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Editorial Team
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Asked: 2026-06-10T06:50:37+00:00

I have this code form the book SCJP: 1. class Mammal { 2. String

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I have this code form the book SCJP:

1.  class Mammal {
2.    String name = "furry ";
3.    String makeNoise() { return "generic noise"; }
4.  }
5.  
6.  class Zebra extends Mammal {
7.    String name = "stripes ";
8.    String makeNoise() { return "bray"; }
9.  }
10.
11. public class ZooKeeper {
12.   public static void main(String[] args) {
13.     new ZooKeeper().go();
14.   }
15.   
16.   void go() {
17.     Mammal m = new Zebra();
18.     System.out.println(m.name + m.makeNoise());
19.   }
20. }

The result from running this code is “furry bray”.

Question 1

I don’t understand why line 17 is not :
Zebra zebra2 = new Zebra();

What is the purpose in each of the following cases, when to use which?

Mammal zebra1 = new Zebra();

vs

Zebra zebra2 = new Zebra();

Question 2

Why is the variable name = "stripes" from the Zebra class overridden by the name = "furry" from the Mammal class? I expect the opposote: the variable from subclass will override that from superclass.

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1 Answer

  1. Editorial Team

    Answer 1: Depends on what you want. In this case it is not that obvious. But sometimes you just want an object to be off class 1 or class 2. imagine you have a method in zebra and not in mammal and you want tot use that method, then your object needs to be off class zebra and not mammal. But if there is a third-party class with a method only accepting mammal variables, then you must pass a mammal object.

    Answer2: m.name is the variable of mammal (cannot be overridden) and m.makeNoise() is a method and this is overriden in Zebra.

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