CODEWITHSUNDEEP
c++
rad123
rad123

Reputation: 187

derived class has same member variable name as base class

#include<iostream>
using namespace std;

class A
{
protected:
    int m_nValue;
public:
    A(int nValue):m_nValue(nValue)
    {
        cout << m_nValue << endl;
    }
};
class B: public A
{
public:
    B(int nValue): A(m_nValue)
    {
        cout << m_nValue << endl;
    }
    int getValue()
    {
        return m_nValue;
    }
};
int main()
{
    B b(4);
    cout << b.getValue() << endl;
    return 0;
}

Here, in the above program I am not declaring m_nValue again in the Derived class. In the output, I see only junk values getting displayed rather than displaying value "4". Please explain this.

Upvotes: 1

Views: 98

Answers (3)

songyuanyao
songyuanyao

Reputation: 172924

You're trying to initialize m_nValue with m_nValue itself. The parameter nValue (passed in the value 4) is not used at all. That's why m_nValue has only garbage value.

You might want

B(int nValue): A(nValue)
{
    cout << m_nValue << endl;
}

Upvotes: 3

Jack
Jack

Reputation: 133577

Your constructor is bugged B(int nValue): A(m_nValue). You are not using nValue to pass it to A constructor but the instance variable which is uninitialized.

Upvotes: 2

AlphaRL
AlphaRL

Reputation: 1629

B(int nValue): A(nValue)

You should initialize A with nValue rather than m_nValue

Upvotes: 2

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