Why I need to set a variable to some value in base class when I declare a derived class?

Question

I dont understand why, when i try to create a object of a base class, i can, but only without declaring derived class, and when declare Derived i need to set argument value to "= 0" in base class constructor?

#include <cstdlib>

using namespace std;

class Base {
   public:
       int m_nValue;

       Base(int nValue=0) //<--- why here i need = 0 ?
           : m_nValue(nValue)
       {}
};

class Derived: public Base {
public:
    double m_dValue;

    Derived(double dValue) 
        : m_dValue(dValue)
    {}
};


int main(int argc, char** argv) {
  Base cBase(5); // use Base(int) constructor
  return 0;
}

Answer 1

No you don't. The problem is that when you subclass, you never give any value to the base class ctor, something like this:

Derived(double dValue) : Base(0), m_dValue(dValue) {}

So the compiler looks at the base class and search either: a ctor with no argument, or a ctor with default values (he doesn't look at the default ctor as defining a ctor removes the default ctor). Either:

class Base {
public:    
    int m_nValue;   
    Base() : m_nValue(0) {}
    Base(int nValue) : m_nValue(nValue) {}
};

or

class Base {
public:    
    int m_nValue;   
    Base(int nValue=0) : m_nValue(nValue) {}
};

Answer 2

Base(int nValue = 0) will also act as the default constructor, since a default value for nValue has been supplied. Currently, your Derived constructor will call this constructor implicitly, with nValue set to 0.

Formally you don't actually need to do that - but that would mean that the user of your class would have to supply an explicit value for nValue and you would have to write a default constructor . You could achieve the latter by writing Base() = default; although then the member m_nValue will not be initialised unless you use the C++11 curly brace notation (eg Base b{}; ). (Note that the behaviour on reading an uninitialised variable is undefined .)