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Inheritance & Method Resolution

Understand how Java resolves which method to call when classes form an inheritance hierarchy.

A subclass gets the superclass's fields and methods. Java allows one superclass and any number of interfaces.

A family recipe: the child inherits grandma's recipe but can modify it (override). When you ask the child to cook, they use their version.

Key Concepts

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The rule people get wrong: overriding is decided at runtime, overloading at compile time.
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    class Animal { void speak() { print("..."); } }
    class Dog extends Animal { void speak() { print("Woof"); } }
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    Animal a = new Dog();
    a.speak();            // "Woof" -- the OBJECT decides
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The reference says Animal, but the object is a Dog, and overriding follows the object. This is dynamic dispatch, and the JVM does it through a table of method pointers per class.
AnimalDog
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Overloading is the opposite. It picks a signature from the reference type:
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    void feed(Animal x) { print("animal"); }
    void feed(Dog x)    { print("dog"); }
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    Animal a = new Dog();
    feed(a);              // "animal" -- the REFERENCE decides
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Nothing is looked up at runtime here. The compiler chose feed(Animal) and that is what runs.
feed(Animal)
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Constructors chain upward. Every constructor calls a superclass constructor first:
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    class Dog extends Animal {
      Dog(String name) {
        super(name);      // implicit super() if you omit it
      }
    }
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If the superclass has no no-argument constructor, you must call super(...) yourself, or it will not compile.
super(...)
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super reaches the parent's version of a method you have overridden:
super
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    void speak() { super.speak(); print("Woof"); }
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final blocks it. On a method, no subclass can override. On a class, nothing can extend it, which is why String is final.
finalString
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Deep hierarchies hurt. Past about three levels, a change in the base class breaks children in ways nobody predicted. Prefer composition unless the relationship really is "is-a".
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What the interviewer is probing.1. "A variable is declared Animal and holds a Dog. Which speak() runs, and which feed() overload?" Probing: overriding versus overloading, the perennial question. Stalls: "Both use the Dog." Moves up: overriding follows the object so Dog.speak() runs; overloading is chosen by the compiler from the reference type, so feed(Animal) is selected.
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2. "What happens if the superclass has no no-argument constructor?" Probing: constructor chaining. Stalls: "It still works." Moves up: the implicit super() call fails to compile, so the subclass must call super(...) explicitly as its first statement.
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3. "Why is a deep inheritance hierarchy a problem?" Probing: whether you have felt the fragile base class issue. Stalls: "It gets confusing." Moves up: subclasses depend on how the parent is implemented, not just what it promises, so a parent change breaks children in ways nobody predicted.
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4. "Can a constructor be overridden?" Probing: whether you know constructors are not inherited. Stalls: "Yes, by redefining it." Moves up: no — constructors are not inherited, so a same-named constructor in a subclass is simply that subclass's own.
Use cases:OOPPolymorphismFundamentals