CODEWITHSUNDEEP
c++inheritancepolymorphism
Anoop Kumar
Anoop Kumar

Reputation: 137

Virtual table for the derived class which is not having any virtual function except parent class a virtual function

Is virtual table will be created for the derived class which is not having any virtual function except parent class a virtual function which is not overridden by derived class.

for ex:

class A{
public:
    virtual void show();

};

class B : public A
{

};

How about the virtual table of class B.

Upvotes: 6

Views: 1923

Answers (3)

dmytro.poliarush
dmytro.poliarush

Reputation: 383

Here is what gdb says when compiled with g++ (Ubuntu 8.2.0-1ubuntu2~18.04) 8.2.0

15  class A
16  {
17      public:
18          virtual void show(){}
19  };  
20  
21  class B:public A
22  {
23  };  
24  
(gdb) l
25  int main()
26  {
27      A a;
28      B b;
29  }
(gdb) p a 
$5 = {_vptr.A = 0x55555575f5c8 <vtable for A+16>}
(gdb) p b 
$6 = {<A> = {_vptr.A = 0x55555575f5b0 <vtable for B+16>}, <No data fields>}
(gdb)

Thus, at least in this case, we can conclude that base and derived classes have distinct vtables.

Upvotes: 2

psliwa
psliwa

Reputation: 1092

You can check it by looking into object's contents. I wrote this simple program that prints the contents of the base class, derived class and a class that is the same as the base class but with a normal method instead of the virtual one:

#include <iostream>
#include <string>
#include <iomanip>

using namespace std;

class Base {
public:
    virtual void show() {}
};

class Derived : public Base
{ };

class NonVirtual {
public:
    void show() {}
};

struct Test
{
    int data1, data2;
};

template <typename T>
void showContents(T* obj, string name)
{
    Test* test = new Test{};
    test = reinterpret_cast<Test*>(obj);
    cout << name << ": " << hex << "0x" << test->data1 << " " << "0x" << test->data2 << endl;
    delete test;
}

int main()
{
    Base* base = new Base{};
    Derived* derived = new Derived{};
    NonVirtual* nonVirtual = new NonVirtual{};

    showContents(base, "Base");
    showContents(derived, "Derived");
    showContents(nonVirtual, "NonVirtual");

    delete base;
    delete derived;
    delete nonVirtual;
}

Live demo

The result of running the above program after compiling it with cpp.sh (I'm not sure what compiler is used there):

Base: 0x4013e0 0x0
Derived: 0x401400 0x0
NonVirtual: 0x0 0x0

so I'd expect it to mean that there has been a virtual table created for the Derived object indeed (at least for this compiler - as required behavior is not defined in the C++ standard).

Upvotes: 0

CyberGuy
CyberGuy

Reputation: 2813

There is no standard answer for your question. Its really depending on the compiler version. There is no standard ABI specified in C++. If you are interested deeper please take a look at "Itanium C++ ABI" or try to find the answer on your own by looking into the asembler code.

There was even a proposal to define portable ABI for C++

http://www.open-std.org/jtc1/sc22/wg21/docs/papers/2014/n4028.pdf

Upvotes: 2

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