What will be the output?interface A{ public void method1();}class One implements A{ public void method1(){ System.out.println("Class One method1"); }}class Two extends One{ public void method1(){ System.out.println("Class Two method1"); }}public class Test extends Two{ public static void main(String[] args){ A a = new Two(); a.method1(); }} Class One method1Class Two method1Nothing will be printed Compilation Error
Question
What will be the output?interface A{ public void method1();}class One implements A{ public void method1(){ System.out.println("Class One method1"); }}class Two extends One{ public void method1(){ System.out.println("Class Two method1"); }}public class Test extends Two{ public static void main(String[] args){ A a = new Two(); a.method1(); }} Class One method1Class Two method1Nothing will be printed Compilation Error
Solution
The output will be "Class Two method1".
Here's the step by step explanation:
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An interface A is declared with a method named method1().
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Class One implements interface A and provides an implementation for method1(). When this method is called, it will print "Class One method1".
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Class Two extends Class One and overrides the method1(). When this method is called, it will print "Class Two method1".
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In the main method of the Test class, an instance of Class Two is created but the reference is of type interface A.
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When method1() is called on this reference, the overridden version in Class Two is executed because the actual object is of Class Two. Hence, "Class Two method1" is printed.
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