Private access modifier in Java - Step-by-Step Execution

Concept Flow - Private access modifier

Class with private member

↓

Access private member inside class?

No→Error if accessed outside

Yes↓

Access allowed, use member

↓

End

The private modifier restricts access to class members only within the class itself.

Execution Sample

Java

class Box {
  private int size = 5;
  int getSize() {
    return size;
  }
}

public class Main {
  public static void main(String[] args) {
    Box b = new Box();
    System.out.println(b.getSize());
  }
}

This code shows a private variable accessed inside the class via a public method.

Execution Table

Step	Action	Evaluation	Result
1	Create Box object b	Box object with size=5	b.size is private, stored as 5
2	Call b.getSize()	Inside getSize(), access private size	Returns 5
3	Print returned value	Output value	5
4	Try to access b.size directly (not shown in code)	Compile error	Cannot access private field
5	End	-	-

💡 Execution stops after printing 5; direct access to private field is not allowed.

Variable Tracker

Variable	Start	After Step 1	After Step 2	Final
b.size	undefined	5	5	5

Key Moments - 2 Insights

Why can't we access b.size directly outside the class?

Because size is declared private, it can only be accessed inside the Box class, as shown in step 4 of the execution_table where direct access causes a compile error.

How does the getSize() method access the private variable?

The getSize() method is inside the Box class, so it can access the private variable size directly, as shown in step 2 of the execution_table.

Visual Quiz - 3 Questions

Look at the execution_table, what value does getSize() return at step 2?

A 0

B 5

C Compile error

D null

💡 Hint

Check the 'Result' column at step 2 in the execution_table.

At which step does the program print the value 5?

A Step 1

B Step 2

C Step 3

D Step 4

💡 Hint

Look for the 'Print returned value' action in the execution_table.

If size was not private, what would happen at step 4?

A Direct access to b.size would be allowed

B Program would crash

C Compile error still occurs

D size would become null

💡 Hint

Step 4 shows a compile error due to private access; removing private allows direct access.

Concept Snapshot

private modifier:
- Used to restrict access to class members
- Members are accessible only inside their own class
- Outside access causes compile errors
- Use public methods to access private members
- Helps protect data and enforce encapsulation

Full Transcript

This example shows how the private access modifier works in Java. A class Box has a private variable size set to 5. Inside the class, a method getSize() returns this private variable. When we create an object b of Box and call b.getSize(), it returns 5. However, trying to access b.size directly outside the class causes a compile error because size is private. This protects the variable from outside access. To get the value, we use a public method inside the class. This is how private access modifier helps keep data safe and controlled.

Practice

1. What does the private access modifier do in Java?

easy

A. It restricts access to the member only within its own class.

B. It allows access from any class in the same package.

C. It allows access from any class in any package.

D. It allows access only from subclasses.

Solution

1. Step 1: Understand the meaning of private

   The private modifier restricts access to the member so that only code inside the same class can use it.

2. Step 2: Compare with other access levels

   Other modifiers like public or protected allow wider access, but private is the most restrictive.

3. Final Answer:

   It restricts access to the member only within its own class. -> Option A

4. Quick Check:

   Private means only inside class [OK]

Hint: Private means only accessible inside the same class [OK]

Common Mistakes:

• Confusing private with protected or public
• Thinking private allows subclass access
• Assuming private allows package access

2. Which of the following is the correct way to declare a private integer variable named count in a Java class?

easy

A. private count int;

B. int private count;

C. private int count;

D. count private int;

Solution

1. Step 1: Recall Java variable declaration syntax

   In Java, the access modifier comes first, then the type, then the variable name.

2. Step 2: Check each option

   private int count; follows the correct order: private int count;. Others have wrong order or syntax.

3. Final Answer:

   private int count; -> Option C

4. Quick Check:

   Access modifier + type + name [OK]

Hint: Access modifier comes before type and variable name [OK]

Common Mistakes:

• Placing 'private' after the type or variable name
• Mixing order of keywords
• Missing semicolon

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

class Test {
  private int number = 5;
  public void printNumber() {
    System.out.println(number);
  }
}
public class Main {
  public static void main(String[] args) {
    Test t = new Test();
    t.printNumber();
  }
}

medium

A. Runtime error

B. Compilation error

C. 0

D. 5

Solution

1. Step 1: Understand private member access inside class

   The variable number is private but accessed inside its own class method printNumber(), which is allowed.

2. Step 2: Check main method output

   The main method creates an object and calls printNumber(), which prints the value 5.

3. Final Answer:

   5 -> Option D

4. Quick Check:

   Private accessible inside class method [OK]

Hint: Private fields can be used inside their own class methods [OK]

Common Mistakes:

• Thinking private prevents access even inside class
• Expecting compilation error due to private
• Confusing default value with actual initialization

4. Identify the error in the following Java code:

class Person {
  private String name = "Alice";
}
public class Main {
  public static void main(String[] args) {
    Person p = new Person();
    System.out.println(p.name);
  }
}

medium

A. Missing constructor in Person class.

B. Cannot access private field name from outside the class.

C. Syntax error in class declaration.

D. No error; code runs fine.

Solution

1. Step 1: Check access to private field from outside class

   The field name is private in Person class, so it cannot be accessed directly from Main class.

2. Step 2: Understand Java access rules

   Accessing p.name from Main causes a compilation error due to private access.

3. Final Answer:

   Cannot access private field name from outside the class. -> Option B

4. Quick Check:

   Private fields block outside access [OK]

Hint: Private fields can't be accessed outside their class [OK]

Common Mistakes:

• Assuming private fields are accessible everywhere
• Thinking missing constructor causes error here
• Ignoring access modifiers

5. You want to keep a user's password private in a Java class but allow other classes to check if a password matches. Which is the best way to do this?

hard

A. Make the password field private and provide a public method to check the password.

B. Make the password field public so other classes can read it directly.

C. Make the password field protected and access it from other classes.

D. Make the password field private and provide a public method to return the password.

Solution

1. Step 1: Protect sensitive data with private fields

   Passwords should be private to prevent direct access and protect security.

2. Step 2: Provide controlled access via public method

   A public method can check if a given password matches without revealing the actual password.

3. Final Answer:

   Make the password field private and provide a public method to check the password. -> Option A

4. Quick Check:

   Private data + public check method = secure design [OK]

Hint: Keep data private; use public methods to interact safely [OK]

Common Mistakes:

• Making password public exposes sensitive data
• Returning password directly breaks security
• Using protected exposes data to subclasses unnecessarily