Exception Handling with Method Overriding in Java | When a superclass method (overridden method) declares that it can throw an exception then subclass method (overriding method) must also declare that it can throw the same kind of exception or a sub type of that exception.

To handle the exception while you overriding a method in Java, you will have to follow three important rules. They are as follows:

1. If an overridden method does not throw an exception using throws clause then

The overriding method can not throw any checked or compile-time exception.

The overriding method can throw any unchecked or runtime exception.

2. If an overridden method throws an unchecked or runtime exception then

The overriding method can throw any unchecked or runtime exception.

The overriding method can throw the same exception which is thrown by the overridden method.

The overriding method can ignore the method level exception.

3. If the superclass method throws checked or compile-time exception then

Subclass method can throw an exception which is a subclass of the superclass method’s exception.

Subclass method cannot throw the exception which is a superclass of the superclass method’s exception.

Subclass method can throw any unchecked or runtime exception.

Let’s take different types of example programs based on these rules. Program source code 1: package overriding; public class Parent { // Overridden method is not throwing an exception. void msg() { System.out.println("msg-Parent"); } } import java.io.IOException; public class Child extends Parent { void msg() throws IOException // Compile-time error because the overriding method is throwing a checked exception. { System.out.println("msg-Child"); } public static void main(String[] args) throws IOException { Parent p = new Child(); p.msg(); Child c = new Child(); c.msg(); } } Output: Unresolved compilation problem: Exception IOException is not compatible with throws clause in Parent.msg() In the above example program, if overriding method throws an unchecked exception, there will be no compile-time error. Look at the program source code.



Program source code 2: package overriding; public class Parent { // Overridden method is not throwing an exception. void msg() { System.out.println(“msg-Parent”); } } import java.io.IOException; public class Child extends Parent { void msg() throws ArithmeticException // No compile-time error because the overriding method is throwing an unchecked exception. { System.out.println(“msg-Child”); } public static void main(String[] args) throws IOException { Parent p = new Child(); p.msg(); Child c = new Child(); c.msg(); } } Output: msg-Child msg-Child Program source code 3: package overriding; public class Parent { // Overridden method is throwing an unchecked exception. void msg() throws ArithmeticException { System.out.println("msg-Parent"); } } public class Child extends Parent { void msg() throws ClassCastException // No Compile-time error because the overriding method is throwing an unchecked exception. { System.out.println("msg-Child"); } public static void main(String[] args) { Parent p = new Child(); p.msg(); Child c = new Child(); c.msg(); } } Output: msg-Child msg-Child Q. What will be the error in the following program?

Assumption X is superclass and Y is subclass. 1. In superclass public void m1() throws IOException { System.out.println("Hello"); } In subclass public void m1() throws Exception { System.out.println("Hi"); } A. Compile-time error because the overriding method is throwing an exception which is the superclass of the superclass method’ exception. i.e, Exception is the superclass of IOExecption. 2. In superclass public void m1() throws Throwable { System.out.println("Parent"); } In subclass public void m1() throws Exception { System.out.println("Child"); } A. No compile-time error because Throwable is the superclass of all exceptions. 3. In base class public void m1() throws Exception { System.out.println("Base"); } In derived class void m1() throws InterrurptedException { System.out.println("Child"); } A. Compile-time error because you cannot reduce the visibility of inherited method from Parent. Access modifiers must be bigger or same. 4. In base class protected void m1() throws Exception { System.out.println("Base"); } In derived class public void m1() throws Exception { System.out.println("Child"); } A. No error. 5. In superclass protected void m1(char c) throws Throwable { System.out.println("Parent"); } In subclass void m1(char c) { System.out.println("Child"); } A. Error because we cannot reduce the visibility while overriding method.



Hope that this tutorial has covered almost all important rules of exception handling with method overriding in Java with example programs. I hope that you will have understood nicely this topic and enjoyed it.

Thanks for reading!!!

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Look at the below figure to understand better.