Java Execption handling 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 ExceptionDemo { 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 :
Arithmetic Exception: You can't divide an integer by 0
In the above example I’ve divided an integer by a zero and due to which
ArithmeticException
is thrown.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 ExceptionDemo1 { 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.
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 ExceptionDemo2 { 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
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.
class ExceptionDemo3 { 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.
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:
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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