Constructor calling order in C++ - Practice Problems & Coding Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

Constructor calling order in single inheritance

What is the output of this C++ program showing constructor calls in single inheritance?

C++

#include <iostream>
using namespace std;

class Base {
public:
    Base() { cout << "Base constructor\n"; }
};

class Derived : public Base {
public:
    Derived() { cout << "Derived constructor\n"; }
};

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

A

Derived constructor

B

Derived constructor
Base constructor

C

Base constructor

D

Base constructor
Derived constructor

💡 Hint

Remember base class constructors run before derived class constructors.

In C++, when an object of a derived class is created, the base class constructor runs first, then the derived class constructor.

❓ Predict Output

intermediate

Constructor calling order in multiple inheritance

What is the output of this C++ program demonstrating constructor calls in multiple inheritance?

C++

#include <iostream>
using namespace std;

class A {
public:
    A() { cout << "A constructor\n"; }
};

class B {
public:
    B() { cout << "B constructor\n"; }
};

class C : public A, public B {
public:
    C() { cout << "C constructor\n"; }
};

int main() {
    C c;
    return 0;
}

A

C constructor
A constructor
B constructor

B

B constructor
A constructor
C constructor

C

A constructor
B constructor
C constructor

D

A constructor
C constructor
B constructor

💡 Hint

Base classes are constructed in the order they are declared in the class header.

In multiple inheritance, base classes are constructed in the order they appear in the derived class declaration, then the derived class constructor runs.

❓ Predict Output

advanced

Constructor calling order with virtual inheritance

What is the output of this C++ program illustrating constructor calls with virtual inheritance?

C++

#include <iostream>
using namespace std;

class Base {
public:
    Base() { cout << "Base constructor\n"; }
};

class A : virtual public Base {
public:
    A() { cout << "A constructor\n"; }
};

class B : virtual public Base {
public:
    B() { cout << "B constructor\n"; }
};

class Derived : public A, public B {
public:
    Derived() { cout << "Derived constructor\n"; }
};

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

A

Base constructor
A constructor
B constructor
Derived constructor

B

A constructor
B constructor
Base constructor
Derived constructor

C

Base constructor
B constructor
A constructor
Derived constructor

D

A constructor
Base constructor
B constructor
Derived constructor

💡 Hint

Virtual base class constructors are called before any derived class constructors.

With virtual inheritance, the virtual base class constructor is called only once before any non-virtual base classes or derived class constructors.

❓ Predict Output

advanced

Constructor calling order with member objects

What is the output of this C++ program showing constructor calls including member objects?

C++

#include <iostream>
using namespace std;

class Member {
public:
    Member() { cout << "Member constructor\n"; }
};

class Container {
    Member m;
public:
    Container() { cout << "Container constructor\n"; }
};

int main() {
    Container c;
    return 0;
}

A

Member constructor
Container constructor

B

Container constructor
Member constructor

C

Member constructor

D

Container constructor

💡 Hint

Member objects are constructed before the container class constructor body runs.

When an object contains member objects, those members are constructed first, then the container's constructor runs.

🧠 Conceptual

expert

Order of constructor calls in complex inheritance

Given this class hierarchy with virtual and non-virtual inheritance, which sequence correctly shows the constructor call order when creating an object of class D?

C++

class A { public: A() { cout << "A"; } };
class B : virtual public A { public: B() { cout << "B"; } };
class C : virtual public A { public: C() { cout << "C"; } };
class D : public B, public C { public: D() { cout << "D"; } };

A B A C D

B A B C D

C A C B D

D B C A D

💡 Hint

Virtual base class A is constructed once before B and C.

Virtual base A is constructed first, then B and C in declaration order, then D.

Practice

1. In C++, when creating an object of a derived class, which constructor is called first?

easy

A. Member objects' constructors

B. Derived class constructor

C. Base class constructor

D. Destructor of the base class

Solution

1. Step 1: Understand constructor order in inheritance

   When an object of a derived class is created, the base class constructor runs first to initialize the base part.

