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Home/ Questions/Q 3495632
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Editorial Team
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Editorial Team
Asked: May 18, 20262026-05-18T12:09:01+00:00 2026-05-18T12:09:01+00:00

class Base { public: Base() { cout<<base class<<endl; fun(); } virtual void fun(){cout<<fun of

  • 0
class Base
{
public:
Base()
{
cout<<"base class"<<endl;
fun();
}
virtual void fun(){cout<<"fun of base"<<endl;}
};

class Derive:public Base
{
public:
Derive()
{
cout<<"derive class"<<endl;
fun();
}
void fun(){ cout<<"fun of derive"<<endl;}
};

void main()
{
Derive d;
}

The output is:

base class
fun of base
derive class
fun of derive

Why the second line is not fun of derive?

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1 Answer

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  1. Editorial Team
    Editorial Team
    2026-05-18T12:09:01+00:00Added an answer on May 18, 2026 at 12:09 pm

    When you call fun() in the base class constructor, the derived class has not yet been constructed (in C++, classes a constructed parent first) so the system doesn’t have an instance of Derived yet and consequently no entry in the virtual function table for Derived::fun().

    This is the reason why calls to virtual functions in constructors are generally frowned upon unless you specifically want to call the implementation of the virtual function that’s either part of the object currently being instantiated or part of one of its ancestors.

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