Primitive vs reference storage in Java - Hands-On Comparison

Primitive vs Reference Storage in Java

📖 Scenario: Imagine you are organizing a small library. You want to keep track of the number of books and the details of each book separately.

🎯 Goal: You will create variables to store the number of books (a simple number) and the details of each book (like title and author) using objects. This will help you understand how Java stores simple values and objects differently.

📋 What You'll Learn

Create a primitive variable to store the number of books

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

Create an object of Book with specific values

Print the number of books and the book details

💡 Why This Matters

🌍 Real World

Understanding how Java stores simple data and objects helps when managing data in apps, games, or websites.

💼 Career

Knowing primitive vs reference storage is essential for Java developers to write efficient and bug-free code.

1

Create a primitive variable for the number of books

Create an int variable called numberOfBooks and set it to 3.

Java

# Your code here

Hint

Use int to store whole numbers like the count of books.

2

Create a Book class with title and author fields

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

Java

public class Library {
    public static void main(String[] args) {
        int numberOfBooks = 3;
    }
}

// Your code here

Hint

Define a class outside Library with two String fields.

3

Create a Book object with specific title and author

Inside the main method, create a Book object called myBook. Set its title to "Java Basics" and author to "Alice".

Java

public class Library {
    public static void main(String[] args) {
        int numberOfBooks = 3;
        // Your code here
    }
}

class Book {
    String title;
    String author;
}

Hint

Create the object with new Book() and assign values to fields using dot notation.

4

Print the number of books and book details

Use System.out.println to print the numberOfBooks, then print the title and author of myBook each on a new line.

Java

public class Library {
    public static void main(String[] args) {
        int numberOfBooks = 3;
        Book myBook = new Book();
        myBook.title = "Java Basics";
        myBook.author = "Alice";
        // Your code here
    }
}

class Book {
    String title;
    String author;
}

Hint

Print the primitive variable and then the object's fields using System.out.println.

Practice

1. Which of the following correctly describes how primitive types are stored in Java?

easy

A. They store values as strings internally.

B. They store the actual value directly in the variable.

C. They store the memory address of the variable.

D. They store a reference to an object in memory.

Solution

1. Step 1: Understand primitive storage

   Primitive types like int, double, and boolean store the actual data value directly in the variable's memory location.

2. Step 2: Compare with reference types

   Reference types store addresses pointing to objects, not the actual data value.

3. Final Answer:

   They store the actual value directly in the variable. -> Option B

4. Quick Check:

   Primitive = actual value [OK]

Hint: Primitives hold values; references hold addresses [OK]

Common Mistakes:

• Thinking primitives store memory addresses
• Confusing reference storage with primitive storage
• Assuming primitives store objects

2. Which of the following is the correct way to declare a reference variable for a String in Java?

easy

A. int s = new String();

B. boolean s = new String();

C. String s = 10;

D. String s = new String();

Solution

1. Step 1: Check variable type and assignment

   String is a reference type, so it can be assigned with new String() to create a new object.

2. Step 2: Verify type compatibility

   int and boolean are primitive types and cannot be assigned a String object.

3. Final Answer:

   String s = new String(); -> Option D

4. Quick Check:

   Reference declaration matches object creation [OK]

Hint: Reference types use 'new' with matching type [OK]

Common Mistakes:

• Assigning objects to primitive types
• Using incompatible types in declaration
• Missing 'new' keyword for object creation

3. What will be the output of the following code?

int a = 5;
int b = a;
b = 10;
System.out.println(a);

medium

A. 5

B. 10

C. 0

D. Compilation error

Solution

1. Step 1: Understand primitive assignment

   Variable b copies the value of a (which is 5). Changing b to 10 does not affect a.

2. Step 2: Print value of a

   Since a was not changed, printing a outputs 5.

3. Final Answer:

   5 -> Option A

4. Quick Check:

   Primitive copy keeps original value [OK]

Hint: Changing copied primitive doesn't affect original [OK]

Common Mistakes:

• Assuming a changes when b changes
• Confusing reference behavior with primitive
• Expecting compilation error

4. Consider the code below. What is the error and how to fix it?

String s1 = "hello";
String s2 = s1;
s2 = "world";
System.out.println(s1);

medium

A. Output is 'world' because s2 changes s1.

B. Error: Strings cannot be assigned like this.

C. No error; output is 'hello'.

D. Compilation error due to missing new keyword.

Solution

1. Step 1: Analyze reference assignment

   s2 points to the same String object as s1 initially, but assigning s2 = "world" makes s2 point to a new String object.

2. Step 2: Check effect on s1

   s1 still points to "hello" because Strings are immutable and s2 reassignment does not affect s1.

3. Final Answer:

   No error; output is 'hello'. -> Option C

4. Quick Check:

   Reference reassignment doesn't change original object [OK]

Hint: Reassigning reference doesn't change original object [OK]

Common Mistakes:

• Thinking s1 changes when s2 changes
• Expecting compilation error for String assignment
• Confusing mutable and immutable objects

5. Given the following code, what will be the output?

class Box {
  int value;
  Box(int v) { value = v; }
}

Box b1 = new Box(5);
Box b2 = b1;
b2.value = 10;
System.out.println(b1.value);

hard

A. 10

B. 5

C. Compilation error

D. NullPointerException

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 through b2

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

3. Step 3: Print b1.value

   Since b1 points to the same object, b1.value is now 10.

4. Final Answer:

   10 -> Option A

5. Quick Check:

   Reference change affects original object [OK]

Hint: Object references share same data; change affects all [OK]

Common Mistakes:

• Assuming b1.value stays 5 after b2 changes it
• Confusing primitive and reference behavior
• Expecting compilation or runtime errors