c++ - name hiding and fragile base problem

Question

I've seen it stated that C++ has name hiding for the purposes of reducing the fragile base class problem. However, I definitely don't see how this helps. If the base class introduces a function or overload that previously did not exist, it might conflict with those introduced by the derived class, or unqualified calls to global functions or member functions- but what I don't see is how this is different for overloads. Why should overloads of virtual functions be treated differently to, well, any other function?

Edit: Let me show you a little more what I'm talking about.

struct base {
    virtual void foo();
    virtual void foo(int);
    virtual void bar();
    virtual ~base();
};
struct derived : base {
    virtual void foo();
};

int main() {
    derived d;
    d.foo(1); // Error- foo(int) is hidden
    d.bar(); // Fine- calls base::bar()
}

Here, foo(int) is treated differently to bar(), because it's an overload.

Answer 1

I'll assume that by "fragile base class", you mean a situation where changes to the base class can break code that uses derived classes (that being the definition I found on Wikipedia). I'm not sure what virtual functions have to do with this, but I can explain how hiding helps avoid this problem. Consider the following:

struct A {};

struct B : public A
{
    void f(float);
};

void do_stuff()
{
    B b;
    b.f(3);
}

The function call in do_stuff calls B::f(float).

Now suppose someone modifies the base class, and adds a function void f(int);. Without hiding, this would be a better match for the function argument in main; you've either changed the behaviour of do_stuff (if the new function is public), or caused a compile error (if it's private), without changing either do_stuff or any of its direct dependencies. With hiding, you haven't changed the behaviour, and such breakage is only possible if you explicitly disable hiding with a using declaration.