Access specifiers in C++ - Interactive Code Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to make the member variable accessible only within the class.

C++

class MyClass {
private:
    int [1];
};

Drag options to blanks, or click blank then click option'

Adata

Bvalue

Cnumber

Dcount

Attempts:

3 left

💡 Hint

Common Mistakes

Using reserved keywords as variable names

Leaving the variable name blank

The member variable name can be any valid identifier. Here, value is used as the variable name.

2fill in blank

medium

Complete the code to declare a public member function that returns the private variable.

C++

class MyClass {
private:
    int value;
public:
    int [1]() {
        return value;
    }
};

Drag options to blanks, or click blank then click option'

AreturnValue

BsetValue

Cvalue

DgetValue

Attempts:

3 left

💡 Hint

Common Mistakes

Using a setter name for a getter function

Using the variable name as function name

The function name getValue clearly indicates it returns the value of the private variable.

3fill in blank

hard

Fix the error in the code by choosing the correct access specifier to allow derived classes to access the base class member.

C++

class Base {
[1]:
    int protectedVar;
};

class Derived : public Base {
public:
    void show() {
        std::cout << protectedVar << std::endl;
    }
};

Drag options to blanks, or click blank then click option'

Aprotected

Bpublic

Cinternal

Dprivate

Attempts:

3 left

💡 Hint

Common Mistakes

Using private instead of protected

Using internal which is not a C++ specifier

The protected access specifier allows derived classes to access the member variable.

4fill in blank

hard

Fill both blanks to declare a class with a private variable and a public setter function.

C++

class Sample {
[1]:
    int data;
[2]:
    void setData(int d) {
        data = d;
    }
};

Drag options to blanks, or click blank then click option'

Aprivate

Bprotected

Cpublic

Dinternal

Attempts:

3 left

💡 Hint

Common Mistakes

Making variables public

Using protected for the setter function

The variable data should be private to restrict direct access, and the setter function should be public to allow controlled modification.

5fill in blank

hard

Fill all three blanks to create a class with a private variable, a public getter, and a protected setter.

C++

class Data {
[1]:
    int value;
[2]:
    int getValue() {
        return value;
    }
[3]:
    void setValue(int v) {
        value = v;
    }
};

Drag options to blanks, or click blank then click option'

Aprivate

Bpublic

Cprotected

Dinternal

Attempts:

3 left

💡 Hint

Common Mistakes

Making setter public

Making variable public

The variable value is private to protect it. The getter is public to allow reading the value. The setter is protected to allow modification only within the class and derived classes.

Practice

1. Which access specifier allows members of a class to be accessed from anywhere in the program?

easy

A. protected

B. private

C. public

D. internal

Solution

1. Step 1: Understand access specifiers

   Public members are accessible from any part of the program, unlike private or protected.

2. Step 2: Compare with other specifiers

   Private restricts access to inside the class only; protected allows access in derived classes.

3. Final Answer:

   public -> Option C

4. Quick Check:

   Access anywhere = public [OK]

Hint: Public means accessible everywhere, private means hidden [OK]

Common Mistakes:

• Confusing private with public
• Thinking protected is accessible everywhere
• Assuming internal is a C++ specifier

2. Which of the following is the correct way to declare a private member variable int age inside a class?

easy

A. private int age;

B. private:
int age;

C. int age private;

D. int private age;

Solution

1. Step 1: Recall C++ syntax for access specifiers

   Access specifiers are labels followed by a colon, then members are declared below.

2. Step 2: Identify correct syntax

   private: int age; uses 'private:' followed by 'int age;' which is correct syntax.

3. Final Answer:

   private: int age; -> Option B

4. Quick Check:

   Access specifier label + colon = correct [OK]

Hint: Access specifiers end with colon, members declared below [OK]

Common Mistakes:

• Placing access specifier before type
• Missing colon after access specifier
• Trying to put access specifier after variable

3. What will be the output of this code?

class Test {
private:
    int x = 5;
public:
    int getX() { return x; }
};

int main() {
    Test t;
    // std::cout << t.x << std::endl; // Line A
    std::cout << t.getX() << std::endl;
    return 0;
}

medium

A. 5

B. Compilation error at Line A

C. 0

D. Runtime error

Solution

1. Step 1: Understand private member access

   The variable x is private, so direct access like t.x is not allowed (Line A is commented out).

2. Step 2: Use public getter method

   The method getX() is public and returns x's value, so std::cout << t.getX() prints 5.

3. Final Answer:

   5 -> Option A

4. Quick Check:

   Private hidden, public getter shows 5 [OK]

Hint: Private members need public methods to access outside [OK]

Common Mistakes:

• Trying to access private variable directly
• Expecting 0 as default value
• Confusing compilation and runtime errors

4. Identify the error in this code snippet:

class Base {
protected:
    int value = 10;
};

class Derived : public Base {
public:
    void show() {
        std::cout << value << std::endl;
    }
};

int main() {
    Derived d;
    d.show();
    std::cout << d.value << std::endl;
    return 0;
}

medium

A. Error: Cannot access protected member 'value' from main using object

B. Error: Cannot access protected member 'value' inside Derived class

C. Error: Missing semicolon after class Derived

D. No error, code runs and prints 10 twice

Solution

1. Step 1: Check access to protected member inside Derived

   Derived class can access protected members of Base, so d.show() works and prints 10.

2. Step 2: Check access to protected member from main

   Main function cannot access protected members directly via object, so std::cout << d.value causes compilation error.

3. Final Answer:

   Error: Cannot access protected member 'value' from main using object -> Option A

4. Quick Check:

   Protected accessible in subclass, not outside [OK]

Hint: Protected members hidden outside class and subclasses [OK]

Common Mistakes:

• Thinking protected is public
• Assuming main can access protected members
• Missing error on direct access in main

5. You want to design a class where some data is hidden from everyone except derived classes, and some functions are accessible everywhere. Which access specifiers should you use for data and functions respectively?

hard

A. Data: private, Functions: public

B. Data: protected, Functions: private

C. Data: public, Functions: protected

D. Data: protected, Functions: public

Solution

1. Step 1: Determine data access for derived classes only

   Protected data is hidden from outside but accessible in derived classes, fitting the requirement.

2. Step 2: Determine function access for everyone

   Functions accessible everywhere must be public.

3. Final Answer:

   Data: protected, Functions: public -> Option D

4. Quick Check:

   Protected data + public functions = correct design [OK]

Hint: Protected hides data, public shares functions [OK]

Common Mistakes:

• Using private data blocks derived classes
• Making functions private hides them outside
• Confusing public and protected roles