11.4 — Constructors and initialization of derived classes

In the past two lessons, we’ve explored some basics around inheritance in C++ and the order that derived classes are initialized. In this lesson, we’ll take a closer look at the role of constructors in the initialization of derived classes. To do so, we will continue to use the simple Base and Derived class we developed in the previous lesson:

With non-derived classes, constructors only have to worry about their own members. For example, consider Base. We can create a Base object like this:

Here’s what actually happens when base is instantiated:

  1. Memory for base is set aside
  2. The appropriate Base constructor is called
  3. The initialization list initializes variables
  4. The body of the constructor executes
  5. Control is returned to the caller

This is pretty straightforward. With derived classes, things are slightly more complex:

Here’s what actually happens when derived is instantiated:

  1. Memory for derived is set aside (enough for both the Base and Derived portions)
  2. The appropriate Derived constructor is called
  3. The Base object is constructed first using the appropriate Base constructor. If no base constructor is specified, the default constructor will be used.
  4. The initialization list initializes variables
  5. The body of the constructor executes
  6. Control is returned to the caller

The only real difference between this case and the non-inherited case is that before the Derived constructor can do anything substantial, the Base constructor is called first. The Base constructor sets up the Base portion of the object, control is returned to the Derived constructor, and the Derived constructor is allowed to finish up its job.

Initializing base class members

One of the current shortcomings of our Derived class as written is that there is no way to initialize m_id when we create a Derived object. What if we want to set both m_cost (from the Derived portion of the object) and m_id (from the Base portion of the object) when we create a Derived object?

New programmers often attempt to solve this problem as follows:

This is a good attempt, and is almost the right idea. We definitely need to add another parameter to our constructor, otherwise C++ will have no way of knowing what value we want to initialize m_id to.

However, C++ prevents classes from initializing inherited member variables in the initialization list of a constructor. In other words, the value of a variable can only be set in an initialization list of a constructor belonging to the same class as the variable.

Why does C++ do this? The answer has to do with const and reference variables. Consider what would happen if m_id were const. Because const variables must be initialized with a value at the time of creation, the base class constructor must set its value when the variable is created. However, when the base class constructor finishes, the derived class constructors initialization lists are then executed. Each derived class would then have the opportunity to initialize that variable, potentially changing its value! By restricting the initialization of variables to the constructor of the class those variables belong to, C++ ensures that all variables are initialized only once.

The end result is that the above example does not work because m_id was inherited from Base, and only non-inherited variables can be changed in the initialization list.

However, inherited variables can still have their values changed in the body of the constructor using an assignment. Consequently, new programmers often also try this:

While this actually works in this case, it wouldn’t work if m_id were a const or a reference (because const values and references have to be initialized in the initialization list of the constructor). It’s also inefficient because m_id gets assigned a value twice: once in the initialization list of the Base class constructor, and then again in the body of the Derived class constructor. And finally, what if the Base class needed access to this value during construction? It has no way to access it, since it’s not set until the Derived constructor is executed (which pretty much happens last).

So how do we properly initialize m_id when creating a Derived class object?

In all of the examples so far, when we instantiate a Derived class object, the Base class portion has been created using the default Base constructor. Why does it always use the default Base constructor? Because we never told it to do otherwise!

Fortunately, C++ gives us the ability to explicitly choose which Base class constructor will be called! To do this, simply add a call to the base class Constructor in the initialization list of the derived class:

Now, when we execute this code:

The base class constructor Base(int) will be used to initialize m_id to 5, and the derived class constructor will be used to initialize m_cost to 1.3!

Thus, the program will print:

Id: 5
Cost: 1.3

In more detail, here’s what happens:

  1. Memory for derived is allocated.
  2. The Derived(double, int) constructor is called, where cost = 1.3, and id = 5
  3. The compiler looks to see if we’ve asked for a particular Base class constructor. We have! So it calls Base(int) with id = 5.
  4. The base class constructor initialization list sets m_id to 5
  5. The base class constructor body executes, which does nothing
  6. The base class constructor returns
  7. The derived class constructor initialization list sets m_cost to 1.3
  8. The derived class constructor body executes, which does nothing
  9. The derived class constructor returns

This may seem somewhat complex, but it’s actually very simple. All that’s happening is that the Derived constructor is calling a specific Base constructor to initialize the Base portion of the object. Because m_id lives in the Base portion of the object, the Base constructor is the only constructor that can initialize that value.