2. Step 2: Identify the first constructor called

   After the base class constructor, member objects and then the derived class constructor run.

3. Final Answer:

   Base class constructor -> Option C

4. Quick Check:

   Base constructor runs first [OK]

Hint: Base constructor always runs before derived [OK]

Common Mistakes:

• Thinking derived constructor runs first
• Confusing member object construction order
• Assuming destructors run during construction

2. Which of the following is the correct syntax to call a base class constructor from a derived class constructor in C++?

easy

A. Derived() { Base(); }

B. Derived() : Base {}

C. Derived() -> Base() {}

D. Derived() : Base() {}

Solution

1. Step 1: Recall base constructor call syntax

   In C++, the base class constructor is called in the initializer list of the derived constructor using a colon and parentheses.

2. Step 2: Match correct syntax

   Derived() : Base() {} uses the correct syntax: Derived() : Base() {}. Others are invalid or incorrect.

3. Final Answer:

   Derived() : Base() {} -> Option D

4. Quick Check:

   Use colon and parentheses for base constructor call [OK]

Hint: Use colon and parentheses to call base constructor [OK]

Common Mistakes:

• Calling base constructor inside derived constructor body
• Using arrow (->) instead of colon
• Omitting parentheses after base class name

3. Consider the following code:

class A {
public:
  A() { std::cout << "A "; }
};
class B {
public:
  B() { std::cout << "B "; }
};
class C : public A {
  B b;
public:
  C() { std::cout << "C "; }
};

int main() {
  C obj;
  return 0;
}

What is the output when this program runs?

medium

A. A B C

B. B A C

C. C A B

D. A C B

Solution

1. Step 1: Determine constructor call order

   First, base class A constructor runs printing "A ". Then member object b of class B is constructed printing "B ". Finally, derived class C constructor runs printing "C ".

2. Step 2: Combine outputs in order

   The output is "A B C " exactly in that sequence.

3. Final Answer:

   A B C -> Option A

4. Quick Check:

   Base, members, derived = A B C [OK]

Hint: Base then members then derived constructor order [OK]

Common Mistakes:

• Assuming members construct before base
• Mixing order of output strings
• Ignoring member object construction

4. Given this code snippet:

class Base {
public:
  Base() { std::cout << "Base "; }
};
class Derived : public Base {
public:
  Derived() { std::cout << "Derived "; }
};

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

What is the output when this program runs?

medium

A. No output

B. Derived Base

C. Base Derived

D. Compilation error

Solution

1. Step 1: Analyze object declaration

   The line 'Derived d();' is a function declaration, not an object instantiation, so no constructor runs.

2. Step 2: Understand program output

   Since no object is created, no constructors run and no output is printed.

3. Final Answer:

   No output -> Option A

4. Quick Check:

   Function declaration, no object created [OK]

Hint: Watch out for most vexing parse: 'Derived d();' is a function [OK]

Common Mistakes:

• Thinking 'Derived d();' creates an object
• Expecting constructor output
• Ignoring C++ parsing rules

5. You have a class hierarchy:

class X {
public:
  X() { std::cout << "X "; }
};
class Y {
public:
  Y() { std::cout << "Y "; }
};
class Z : public X {
  Y y1, y2;
public:
  Z() { std::cout << "Z "; }
};

int main() {
  Z obj;
  return 0;
}

What is the exact output when this program runs?

hard

A. Y Y X Z

B. X Y Y Z

C. X Z Y Y

D. Y X Y Z

Solution

1. Step 1: Identify constructor call order

   First, base class X constructor runs printing "X ". Then member objects y1 and y2 of class Y are constructed in declaration order, printing "Y " twice.

2. Step 2: Derived class constructor runs last

   Finally, derived class Z constructor runs printing "Z ".

3. Final Answer:

   X Y Y Z -> Option B

4. Quick Check:

   Base, members in order, derived = X Y Y Z [OK]

Hint: Members construct in declaration order after base [OK]

Common Mistakes:

• Mixing member construction order
• Assuming derived runs before members
• Ignoring multiple member objects