Access specifiers in C++ - Practice Problems & Coding Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

Output of accessing private member in derived class

What is the output of this C++ code?

C++

#include <iostream>
using namespace std;

class Base {
private:
    int x = 10;
public:
    int getX() { return x; }
};

class Derived : public Base {
public:
    void printX() {
        // cout << x << endl; // line A
        cout << getX() << endl;
    }
};

int main() {
    Derived d;
    d.printX();
    return 0;
}

A 0

B Compilation error due to accessing private member x

C 10

D Runtime error

💡 Hint

Private members are not accessible directly in derived classes, but public methods can access them.

The private member 'x' is not accessible directly in Derived, so line A is commented out. Instead, the public method getX() returns the value 10, which is printed.

❓ Predict Output

intermediate

Output of protected member access in derived class

What will this program print?

C++

#include <iostream>
using namespace std;

class Parent {
protected:
    int val = 5;
};

class Child : public Parent {
public:
    void show() {
        cout << val << endl;
    }
};

int main() {
    Child c;
    c.show();
    return 0;
}

A Compilation error: val is inaccessible

B 5

C 0

D Runtime error

💡 Hint

Protected members are accessible in derived classes.

The protected member 'val' is accessible in the derived class Child, so printing val outputs 5.

❓ Predict Output

advanced

Accessing private member from outside class

What error does this code produce?

C++

#include <iostream>
using namespace std;

class Sample {
private:
    int secret = 42;
};

int main() {
    Sample s;
    cout << s.secret << endl;
    return 0;
}

A Compilation error: 'secret' is private within this context

B 42

C Runtime error

D 0

💡 Hint

Private members cannot be accessed outside their class.

The member 'secret' is private and cannot be accessed directly from main, causing a compilation error.

❓ Predict Output

advanced

Output of mixed access specifiers in inheritance

What is the output of this program?

C++

#include <iostream>
using namespace std;

class A {
public:
    int x = 1;
protected:
    int y = 2;
private:
    int z = 3;
};

class B : public A {
public:
    void print() {
        cout << x << ' ' << y << ' ';
        // cout << z << ' '; // line X
    }
};

int main() {
    B b;
    b.print();
    return 0;
}

A Compilation error due to accessing private member z

B 1 2 3

C Runtime error

D 1 2

💡 Hint

Private members are not accessible in derived classes.

The public member x and protected member y are accessible in B, but private member z is not. The commented line X would cause a compilation error if uncommented.

🧠 Conceptual

expert

Number of accessible members in derived class

Given the class below, how many members are accessible directly in class Derived?

C++

class Base {
public:
    int a;
protected:
    int b;
private:
    int c;
};

class Derived : public Base {
    void func() {
        // Which members can be accessed here?
    }
};

A 2 members: a and b

B 3 members: a, b, and c

C 1 member: a only

D 0 members

💡 Hint

Private members are not accessible in derived classes.

In public inheritance, public and protected members of Base are accessible in Derived. Private members are not accessible.

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