Inheritance

Question 1

What will the following code print when compiled and run?
… main

List list = new ArrayList<>();

StringBuilder sb = new StringBuilder("mrx");         
String s = sb.toString();         
list.add(s);         
System.out.println(s.getClass());         System.out.println(list.getClass());

Answer 1

class java.lang.String 
class java.util.ArrayList

The getClass method always returns the Class object for the actual object on which the method is called irrespective of the type of the reference. 
Since s refers to an object (and not the reference) of class String, s.getClass returns Class object for String and similarly list.getClass returns Class object for ArrayList.

Question 2

Consider the following code:   
interface Flyer { 
   String getName(); 
}  
class Bird implements Flyer {     
   public String name;    
   public Bird(String name) {         
      this.name = name;    
   }     
  public String getName(){ 
    return name; 
  } 
}  
class Eagle extends Bird {     
    public Eagle(String name){         
                     super(name);     
     } 
}  
public class TestClass {     
    public static void main(String[] args) throws Exception {         
       Flyer f = new Eagle("American Bald Eagle");         
        //PRINT NAME HERE    
    } 
} 

Which of the following lines of code will print the name of the Eagle object?

System.out.println(f.name);
System.out.println(f.getName());
System.out.println(((Eagle)f).name); 
System.out.println(((Bird)f).getName());
System.out.println(Eagle.name); 
System.out.println(Eagle.getName(f));

Answer 2

While accessing a method or variable, the compiler will only allow you to access a method or variable that is visible through the class of the reference.

When you try to use f.name, the class of the reference f is Flyer and Flyer has no field named “name”, thus, it will not compile.

But when you cast f to Bird (or Eagle), the compiler sees that the class Bird (or Eagle, because Eagle inherits from Bird) does have a field named “name” so ((Eagle)f).name or ((Bird)f).name will work fine.

f.getName() will work because Flyer does have a getName() method.

interface Flyer { 
   String getName(); 
}  
class Bird implements Flyer {     
   public String name;    
   public Bird(String name) {         
      this.name = name;    
   }     
  public String getName(){ 
    return name; 
  } 
}  
class Eagle extends Bird {     
    public Eagle(String name){         
                     super(name);     
     } 
}  
public class TestClass {     
    public static void main(String[] args) throws Exception {         
       Flyer f = new Eagle("American Bald Eagle");         
        //PRINT NAME HERE    
    } 
}