Enhanced For Loop - Array of Objects

Question

Ok so I have a class called Dog() which takes two parameters, a string and an integer.
This class has a method called bark(), which prints a string depending on the integer passed into the Dog() constructor.

I also have a class called Kennel() which creates an array of 5 Dog()s... looks like this:

public class Kennel
{
    Dog[] kennel = new Dog[5];
    public Kennel()
    {
        kennel[0] = new Dog("Harold",1);
        kennel[1] = new Dog("Arnold",2);
        kennel[2] = new Dog("Fido",3);
        kennel[3] = new Dog("Spot",4);
        kennel[4] = new Dog("Rover",5);
    }
}

For starters, this works, but seems wrong. Why do I have to start with Dog[] ... new Dog[5]? Maybe stupid question... I'm new to this.

Anyway... What I have been asked to do is use the "enhanced" for loop to iterate through the array calling bark().

So with a traditional for loop it would look like this:

for (i=0;i<kennel.length;i++)
{
    kennel[i].bark();
}

Simple stuff, right? But how do I implement this using the for(type item : array) syntax?

Its a constructor for the Dog class. See Java Constructors — Tristian
– Tristian, Commented Mar 2, 2012 at 9:22

Tristian · Accepted Answer · 2012-03-02 09:20:28Z

28

Just use it in the for each

for(Dog d : kennel) {
    d.bark();
}

answered Mar 2, 2012 at 9:20

Tristian

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1 Comment

MHz Over a year ago

Haha. You sir are a legend. A class can be a type. Who knew? I guess that explains the creation of the array too. An array of type Dog. Brilliant. Thank you.

adarshr · Accepted Answer · 2012-03-02 09:20:51Z

10

Here's how you do it using enhanced for loop.

for(Dog dog : kennel) {
    dog.bark();
}

For your other question, if you're going to be using arrays, you'll have to declare the size before you start adding elements to it. One exception, however is if you are doing both initialization and declaration in the same line. For example:

Dog[] dogs = {new Dog("Harold", 1), new Dog("Arnold", 2)};

answered Mar 2, 2012 at 9:20

adarshr

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1 Comment

MHz Over a year ago

Thanks brethren! Works like a charm. Although I meant why am I using Dog[] arrayName rather than Foo[] arrayName. You know?

AKS · Accepted Answer · 2012-09-25 18:36:22Z

0

About your second question:"Why do I have to start with Dog[] ... new Dog[5]?"

Its because of same logic you have to put Dog dog=new Dog(); ----(1) That's why Dog[] dogArray=new Dog[5]; ---(2)

If you don't have problem with the first one then why crib about the second one.

answered Sep 25, 2012 at 18:36

AKS

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