Note that it doesn’t matter where in the Derived constructor initialization list the Base constructor is called -- it will always execute first.

Now we can make our members private

Now that you know how to initialize base class members, there’s no need to keep our member variables public. We make our members variables private again, as they should be.

As a quick refresher, public members can be accessed by anybody. Private members can only be accessed by member functions of the same class. Note that this means derived classes can not access private members of the base class directly! Derived classes will need to use access functions to access private members of the base class.


In the above code, we’ve made m_id and m_cost private. This is fine, since we use the relevant constructors to initialize them, and use a public accessor to get the values.

This prints, as expected:

Id: 5
Cost: 1.3

We’ll talk more about access specifiers in the next lesson.

Another example

Let’s take a look at another pair of class we’ve previously worked with:

As we’d previously written it, BaseballPlayer only initializes its own members and does not specify a Person constructor to use. This means every BaseballPlayer we create is going to use the default Person constructor, which will initialize the name to blank and age to 0. Because it makes sense to give our BaseballPlayer a name and age when we create them, we should modify this constructor to add those parameters.

Here’s our updated classes that use private members, with the BaseballPlayer class calling the appropriate Person constructor to initialize the inherited Person member variables:

Now we can create baseball players like this:

This outputs:

Pedro Cerrano

As you can see, the name and age from the base class were properly initialized, as was the number of home runs from the derived class.

Inheritance chains

Classes in an inheritance chain work in exactly the same way.

In this example, class C is derived from class B, which is derived from class A. So what happens when we instantiate an object of class C?

First, main() calls C(int, double, char). The C constructor calls B(int, double). The B constructor calls A(int). Because A does not inherit from anybody, this is the first class we’ll construct. A is constructed, prints the value 5, and returns control to B. B is constructed, prints the value 4.3, and returns control to C. C is constructed, prints the value ‘R’, and returns control to main(). And we’re done!

Thus, this program prints:

A: 5
B: 4.3
C: R

It is worth mentioning that constructors can only call constructors from their immediate parent/base class. Consequently, the C constructor could not call or pass parameters to the A constructor directly. The C constructor can only call the B constructor (which has the responsibility of calling the A constructor).


When a derived class is destroyed, each destructor is called in the reverse order of construction. In the above example, when c is destroyed, the C destructor is called first, then the B destructor, then the A destructor.


When constructing a derived class, the derived class constructor is responsible for determining which base class constructor is called. If no base class constructor is specified, the default base class constructor will be used. In that case, if no default base class constructor can be found (or created by default), the compiler will display an error. The classes are then constructed in order from most base to most derived.

At this point, you now understand enough about C++ inheritance to create your own inherited classes!

Quiz time!

1) Let’s implement our Fruit example that we talked about in our introduction to inheritance. Create a Fruit base class that contains two private members: a name (std::string), and a color (std::string). Create an Apple class that inherits Fruit. Apple should have an additional private member: fiber (double). Create a Banana class that also inherits Fruit. Banana has no additional members.

The following program should run:

And print the following:

Apple(Red delicious, red, 4.2)
Banana(Cavendish, yellow)

Hint: Because a and b are const, you’ll need to mind your const’s. Make sure your parameters and functions are appropriately const.

Show Solution

11.5 -- Inheritance and access specifiers
11.3 -- Order of construction of derived classes

72 comments to 11.4 — Constructors and initialization of derived classes

  • Hi,

    in the solution here:

    I am just concerned about memory wastage if we add more fruit and need to keep redefining the Base class....

    • Alex

      I got 56, 64, 56. Banana doesn't take any additional memory over Fruit, and Apple only does because it has an added double member.

