Reference data types in Java - Mini Project: Build & Apply

Understanding Reference Data Types in Java

📖 Scenario: You are creating a simple program to store and display information about a book. This will help you understand how reference data types work in Java.

🎯 Goal: Build a Java program that creates a Book object with a title and author, then prints the book details.

📋 What You'll Learn

Create a class called Book with two String fields: title and author.

Create an instance of Book with specific title and author values.

Use a variable to hold the Book object reference.

Print the book's title and author using the object reference.

💡 Why This Matters

🌍 Real World

Reference data types like objects are used to model real-world things in programs, such as books, people, or products.

💼 Career

Understanding how to create and use objects is fundamental for Java programming jobs, especially in software development and application design.

1

Create the Book class with fields

Write a class called Book with two public String fields named title and author. Do not add any methods yet.

Java

public class Book {
    // Your code here
}

Hint

Use public String title; and public String author; inside the class.

2

Create a Book object with title and author

In the Main class, create a variable called myBook of type Book. Then create a new Book object and assign it to myBook. Set myBook.title to "The Java Journey" and myBook.author to "Alex Reader".

Java

public class Book {
    public String title;
    public String author;
}

public class Main {
    public static void main(String[] args) {
        // Create a Book object called myBook
        // Your code here
    }
}

Hint

Use Book myBook = new Book(); to create the object, then set fields with myBook.title = ....

3

Use the myBook reference to access fields

Add a String variable called bookInfo that combines myBook.title and myBook.author in this format: "Title: The Java Journey, Author: Alex Reader".

Java

public class Book {
    public String title;
    public String author;
}

public class Main {
    public static void main(String[] args) {
        Book myBook = new Book();
        myBook.title = "The Java Journey";
        myBook.author = "Alex Reader";
        
        // Create bookInfo string combining title and author
        // Your code here
    }
}

Hint

Use string concatenation with +"Title: " + myBook.title + ", Author: " + myBook.author.

4

Print the book information

Use System.out.println to print the bookInfo variable.

Java

public class Book {
    public String title;
    public String author;
}

public class Main {
    public static void main(String[] args) {
        Book myBook = new Book();
        myBook.title = "The Java Journey";
        myBook.author = "Alex Reader";
        
        String bookInfo = "Title: " + myBook.title + ", Author: " + myBook.author;
        // Print bookInfo
        // Your code here
    }
}

Hint

Use System.out.println(bookInfo); to display the text.

Practice

1.

Which of the following is a reference data type in Java?

int, double, String, boolean

easy

A. String

B. int

C. double

D. boolean

Solution

1. Step 1: Identify primitive vs reference types

   int, double, and boolean are primitive types storing actual values.

2. Step 2: Recognize String as a reference type

   String stores a reference to an object containing text data.

3. Final Answer:

   String -> Option A

4. Quick Check:

   Reference type = String [OK]

Hint: Remember: String is an object, others are primitives [OK]

Common Mistakes:

• Confusing String as primitive
• Thinking int is a reference type
• Assuming all data types are reference types

2.

Which of the following is the correct way to create a new String object in Java?

?

easy

A. String s = new string("hello");

B. String s = new String("hello");

C. String s = String("hello");

D. String s = new String["hello"];

Solution

1. Step 1: Check correct constructor syntax

   Use 'new' keyword with class name and parentheses: new String("hello")

2. Step 2: Verify case sensitivity and brackets

   Class names are case-sensitive (String, not string), and arrays use [] differently.

3. Final Answer:

   String s = new String("hello"); -> Option B

4. Quick Check:

   Correct object creation syntax = String s = new String("hello"); [OK]

Hint: Use 'new ClassName()' with exact case for objects [OK]

Common Mistakes:

• Using lowercase 'string' instead of 'String'
• Missing 'new' keyword
• Using array brackets [] incorrectly

3.

What will be the output of the following code?

String a = "hello";
String b = a;
b = b + " world";
System.out.println(a);

medium

A. world

B. hello world

C. hello

D. Compilation error

Solution

1. Step 1: Understand reference assignment

   Variable b initially points to the same String object as a ("hello").

2. Step 2: String concatenation creates new object

   b = b + " world" creates a new String object; a still points to original "hello".

3. Final Answer:

   hello -> Option C

4. Quick Check:

   Strings are immutable; original stays same [OK]

Hint: String concatenation creates new object, original unchanged [OK]

Common Mistakes:

• Assuming b changes a's value
• Thinking references share mutable data
• Expecting combined string in a

4.

Find the error in the following code snippet:

String[] arr = new String[3];
arr[0] = "Java";
arr[1] = "Python";
arr[3] = "C++";

medium

A. No error, code runs fine

B. Syntax error in array declaration

C. Missing initialization of array elements

D. Array index out of bounds error

Solution

1. Step 1: Check array size and indices

   Array size is 3, valid indices are 0, 1, 2.

2. Step 2: Identify invalid index usage

   arr[3] accesses index 3 which is outside valid range, causing runtime error.

3. Final Answer:

   Array index out of bounds error -> Option D

4. Quick Check:

   Array indices start at 0, max index = size-1 [OK]

Hint: Array indices go from 0 to length-1 only [OK]

Common Mistakes:

• Using index equal to array length
• Confusing syntax error with runtime error
• Assuming arrays auto-expand

5.

Given the class below, what will be the output?

class Box {
  int length;
  Box(int length) { this.length = length; }
}

public class Test {
  public static void main(String[] args) {
    Box b1 = new Box(5);
    Box b2 = b1;
    b2.length = 10;
    System.out.println(b1.length);
  }
}

hard

A. 10

B. 5

C. Compilation error

D. Runtime error

Solution

1. Step 1: Understand reference assignment for objects

   b2 is assigned the reference of b1, so both point to the same Box object.

2. Step 2: Modify object property via one reference

   Changing b2.length to 10 changes the single Box object's length property.

3. Step 3: Print property via original reference

   Printing b1.length shows 10 because b1 and b2 refer to the same object.

4. Final Answer:

   10 -> Option A

5. Quick Check:

   Object references share same data; change reflects everywhere [OK]

Hint: Object references point to same data; changes affect all refs [OK]

Common Mistakes:

• Thinking b1 and b2 are separate objects
• Expecting original value after change
• Confusing primitives with objects