1. Sometimes we may get the following error message in an 'if' (selection) structure.

incompatible types

found : int

required: boolean

This is because, we have not specified a boolean expression as the condition of the 'if'; instead we might have written such an expression which is may be int, float, String or other types but not boolean.

See the code below:

int a=10,b=10;

if(a=b)
{
System.out.println("a and b are equal");
}

Check the condition is a=b. Is it a boolean expression? By using a single = (equal sign), value of b is assigned to a which is an integer. That's not correct. We must use == (double equal sign) for comparing two variables for equality.

Correct code is the below:

int a=10,b=10;

if(a==b)
{
System.out.println("a and b are equal");
}

2. Sometimes it happens because of assigning one type of value to a variable of other types. See the code below:

String text='a';
char ch="a";

In this case the error message will be:

incompatible types

found : char

required: java.lang.String

String text='a';

incompatible types

found : java.lang.String

required: char

char ch="a";

Here we have not assigned correct value for the variables text and ch. In the first line, we have assigned a char value to a String variable. 'a' is a char value because it is written within ' ' (single quotation mark) . In the second line, we have assigned String value "a" to a char type variable. Note that String value is written within " " (double quotation mark). So we tried to assign char value to a String variable and a String value to a char variable. The correct code is:

String text="a";
char ch='a';

If you see this error, check that the correct type of value or variable is assigned.

cyclic inheritance involving ...

2009-08-28T22:37:00.007+07:00

public class Person extends Employee
{
private String firstName;
private String lastName;

public Person(String firstName,String lastName)
{
this.firstName=firstName;
this.lastName=lastName;
}

}

Person is a class which inherits Employee and see the code below where Employee also inherit the Class Person.

public class Employee extends Person
{
private double salary;

public Employee(String firstName,String lastName,double salary)
{
super(firstName,lastName);
this.salary=salary;
}
}

In the above code we will see the error message "cyclic inheritance involving Person

public class Person extends Employee"

Actually Person should be the super class here and Employee is the subclass. But here the relationsship is such that Employee is the subclass of Person, again Person is the subclass of Employee. It's a cyclic inheritance which is not possible.

The solution of this problem is identifying the proper relationship i.e, finding out the super class and the subclass. In our case Person is the super class, so it should not try to inherit Employee and Employee is the subclass which will inherit Person. Correction is ommiting "extends Employee" from the Person class signature as below:

public class Person
{
private String firstName;
private String lastName;

public Person(String firstName,String lastName)
{
this.firstName=firstName;
this.lastName=lastName;
}

}

public class Employee extends Person
{
private double salary;

public Employee(String firstName,String lastName,double salary)
{
super(firstName,lastName);
this.salary=salary;
}

}

class, interface, or enum expected

2009-08-19T14:19:00.009+07:00

You see this error message because of any one of the followings:

1. You misspelled the keyword class (fully lower case) or you did not write it at all for writing your class.

public Class Test //actual keyword is class
{
public void myMethod()
{
System.out.println("class, interface, or enum expected?");
}
}

In the above example Class (C is in upper case) is written instead of class (fully lower case).

2. Remember you cannot declare variables outside of class body like below.

int variable; //variable must be declared inside the class body
public class Test
{
public void myMethod()
{
System.out.println("class, interface, or enum expected?");
}
}

To solve this declare variables inside the class body as below.

public class Test
{
int variable; //variable must be declared inside the class body
public void myMethod()
{
System.out.println("class, interface, or enum expected?");
}
}

3. All methods are also defined within the method body. If you define any method outside of the class body you will see this error message.

public class Test
{
public void myMethod()
{
System.out.println("method inside the class body");
}
} //end of class

public void anotherMethod()
{
System.out.println("method outside of the class body");
}

In the above code, the method anotherMethod is defined after the end of the class. So this error occurred. Most often it occurs just because of incorrect use of {} (curly braces). Check all { are closed with } in the proper place.

4. If you are defining an interface, check you spelled it correctly in fully lowercase.

public Interface Test //actual keyword is interface
{
public void myMethod();

}

In the above code, keyword interface is misspelled with upper I.

