CODEWITHSUNDEEP
c++inheritance
amrhassan
amrhassan

Reputation: 2445

How can I initialize base class member variables in derived class constructor?

Why can't I do this?

class A
{
public:
    int a, b;
};

class B : public A
{
    B() : A(), a(0), b(0)
    {
    }

};

Upvotes: 187

Views: 234500

Answers (7)

John Dibling
John Dibling

Reputation: 101494

Why can't you do it? Because the language doesn't allow you to initializa a base class' members in the derived class' initializer list.

How can you get this done? Like this:

class A
{
public:
    A(int a, int b) : a_(a), b_(b) {};
    int a_, b_;
};

class B : public A
{
public:
    B() : A(0,0) 
    {
    }
};

Upvotes: 3

Violet Giraffe
Violet Giraffe

Reputation: 33607

Somehow, no one listed the simplest way:

class A
{
public:
    int a, b;
};

class B : public A
{
    B()
    {
        a = 0;
        b = 0;
    }

};

You can't access base members in the initializer list, but the constructor itself, just as any other member method, may access public and protected members of the base class.

Upvotes: 22

NPE
NPE

Reputation: 500853

Leaving aside the fact that they are private, since a and b are members of A, they are meant to be initialized by A's constructors, not by some other class's constructors (derived or not).

Try:

class A
{
    int a, b;

protected: // or public:
    A(int a, int b): a(a), b(b) {}
};

class B : public A
{
    B() : A(0, 0) {}
};

Upvotes: 31

manish srivastava
manish srivastava

Reputation: 39

# include<stdio.h>
# include<iostream>
# include<conio.h>

using namespace std;

class Base{
    public:
        Base(int i, float f, double d): i(i), f(f), d(d)
        {
        }
    virtual void Show()=0;
    protected:
        int i;
        float f;
        double d;
};


class Derived: public Base{
    public:
        Derived(int i, float f, double d): Base( i, f, d)
        {
        }
        void Show()
        {
            cout<< "int i = "<<i<<endl<<"float f = "<<f<<endl <<"double d = "<<d<<endl;
        }
};

int main(){
    Base * b = new Derived(10, 1.2, 3.89);
    b->Show();
    return 0;
}

It's a working example in case you want to initialize the Base class data members present in the Derived class object, whereas you want to push these values interfacing via Derived class constructor call.

Upvotes: 3

Nikos Athanasiou
Nikos Athanasiou

Reputation: 31587

While this is usefull in rare cases (if that was not the case, the language would've allowed it directly), take a look at the Base from Member idiom. It's not a code free solution, you'd have to add an extra layer of inheritance, but it gets the job done. To avoid boilerplate code you could use boost's implementation

Upvotes: 1

Gene Bushuyev
Gene Bushuyev

Reputation: 5538

Aggregate classes, like A in your example(*), must have their members public, and have no user-defined constructors. They are intialized with initializer list, e.g. A a {0,0}; or in your case B() : A({0,0}){}. The members of base aggregate class cannot be individually initialized in the constructor of the derived class.

(*) To be precise, as it was correctly mentioned, original class A is not an aggregate due to private non-static members

Upvotes: -1

In silico
In silico

Reputation: 52197

You can't initialize a and b in B because they are not members of B. They are members of A, therefore only A can initialize them. You can make them public, then do assignment in B, but that is not a recommended option since it would destroy encapsulation. Instead, create a constructor in A to allow B (or any subclass of A) to initialize them:

class A 
{
protected:
    A(int a, int b) : a(a), b(b) {} // Accessible to derived classes
    // Change "protected" to "public" to allow others to instantiate A.
private:
    int a, b; // Keep these variables private in A
};

class B : public A 
{
public:
    B() : A(0, 0) // Calls A's constructor, initializing a and b in A to 0.
    {
    } 
};

Upvotes: 207

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