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Home/ Questions/Q 7515047
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Editorial Team
  • 0
Editorial Team
Asked: May 30, 20262026-05-30T00:40:36+00:00 2026-05-30T00:40:36+00:00

Please take a look on the following example: class Base { protected: int m_nValue;

  • 0

Please take a look on the following example:

class Base
{
protected:
    int m_nValue;

public:
    Base(int nValue)
        : m_nValue(nValue)
    {
    }

    const char* GetName() { return "Base"; }
    int GetValue() { return m_nValue; }
};

class Derived: public Base
{
public:
    Derived(int nValue)
        : Base(nValue)
    {
    }
    Derived( const Base &d ){
        std::cout << "copy constructor\n";
    }

    const char* GetName() { return "Derived"; }
    int GetValueDoubled() { return m_nValue * 2; }
};

This code keeps throwing me an error that there are no default contructor for base class. When I declare it everything is ok. But when i dont, code does not work.

How can I declare a copy constructor in derived class without declaring default contructor in base class?

Thnaks.

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  1. Editorial Team
    Editorial Team
    2026-05-30T00:40:38+00:00Added an answer on May 30, 2026 at 12:40 am

    Call the copy-constructor (which is generated by the compiler) of the base:

    Derived( const Derived &d ) : Base(d)
    {            //^^^^^^^ change this to Derived. Your code is using Base
        std::cout << "copy constructor\n";
    }
    

    And ideally, you should call the compiler generated copy-constructor of the base. Don’t think of calling the other constructor. I think that would be a bad idea.

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