Before we begin, you should ensure that you are familiar with the standard Java arrays. To check that you are familiar I would ask you to solve this very short problem yourself: Write a Java application that reads in a number of strings from the command line and provides a histogram of the numbers of letters in the words passed e.g. if the program was executed as: java ArrayApp This is a test string it would return 1 word with 1 letter, 1 words with 2 letters, 2 words with 4 letters etc. The solution to this problem is here: java/ArrayApp.java

import java.util.Arrays;

public class SortArrays {

	public static void main(String[] args) {
		Arrays.sort(args);
		for(int i=0; i<args.length; i++)
		{
			System.out.println(args[i]);
		}
	}
}

C:\temp\java SortArrays Dog Bird Cat Apple
Apple
Bird
Cat
Dog

public class Student extends Object{

	private String name;
	private int id;
	
	public Student(String name, int id){
	  this.name = name;
	  this.id = id;
	}
	
	public void display(){
	  System.out.println("A student with name: " + name 
	     + " and id " + id);
	}

	public String getName() { return name; }
	public int getID() { return id; }
}

import java.util.Comparator;

public class StudentIDComparator implements Comparator<Student> {

	public int compare(Student a, Student b) {
	  if (a.getID() < b.getID()) return -1;
	  else if (a.getID() > b.getID()) return 1;
	  else return 0;
	}
}

import java.util.Comparator;

public class StudentNameComparator implements Comparator<Student> 
{
	public int compare(Student a, Student b) {
	  return a.getName().compareTo(b.getName());
	}
}

import java.util.Arrays;
import java.util.Comparator;  //Comparator Interface

public class ArraysObjects {
	
	public static void main(String[] args) {
	
		Comparator<Student> byName = new StudentNameComparator();
		Comparator<Student> byID = new StudentIDComparator();
		
		Student[] students = new Student[]{
			new Student("Derek", 1234),
			new Student("Sarah", 1238),
			new Student("Mick",  1237),
			new Student("Adam",  1233),
			new Student("Jane",  1231)};
				
		// Sort by Name
		Arrays.sort(students, byName);
		System.out.println("The students sorted by name:");
		for (int i=0; i<students.length; i++){
			students[i].display();
		}
		
		// Now sort by ID
		Arrays.sort(students, byID);
		System.out.println("\nThe students sorted by id:");
		for (int i=0; i<students.length; i++){
			students[i].display();
		}
	}
}

The students sorted by name:
A student with name: Adam and id 1233
A student with name: Derek and id 1234
A student with name: Jane and id 1231
A student with name: Mick and id 1237
A student with name: Sarah and id 1238

The students sorted by id:
A student with name: Jane and id 1231
A student with name: Adam and id 1233
A student with name: Derek and id 1234
A student with name: Mick and id 1237
A student with name: Sarah and id 1238

I have given the example before of creating a Vector of objects, and as the Object class is the parent of all classes in Java, we are able to store any object of a class in such a Vector. Now, this is a useful feature of the language; however, it does not provide us with compile-time safety with our code. For example, we could accidentally try to convert a String object into a Fish object, but this would only be determined at run time, which will also likely result in a ClassCastException.

Generics in Java provides us with a better way. We can create a typed container of elements. For example, we could create a List of students using:

List<Student> studentList = new ArrayList<Student>();

As opposed to the old syntax:

List studentList = new ArrayList();

Now, this line of code will be the same for either version:

studentList.add(new Student("Derek", "1234", "MEQC"));

However, one advantage of the generics version is that if we try to add an incorrect value to the ArrayList we will get a compile time error; so, for example if we incorrectly added a String object to our studentList using:

studentList.add(new String("Test")); 

it will be allowed for the old syntax, but will give a compile-time error with the generics version. In addition the behaviour is different when we wish to operate on the data that we take off the list. With the old syntax version we would have to write:

	List studentList = new ArrayList();
	studentList.add(new Student("Derek", "1234", "MEQC"));
	Student s = (Student) studentList.get(0);
	System.out.println(s.getName());

Whereas, with the generics version we would have:

	List<Student> studentList = new ArrayList<Student>();
	studentList.add(new Student("Derek", "1234", "MEQC"));
	Student s = studentList.get(0);
	System.out.println(s.getName());

So, we no longer have to cast the object of the Object class to a Student object; rather, the List stores a list of Student objects. Again, the big advantage with the generics version is that the errors will be detected at compile time, which is much less expensive than detecting errors at run time.

With the use of generics also comes an enhanced for loop. For example, if we wished to iterate through a Vector of students we could use the three following syntax examples:

import java.util.*;

public class GenericsTest {

	public static void main(String args[])
	{
	    // Populate a Vector using an array of Students
	    Student[] studentArray = new Student[]{
	      new Student("Derek", "1234", "MEQC"),
	      new Student("Tom", "1235", "ICEX"),
	      new Student("Jane", "1236", "DMEX"),
	      new Student("Jill", "1237", "ECSAX")
	    };
	    
	    Vector<Student> students = new Vector<Student>(10);
	    students.addAll(Arrays.asList(studentArray));
	    
	    // Iterate the usual way
	    System.out.println("The Usual Way:");
	    for (int i=0; i<students.size(); i++)
	    {
	      Student tempStudent = students.elementAt(i);
	      System.out.print(tempStudent.getName() + ", ");
	    }
	    
	    // Iterate using an Enumeration
	    System.out.println("\nUsing an Enumeration:");        
	    Enumeration<Student> e = students.elements();
	    while(e.hasMoreElements())
	    {
	      Student tempStudent = e.nextElement();
	      System.out.print(tempStudent.getName() + ", ");
	    }
	    
	    // Using the "enhanced for"
	    System.out.println("\nUsing the enhanced For:");
	    for(Student tempStudent : students)
	    {
	      System.out.print(tempStudent.getName() + ", ");
	    }
	}
}

This results in the output:

The Usual Way:
Derek, Tom, Jane, Jill, 
Using an Enumeration:
Derek, Tom, Jane, Jill, 
Using the enhanced For:
Derek, Tom, Jane, Jill,

which is exactly the same for all three cases.