5. Same way if you define an enum spell the keyword enum correctly.

public Enum Day //actual keyword is enum
{
SUNDAY, MONDAY, TUESDAY, WEDNESDAY,
THURSDAY, FRIDAY, SATURDAY
}

In the above example Enum is incorrect as the E is in upper case, it will be lower case e.

package system does not exist

2009-05-20T09:54:00.004+06:00

Very simple error ‘package system does not exist’- most probably you have used ‘err’ or ‘in’ or ‘out’ field of the ‘System’ class of ‘java.lang’ package but you have given small s instead of capital S in the class name System.

See the below code:

public class PackageSystemDoesNotExistError
{
public static void main(String args[])
{
system.out.println("package system does not exist error"); // here s is small
}
}

But in the following code:

public class PackageSystemDoesNotExistErrorSolved
{
public static void main(String args[])
{
System.out.println("package system does not exist error solved");//S is capital
}
}

Though System is not a package, as you have written system.out.println – three parts here (system, out and println) a package like statement, Java compiler considers it as package but as this package doesn’t exist, the error message is shown.

So the solution is just change the small s of system to capital S.

non-static method ... cannot be referenced from a static context

2009-05-17T14:06:00.005+06:00

If you try to access a non static member of a class from a static context, you will see ‘non-static method … cannot be referenced from a static context’. For example, you might have tried to call a non static method from a static method. Bear in mind, you can access a static member (variables, methods etc) from both static and non-static context but cannot access non-static member from a static context.

See the following example code:

public class NonStaticMethodCannotBeReferencedFromAStaticContext
{
private double firstNumber;
private double secondNumber;
private double result;

public void setFirstNumber(double firstNubmer)
{
this.firstNumber=firstNumber;
}

public void setSecondNubmer(double secondNumber)
{
this.secondNumber=secondNumber;
}

public double add()
{
return firstNumber+secondNumber;
}

public double substract()
{
return firstNumber-secondNumber;
}

public static void main(String args[])
{
setFirstNumber(10.35);// this is a non-static method but referenced from static main method
setSecondNubmer(20.97);
System.out.println("The addition is "+add());
}

}

For the solution you can:
i) declare the non-static member as static and access them from the static context (not always; be careful, this may create other problems)
ii) Create an instance of the class where the non-static members are and then reference the non-static members by the created instance or object.

Browse http://javatech-stuff.blogspot.com/2007/11/differences-between-java-terms.html for understanding the differences between static and non-static member.

Now see the solution below:

public class NonStaticMethodCannotBeReferencedFromAStaticContextSolved
{
private double firstNumber;
private double secondNumber;
private double result;

public void setFirstNumber(double firstNubmer)
{
this.firstNumber=firstNumber;
}

public void setSecondNubmer(double secondNumber)
{
this.secondNumber=secondNumber;
}

public double add()
{
return firstNumber+secondNumber;
}

public double substract()
{
return firstNumber-secondNumber;
}

public static void main(String args[])
{
NonStaticMethodCannotBeReferencedFromAStaticContextSolved object=new NonStaticMethodCannotBeReferencedFromAStaticContextSolved();
object.setFirstNumber(10.35);// accessed by the instance named object
object.setSecondNubmer(20.97);
System.out.println("The addition is "+object.add());
}

}

illegal start of expression

2009-05-17T14:05:00.003+06:00

If a method is defined inside another method we may see ‘illegal start of expression’ error message. Remember that, we cannot define a method inside another method; we just can call a method inside another method. Sometimes it occurs just because of improper opening and closing of curly braces ( { or } ).

See the following example code:

public class IllegalStartOfExpression
{
public void methodA()
{
System.out.println("This is method A");

/* methodB is defined inside methodA because we didn’t close methodA before defining methodB*/

public void methodB()
{
System.out.println("This is method B");
}
}
}

Now see the solution:

public class IllegalStartOfExpression
{
public void methodA()
{
System.out.println("This is method A");
}

/* methodA is finished then, methodB is being defined*/
public void methodB()
{
System.out.println("This is method B");
}

}

identifier expected

2009-05-17T14:04:00.004+06:00

This error message is shown when the statements are not written in proper place. Most often, mistakenly, we may write the processing task (for example assigning value to a variable; writing a loop etc.) outside of any method. In this situation we may see this type of error message.