  • Arumikaze

    Hello! I have completed the quiz in separate files. Could you please give me feedback on my code and if I did it correctly? Thank you!








    Sorry for the long post.

    • Hi Arumikaze!

      * Default arguments should be in the header. If you write a library, the person using the library does only have the source code of the headers files. But they should know about default arguments.
      * Wherever you're accepting an @std::string as a parameter or returning one, you should do so by const reference.
      * main.cpp:9,12: Uniform initialization
      * Since @Banana's constructor should be a copy of @Fruit's constructor, you can inherit the constructor so you don't have to write it again.

  • Endcity

    I have considered the above example and I think constructor of derived class will execute lastly but member initializer lists of derived class will execute first so that int a can pass through C,B to A. Is that right ?

  • Mau


    thanks for the great tutorial and also a question.

    I got a bit confused by all those constructors being called, but despite that, in the end there is only one object on the call stack, isn't it?

    • nascardriver

      Hi Mau!

      There are no objects on the call stack, there are addresses on the call stack.
      Assuming you're talking about regular objects, there's only one object, no matter how many constructors where called.

  • Ran

    My code is less efficient than Alex's solution because I re-write
    getName() twice in the derived class. A better solution is to put
    getName() in the base class.

    I tried to do Alex's style, but I was struck with the constructor.

    Putting the user-defined initial values in the derived class's
    constructor is what I learned from Alex's solution.

    • nascardriver

      Hi Ran!

      * @m_name is never initialized not used.
      * @Banana doesn't have a color on it's own, it's using the default value, it works, but it's not nice.
      * Uniform initialization is preferred.
      * std::strings should be passed and returned by reference

  • xjuggy

    I initially set my classes and member functions up in a similar fashion to yours, Alex, and all was well.  Then I decided to experiment with the "never reuse code" mantra:


    Thus, I tried to create a Fruit overloaded "<<" operator to handle the getName/getColor portion of the output.

    In the course of doing so, I realized I (a) probably overly complicated my code and (b) ended up with what I deduced was a recursive function call to infinity.

    Question: Is there a way to call a specific version of the overloaded "<<" operator?

    My code which prints Apple until a stack overflow:

    • xjuggy

      Well, I made it to lesson 11.6a, and found the answer there: "static_cast"

      • Alex

        There's a better answer. Remove the overloaded operator<< functions for Apple and Banana. You don't need them.

        Without them, Fruit::operator<&lt will be called, and that function already has everything it needs to print the Fruit's name and color.

  • Ada

    how is this? i used a different main function from the one given in the quiz

    using namespace std;

    class Fruit
            string m_name;
            string m_color;
            Fruit(string name=" ",string color=" ")
              string getName()
               const {return m_name;}
              string getColor()
               const {return m_color;}

    class Apple:public Fruit
            double m_fiber;
            Apple(string name=" ",string color=" ",double fiber=0.0)
              : Fruit(name,color), //call Fruit(string,string) to initialize these fields
              double getFiber()
               const {return m_fiber;}
    class Banana:public Apple
            int m_best;
            Banana(string name=" ",string color=" ",double fiber=0.0,int best=0.0)
              : Apple(name,color,fiber), //call Apple(string,string,double) to initialize these fileds
              int getBest()
               const {return m_best;}

    int main()
        const Apple a("Red delicious", "red", 4.2);

        const Banana b("Cavendish", "yellow");

        return 0;

    Red delicious

  • Alexxx

    About that part:
    Note that it doesn’t matter where in the Derived constructor initialization list the Base constructor is called -- it will always execute first.
    If i do

    then i get warnings

    Worth noting in the same paragraph?

    • Alex

      This warning will occur (for compilers that support it) any time you initialize class members in a different order than in which they are defined in the class (or, in this case, with the base class constructor called first). It's not really specific to construction of derived classes, but rather to all classes.

  • AMG

    Could you please tell me if in my version "const" is redundant in class constructors and access functions.

    Thank you very much.

Leave a Comment

Put all code inside code tags: [code]your code here[/code]