Question

Consider the following code :

#include <iostream>
#include <type_traits>

class Base
{
    public: static int f() {return 42;}
};

class Derived : public Base
{
    protected: int x;
};

class NotDerived
{
    protected: int x;
};

int main()
{
    std::cout<<sizeof(Base)<<std::endl;
    std::cout<<sizeof(Derived)<<std::endl;
    std::cout<<sizeof(NotDerived)<<std::endl;
    return 0;
}

With g++ 4.7 -O3, it prints :

1
4
4

and if I understand that well, it means that Empty Base Class Optimization is enabled.

But my question concerns the runtime overhead : is there any overhead creating (and destructing) a Derived object compared to a NotDerived object due to the fact that Derived should construct/destruct the corresponding Base object ?

Answer 1

While the standard makes no guarantees to that there I would consider a compiler that did something different in those cases slightly defective.

There is literally nothing to be done to initialize the base: no memory has to be initialized, no virtual call mechanism has to be set up. No code should be generated for it.

However, you should always check some assembly in a non-trivial setting if this is really important to you.