Then the solution is to write such statements inside the appropriate methods.

See the example code below:

public class IdentifierExptected
{
int number1,number1,sum;

number1=10; // identifier expected
number2=20; // identifier expected
sum=number1+number2; // identifier expected

/*
The assignment statement must be written inside a method. Variables can also be assigned/ initialized during the declaration.
*/

public void display()
{
System.out.println("Number 1 = "+number1);
System.out.println("Number 2 = "+number2);
System.out.println("Sum = "+sum);
}

}

See the solution now:

public class IdentifierExptectedSolved
{
int number1=10,number2=20,sum;

public void sum()
{
sum=number1+number2;
}

public void display()
{
System.out.println("Number 1 = "+number1);
System.out.println("Number 2 = "+number2);
System.out.println("Sum = "+sum);
}

}

Exception in thread "main" java.lang.NoClassDefFoundError: [ClassName]

2008-12-01T15:28:00.007+06:00

If a class file does not exist or the class is in such a package which is not in the CLASSPATH but you are trying to run that class, then you may see "Exception in thread "main" java.lang.NoClassDefFoundError: [ClassName]".

Sometimes, by mistake, we(the students) try to execute a Java program without compiling it successfully. If a Java class is not compiled successfully, no class file (a file with extension class) is generated. So we need to compile the Java program at first, then it should be run.

The class might be in another directory or package. The class must be in the CLASSPATH for the successful running.

In another situation along with the above error message you may see "CMD.EXE was started with the above path as the current directory. UNC paths are not supported. Defaulting to Windows directory" and "CMD does not support UNC paths as current directories". You may see this when you run a program by a batch file or by using TextPad etc. This error means you are running a class which is located in a remote machine and you have given the address of the class like \\RemotePC\directory\ClassName. You should bring the class to your PC (if required) to run it. Or if you want to call a class from the remote PC you may use RMI.

class [ClassName] is public, should be declared in a file named [ClassName].java

2008-11-30T17:54:00.005+06:00

We get error message "class [ClassName] is public, should be declared in a file named [ClassName].java" if the file name of the Java source program is not same as the public class name.

Solution is to rename the file or the class name to make them same. If a public class name is ABC then the file name of that class must be ABC.java. So always copy the class name and paste it when naming the file, then we will not face this error anymore.

Exception in thread "main" java.lang.NoSuchMethodError: main

2008-11-18T18:32:00.008+06:00

'Exception in thread "main" java.lang.NoSuchMethodError: main' is a run time error message. If the main method is not properly defined, we may see this error message. You know if you want to execute/run a program that program must have a main method of the following signature:

public static void main(String args[])

Generally we do the following mistakes in this regard:

1. May be, you haven't defined the main method at all. Define it at first then run it.

2. The method name is not correctly written. The spelling will be exactly in lower case 'main', if we write 'Main' - it will be an error.

For example:

public class NoSuchMethodErrorMain
{
public static void Main(String args) // M is capital here
{
System.out.println("Java is an Object Oriented Language");
}

}

3. The keyword 'static' is omitted. The main method must be static.

For example,

public class NoSuchMethodErrorMain
{
public void main(String args) // keyword static is missing
{
System.out.println("Java is an Object Oriented Language");
}

}

4. The parameter is not declared. Java main method must have a String array parameter. Sometimes, we forget to put [] after the data type or the parameter name (args, here).

For example:

public class NoSuchMethodErrorMain
{
public static void main() // parameter problem
{
System.out.println("Java is an Object Oriented Language");
}

}

public class NoSuchMethodErrorMain
{
public static void main(String args) // the parameter is not array
{
System.out.println("Java is an Object Oriented Language");
}

}

The correct code will be:

public class NoSuchMethodErrorMainSolution
{
public static void main(String args[])
{
System.out.println("Java is an Object Oriented Language");
}

}

is already defined

2008-11-18T18:14:00.007+06:00

If we declare a variable in a method more than once than we see ' is already defined'. Find out the variable for which the message is shown and check whether you have declared the variable more than one time in a particular block.

