Polymorphism in Java
Which of the following best demonstrates polymorphism in Java?
Quick Answer
The detail that makes an interface reference calling a method the clearest demonstration of polymorphism is that the decision about which actual implementation runs is made at runtime, based on the real class of the object behind the reference, rather than at compile time based on the reference's declared type. Declaring a variable with an interface type and assigning it any object that implements that interface lets the same line of calling code — reference.method() — invoke a completely different method body depending on which concrete class was actually assigned, without the calling code needing to know or care which implementation it's dealing with. This is the essence of polymorphism: one interface, many possible forms of behavior, resolved dynamically rather than fixed in advance. It's a more direct illustration of the concept than something like method overloading, which is resolved at compile time based on argument types and doesn't involve this runtime dispatch at all — a distinction worth holding onto, since overloading and polymorphism get conflated but work through entirely different mechanisms. Recognizing runtime method resolution through a supertype or interface reference, as opposed to compile-time resolution, is the specific skill these polymorphism questions test.
⚠ Common exam trap
Oracle often tests the distinction between compile-time polymorphism (overloading) and runtime polymorphism (overriding with interface/superclass references), so candidates mistakenly choose overloading or overriding alone without the reference context.
Answer choices
Why each option matters
Answer the question above first, then reveal the full breakdown to understand why each option is right or wrong.
Correct answer & explanation
✓
Using an interface reference to call a method on an implementing object
Polymorphism in Java allows an object to take multiple forms, typically achieved through inheritance and interfaces. Option B demonstrates this by using an interface reference to invoke a method on an implementing object, where the actual method executed is determined at runtime based on the object's class, not the reference type.
Answer analysis
Option-by-option breakdown
For each option: why learners choose it and why it is or isn't the right answer here.
- ✗
Overloading a method with different parameter lists
Why it's wrong here
Overloading is compile-time polymorphism, but runtime polymorphism is typically emphasized.
- ✓
Using an interface reference to call a method on an implementing object
Why this is correct
Polymorphism allows one interface to be used for different implementations, as when an interface reference invokes the appropriate method at runtime.
- ✗
Using static methods
Why it's wrong here
Static methods are not polymorphic; they are resolved at compile time.
- ✗
Overriding a method in a subclass
Why it's wrong here
Overriding is part of polymorphism, but using an interface reference is a clearer demonstration of polymorphic behavior.
- ✗
Using final methods
Why it's wrong here
Final methods prevent overriding, which is not polymorphism.
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Same concept, more angles
1 more way this is tested on 1Z0-811
These questions test the same concept from different angles. Work through them to make sure you can recognise it however the exam phrases it.
Variation 1. class Parent { void show() { System.out.print("Parent"); } } class Child extends Parent { void show() { System.out.print("Child"); } } public class Test { public static void main(String[] args) { Parent p = new Child(); p.show(); } } What is the output?
medium- A.Compilation error
- B.No output
- C.Parent
- ✓ D.Child
- E.Runtime error
Why D: Although the reference is Parent, the object is Child, and show() is overridden, so Child's version is called.
JA
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