Java Exception Handling examples

by Chaitanya Singh

in Exception Handling

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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