Why is it allowed to overide non-virtual function?

Question

I'm experimneting with inheritance in C++.

struct A {
    virtual void foo(){ std::cout << "foo()" << std::endl; }
    void bar(){ std::cout << "bar()" << std::endl; }
};

struct B : A{
    void foo(){ std::cout << "derived foo()" << std::endl; }
    void bar(){ std::cout << "derived bar()" << std::endl; }
};

struct C : B {
    void foo(){ std::cout << "derived derived foo()" << std::endl; }
    void bar(){ std::cout << "derived derived bar()" << std::endl; }
};

int main()
{

    B* b = new C();
    b->foo();  //derived derived foo()
    b->bar();  //derived bar()
}

LIVE DEMO

Since, the function foo declared as non-virtual in the struct B I expected that B 's function would be called. But foo which one from C was. Why? I change the "virtual status" of the function in B . Why is it still virtual?

Answer 1

foo() is declared as virtual function in the base class A , so foo() in all the derived class will be virtual too.

From the standard, 10.3$2 Virtual functions [class.virtual] (bold by me)

If a virtual member function vf is declared in a class Base and in a class Derived, derived directly or indirectly from Base, a member function vf with the same name, parameter-type-list (8.3.5), cv-qualification, and ref-qualifier (or absence of same) as Base::vf is declared, then Derived::vf is also virtual ( whether or not it is so declared ) and it overrides Base::vf.

Answer 2

Once virtual always virtual.

Since foo is virtual in A it will be virtual in all classes derived from A - whether or not they get the virtual keyword.

Answer 3

Above mention answers are valid, but can we use override keyword, because

The override special identifier means that the compiler will check the
base class(es) to see if there is a virtual function with this exact
signature. And if there is not, the compiler will indicate an error.

so something like this is struct B :

void bar() override