public class Foo {
public void methodB(String s) { System.out.println("Foo " + s); }
}
public class Bar extends Foo {
public void methodB(String s) { System.out.println("Bar " + s); }
}
public class Baz extends Bar {
public void methodB(String s) { System.out.println("Baz " + s); }
}

public class Daze extends Baz {
private Bar bb = new Bar();
public void methodB(String s) {
bb.methodB(s);
super.methodB(s);
}
}

public class TestClass {
public static void main(String[] args) {
Daze d = new Daze();
d.methodB("Hello");
}
}

What is the result?

Question

public class Foo {
 public void methodB(String s) { System.out.println("Foo " + s); }
}
public class Bar extends Foo {
 public void methodB(String s) { System.out.println("Bar " + s); }
}
public class Baz extends Bar {
 public void methodB(String s) { System.out.println("Baz " + s); }
}

public class Daze extends Baz {
 private Bar bb = new Bar();
 public void methodB(String s) {
 bb.methodB(s);
 super.methodB(s);
 }
}

public class TestClass {
 public static void main(String[] args) {
 Daze d = new Daze();
 d.methodB("Hello");
 }
}

What is the result?

Accepted Answer

B. Baz Hello
Baz Hello

Answer

A. Baz Hello
Foo Hello

Answer

C. Baz Hello

Answer

D. A compilation error occurs in the Daze class.

Option	Why wrong
A (Baz/Foo)	`super.methodB` in `Daze` goes to `Baz`, not `Foo`
B (Baz/Baz)	`bb = new Bar()` dispatches to `Bar.methodB`, not `Baz.methodB`
C (Baz only)	Two method calls produce two lines of output
D (compile error)	The code compiles fine

public class Foo { public void methodB(String s) { System.out.println("Foo " + s); } } public class Bar extends Foo { public void methodB(String s) { System.out.println("Bar " + s); } } public class…

Question

Options

How the community answered

Explanation

What Actually Happens

Why Each Option Fails (as written)

Memory Tip

Community Discussion