Why is the "using" directive still needed in C++11 to bring forward methods from the base class that are overloaded in the derived class

Question

The example below gets the following compiled error:

test.cpp: In function ‘int main(int, char**)’:
test.cpp:26:8: error: no match for call to ‘(Derived) (p1&)’
test.cpp:14:8: note: candidate is:
test.cpp:16:10: note: void Derived::operator()(const p2&)
test.cpp:16:10: note:   no known conversion for argument 1 from ‘p1’ to ‘const p2&’

It was my understanding this was getting changed in C++11 so you weren't required to put the using statement in. Is that not correct? Is there some other way around this?

Example (Compiled with gcc 4.7 using --std=c++11):

#include <iostream>
#include <string>

using namespace std;

struct p1{};
struct p2{};

struct Base
{
    void operator()(const p1&) { cout << "p1" << endl; }
};

struct Derived : public Base
{
    void operator()(const p2&) { cout << "p2" << endl; }
    //Works if I include: using Base::operator();
};

int main(int argc, char** argv)
{
    p1 p;
    p2 pp;
    Derived d;

    d(p);
    d(pp);
}

Answer 1

To the best of my knowledge, no, this has not changed in C++11.

And the reason it has not changed is that this behavior is not an accident. The language works like this by design. There are advantages and disadvantages to it, but it's not something that just happened because the people on the standards committee forgot about it.

And no, there's no way around it. It's just how member lookup works in C++