If we compile the following code we will see 'country is already defined in myMethod()'

public class AlreadyDefinedSolution
{
public void myMethod()
{
String country;
String country="Bangladesh";// declared again
System.out.println(country);

}

}

Now the solution is to remove the declaration 'String' from the assignment statement.

public class AlreadyDefinedSolution
{
public void myMethod()
{
String country;
country="Bangladesh";//declaration removed
System.out.println(country);

}

}

reached end of file while parsing

2008-11-18T18:07:00.006+06:00

The error 'reached end of file while parsing' occurs because of curly braces } problem. You have given either more or less curly braces } than the required number of } braces.

See the code below:

public class ReachedEndOfFileWhileParsing
{
public static void main(String args[])
{
System.out.println("java-error-massages.blogspot.com");

}

Here two curly braces are started but only one is ended. We need to put another } after the display statement here like below:

public class ReachedEndOfFileWhileParsingSolution
{
public static void main(String args[])
{
System.out.println("java-error-massages.blogspot.com");
}// it is now given

}

possible loss of precision

2008-11-09T14:51:00.008+06:00

You may see the error message 'possible loss of precision' for any of the following reasons:

1) You are trying to assign a fractional number into a float variable(why not?). Actually Java compiler considers the fractional number as double by default. You need to explicitly specify the fractional number as float by giving the character 'f' or 'F' after the the number, like float number = 65.37F;

2) Or you are trying to assign the value of a double variable into a float variable. Memory size of double is 8 byte and that of float is 4 byte. If you assign a double value into a float value data may be lost. So be sure about the data type. If you are sure you want to assign the double into float, you need to cast it appropriately, like float floatVariable = (float)doubleVariable;

An example is given below:

public class PossibleLossOfPrecision
{
public static void main(String args[])
{

float floatNumber1=65.37; // error
double doubleNumber=65.37;
float floatNumber2=doubleNumber; // error

System.out.println(floatNumber1);
System.out.println(doubleNumber);
System.out.println(floatNumber2);

}
}

The above code is wrong; write this code as below and the problem will be solved.

public class PossibleLossOfPrecisionSolution
{
public static void main(String args[])
{

float floatNumber1=65.37F; // error solved
double doubleNumber=65.37;
float floatNumber2=(float)doubleNumber; // error solved

System.out.println(floatNumber1);
System.out.println(doubleNumber);
System.out.println(floatNumber2);

}
}

integer number too large

2008-11-09T14:10:00.008+06:00

We know that the maximum value of an int is 2147483647. If we write any whole number more than 2147483647 in our Java program, we may get this error message 'integer number too large' but may be you are trying to store this large number in a long type variable, then what to do? The solution is to put the letter 'l' or 'L' after the number. This happens, because the compiler considers the whole number as int by default and we specify the large number by character 'l' or 'L' as long number. Now the compiler will consider the number as long.

For example, we are trying to compile the following Java code:

public class IntegerNumberTooLarge
{
public static void main(String args[])
{
long number=965478213245;
System.out.println(number);
}
}

We will get 'integer number too large' from the above program. Now write the program as below the problem will be solved.

public class IntegerNumberTooLargeSolution
{
public static void main(String args[])
{
long number=965478213245L;
System.out.println(number);
}
}

; (semi-colon) expected

2008-11-04T23:39:00.006+06:00

Each java statement must ends with ; (semicolon) in Java. If we don’t put ; after any statement we will see ‘;’ expected

Here is an example of this error:

public class SemicolonExpected
{
public static void main(String args[])
{
int x=10,y=10 //no ; (semicolon) is given after this statement
if(x==y)
System.out.println(x +" and "+y+" is equal");
}
}

Here no ; (semicolon) is given after the x,y declaration statement. The correct code is:

public class SemicolonExpectedSolution
{
public static void main(String args[])
{
int x=10,y=10; //; (semicolon) is given now
if(x==y)
System.out.println(x +" and "+y+" is equal");
}
}

variable … might not have been initialized

2008-11-04T23:37:00.005+06:00

We see “variable might not have been initialized” when we try to use the value of a variable before assigning any value to that variable. Each local variable must be assigned a value before further using it.

Here is an example of this error :

public class VariableMightNotHaveBeenInitialized
{
public static void main(String args[])
{
int a,b,c;
a=10;
c=a+b;
System.out.println("The sum is "+c);
}
}

Here variable ‘b’ is added with variable ‘a’ before assigning any value.

