Scoping rules when inheriting - C++

Question

I was reading the C++0x FAQ by Stroustrup and got stuck with this code. Consider the following code

struct A
{
    void f(double)
    {
        std::cout << "in double" << std::endl;
    }
};

struct B : A
{
    void f(int)
    {
        std::cout << "in int" << std::endl;
    }
};


int main()
{
    A a; a.f(10.10);  // as expected, A.f will get called
    B b; b.f(10.10);  // This calls b.f and we lose the .10 here
    return 0;
}

My understanding was when a type is inherited, all protected and public members will be accessible from the derived class. But according to this example, it looks like I am wrong. I was expecting the b.f will call base classes f. I got the expected result by changing the derived class like

struct B : A
{
    using A::f;
    void f(int)
    {
        std::cout << "in int" << std::endl;
    }
};

Questions

1. Why was it not working in the first code?
2. Which section in the C++ standard describes all these scoping rules?

Answer 1

Its because A::f is "hidden" rather than "overloaded" or "overridden". Refer:

http://www.parashift.com/c++-faq-lite/strange-inheritance.html#faq-23.9

Answer 2

The first code works as c++ is designed to work.

Overloading resolution follows a very complicated set of rules. From Stroustrup's c++ bible 15.2.2 "[A]mbiguities between functions from different base classes are not resolved based on argument types."

He goes on to explain the use of "using" as you have described.

This was a design decision in the language.

I tend to follow the Stroustrup book rather than the standard, but I'm sure it is in there.

[Edit]

Here it is (from the standard):

Chapter 13

When two or more different declarations are specified for a single name in the same scope, that name is said to be overloaded.

And then:

13.2 Declaration matching

1 Two function declarations of the same name refer to the same function if they are in the same scope and have equivalent parameter declarations (13.1). A function member of a derived class is not in the same scope as a function member of the same name in a base class.

Answer 3

Search for overload resolution. A similar, but not identical question.