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Java Exception Handling Examples

By Chaitanya Singh

In this tutorial, we will see examples of some of the popular exceptions and how to handle them properly using try-catch block. We will see exception handling of ArithmeticException, ArrayIndexOutOfBoundsException, NumberFormatException, StringIndexOutOfBoundsException and NullPointerException.

If you are new to the concept of exception handling, I highly recommend you to refer this starter guide: Exception handling in Java.

Example 1: Arithmetic exception

This exception occurs when the result of a division operation is undefined. When a number is divided by zero, the result is undefined and that is when this exception occurs.

class JavaExample
{
  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 ("You Shouldn't divide a number by zero");
    }
  }
}

Output of above program:

You Shouldn't divide a number by zero

Example 2: ArrayIndexOutOfBounds Exception

This exception occurs when you try to access an array index that doesn’t exist. For example, If array is having only 5 elements and you are trying to display 7th element then it would throw this exception.

class JavaExample
{
  public static void main(String args[])
  {
    try{
      int a[]=new int[10];
      // This will throw exception as Array has
      // only 10 elements and we are trying to access
      // 12th element.
      a[11] = 9;
    }
    catch(ArrayIndexOutOfBoundsException e){
      System.out.println ("ArrayIndexOutOfBounds Exception occurred");
      System.out.println ("System Message: "+e);
    }
  }
}

Output:
Exception output
In the above example the array is initialized to store only 10 elements indexes 0 to 9. Since we are trying to access element of index 11, the program is throwing this exception.

Example 3: NumberFormat Exception

This exception occurs when a string is parsed to any numeric variable.
For example, the statement int num=Integer.parseInt ("XYZ"); would throw NumberFormatException because String “XYZ” cannot be parsed to int.

class JavaExample
{
  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

  • A string is nothing but an array of string type. This exception occurs when you try to access an index that doesn’t exist, similar to what we have seen in ArrayIndexOutOfBoundsException.
  • Each character of a string object is stored in a particular index starting from 0. For example: In the string “beginnersbook”, the char ‘b’ is stored at index 0, char ‘e’ at index 1 and so on.
  • 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.

In the following example, the scope of the string “beginnersbook” is from index 0 to 12, however we are trying to access the char at index 40, which doesn’t exist, hence the exception occurred.

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

Output:

13
StringIndexOutOfBoundsException.

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

Example 5: NullPointer Exception

Class: Java.lang.NullPointer Exception
This exception occurs when you are trying to perform some operation on an object that references to the null value.

class JavaExample
{
  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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