Correct code is:

public class VariableMightNotHaveBeenInitializedSolution
{
public static void main(String args[])
{
int a,b,c;
a=10;
b=20;
c=a+b;
System.out.println("The sum is "+c);
}
}

missing method body, or declare abstract

2008-11-04T23:36:00.005+06:00

If we put a ; (semicolon) after the method header (i.e., before starting the method body) when we are defining a concrete method (which method has body) then we will see “missing method body, or declare abstract”. Another reason is if we don’t write the keyword ‘abstract’ for an abstract method (a method without body), we’ll also see this message.
To resolve the problem we must remove the ; (semicolon) from the header of the concrete method or if it is an abstract method the keyword ‘abstract’ must be written in the method header.

Here is an example of this error:

public abstract class MissingMethodBodyOrDeclareAbstract
{
public void display(String message);
{
System.out.println(message);
}
public void display();
}

Here, a ; (semicolon) is given after the method header of the first display method (with parameter). And for the second display method, keyword ‘abstract’ is not stated in the header and it doesn’t have any body also. Here it is to be noted that if a class contains an abstract method, the class must also be abstract.

The correct code is:

public abstract class MissingMethodBodyOrDeclareAbstractSolution
{
public void display(String message)
{
System.out.println(message);
}
public abstract void display();
}

else without if

2008-11-04T23:35:00.005+06:00

Programmers may see message “else without if” when the selection (if ... else if ... else) structure is used. This may happen for any of the following reasons:
Semicolon (;) is given after the last parenthesis of the condition
Error in use of the curly braces ( } ), may be the } is not closed before starting the next else if or the else or } is given more.

Here is an example of this error:

public class ElseWithoutIf
{
public static void main(String args[])
{
int a=10,b=5;
if(a==b);
{
System.out.println(a+" is not equal to "+b);
else
{
System.out.println(a+" is equal "+b);
}
}
}

Here we have given ; (semicolon) at the last of the first condition, it’s an error. Again, we haven’t closed the } of the if i.e., else is written within the ‘ if’. So we need to remove the ; and put another } before the else.

Now the correct code is:

public class ElseWithoutIfSolution
{
public static void main(String args[])
{
int a=10,b=5;
if(a==b)
{
System.out.println(a+" is not equal to "+b);
}
else
{
System.out.println(a+" is equal "+b);
}
}
}

cannot find symbol / cannot resolve symbol

2008-11-04T23:30:00.005+06:00

“cannot find symbol” or “cannot resolve symbol” is the most common error message which the programmers get from the compiler. However this error message may mean that the variable/method/class/object used are not declared/defined properly. In most cases, this happens because:

· The variable/object is not declared or the spelling of variable/object declaration & later use of that variable/object is not same.

· The method is not defined or the spelling of the method definition and method calling is not same or the number and/or type of parameters are wrongly given when calling the method or the method where it is defined is not imported properly (in case of class library methods, the package where the method is defined is not imported)

· If the message is shown for a class, the package of that class might not have been imported or the spelling is wrong.

The solution of that problem is to check the spelling at first, Java programmers see this message most often just because of the wrong spelling( be careful about the case also, as Java is a case sensitive language). If the spelling is correct check whether the required package is imported.

Here is an example of this error :

public class CannotFindSymbol
{
public static void main(String args[])
{
int quantity;
Scanner input=new Scanner(System.in);
System.out.print(“Enter the quantity: ”);
quntity=imput.nextInteger();
System.out.println(“Your number is ”+quantity);
}
}

In the above code, there are many lines where we will get this error message.

At first it will say Scanner cannot be resolved/found because the package of this class java.util is not imported.

Also problem in ‘quntity’ as the variable is declared as ‘quantity’

Error in ‘imput’ as the object is created as ‘input’

If you correct all the errors it will show error message for nextIntger as the actual method name is nextInt in class Scanner

Now the correct code is:

import java.util.*;
public class CannotFindSymbolSolution
{
public static void main(String args[])
{
int quantity;
Scanner input=new Scanner(System.in);
System.out.print("Enter the quantity: ");
quantity=input.nextInt();
System.out.println("Your number is "+quantity);
}
}