I have been reading Effective C++ 3rd Edition by Scott Meyers and in one of the chapters he says that any class with virtual functions should almost certainly have a virtual destructor. Now for my code below, the function someFunc does not need a virtual in the derived class. However, I decided to put it there to show semantics and for better readability. Since I put the virtual there, does it mean that the destructor in the derived class has to be virtual?
#include <iostream>
using namespace std;
class base
{
public:
base(){...}
virtual ~base(){...}
virtual someFunc() = 0;
};
class derived1:public base
{
public:
derived1(){...}
~derived1(){...} //Does this need to be virtual?
virtual someFunc(/*Implement the function*/); //I made this virtual just to show meaning
};
int main()
{
base* b;
b=new derived1;
delete b; //Will this cause a memory leak?
}
It already is!
The virtual keyword is implicit for your derived::someFunc, and for your derived::~derived, because both functions' base equivalents are marked virtual.
So, you may write the keyword virtual on both, neither or just one. It doesn't matter: they will both be actually virtual regardless.