Java Exception Handling examples

Here we have shared examples of few most frequently seen exceptions in java. These examples will help you understand Exception handling in Java.

Example 1: Arithmetic exception

Class: Java.lang.ArithmeticException
This is a built-in-class present in java.lang package. This exception occurs when an integer is divided by zero.

class ExceptionDemo1
{
   public static void main(String args[])
   {
      try{
         int num1=30, num2=0;
         int output=num1/num2;
         System.out.println ("Result = " +output);
      }
      catch(ArithmeticException e){
         System.out.println ("Arithmetic Exception: You can't divide an integer by 0");
      }
   }
}

Output of above program:

Arithmetic Exception: You can't divide an integer by 0

Explanation: In the above example I’ve divided an integer by a zero and due to which ArithmeticException is thrown.

Example 2: ArrayIndexOutOfBounds Exception

Class: Java.lang.ArrayIndexOutOfBoundsException

This is a built in class present in java.lang package. This exception occurs when the referenced element does not exist in the array. For e.g. If array is having only 5 elements and we are trying to display 7th element then it would throw this exception.

Example:

class ExceptionDemo2
{
   public static void main(String args[])
   {
      try{
        int a[]=new int[10];
        //Array has only 10 elements
        a[11] = 9;
      }
      catch(ArrayIndexOutOfBoundsException e){
         System.out.println ("ArrayIndexOutOfBounds");
      }
   }
}

Output:

ArrayIndexOutOfBounds

In the above example the array is initialized to store only 10 elements indexes 0 to 9. Since we are invoking index 11 that’s why it is throwing this exception.

Example 3: NumberFormat Exception

Class: Java.lang.NumberFormatException

The object of the above built-in class gets created whenever a string is parsed to any numeric variable.
For E.g. The statement int num=Integer.parseInt ("XYZ") ; would throw NumberFormatException because String “XYZ” cannot be parsed to int.

Complete Code:

class ExceptionDemo3
{
   public static void main(String args[])
   {
      try{
	 int num=Integer.parseInt ("XYZ") ;
	 System.out.println(num);
      }catch(NumberFormatException e){
	  System.out.println("Number format exception occurred");
       }
   }
}

Output:

Number format exception occurred

Example 4: StringIndexOutOfBound Exception

Class: Java.lang.StringIndexOutOfBoundsException

  • An object of this class gets created whenever an index is invoked of a string, which is not in the range.
  • Each character of a string object is stored in a particular index starting from 0.
  • To get a character present in a particular index of a string we can use a method charAt(int) of java.lang.String where int argument is the index.

E.g.

class ExceptionDemo4
{
   public static void main(String args[])
   {
      try{
	 String str="easysteps2buildwebsite";
	 System.out.println(str.length());;
	 char c = str.charAt(0);
	 c = str.charAt(40);
	 System.out.println(c);
      }catch(StringIndexOutOfBoundsException e){
	  System.out.println("StringIndexOutOfBoundsException!!");
       }
   }
}

Output:

22
StringIndexOutOfBoundsException!!

Exception occurred because the referenced index was not present in the String.

Example 5: NullPointer Exception

Class: Java.lang.NullPointer Exception
An object of this class gets created whenever a member is invoked with a “null” object.
Example:

package beginnersbook.com;
class Exception2
{
public static void main(String args[])
{
	try{
		String str=null;
		System.out.println (str.length());
	}catch(NullPointerException e){
		System.out.println("NullPointerException..");
	}
}
}

Output:

NullPointerException..

Here, length() is the function, which should be used on an object. However in the above example String object str is null so it is not an object due to which NullPointerException occurred.

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