应该将基类的函数模板赋值给指向成员函数的指针

Question

Should the following code compile in C++98/03?

struct Base
{
    template <typename T> void func () { }
    void norm() {  }
};

struct Derived : public Base { };

template <typename U>
struct Usage
{
    typedef void (U::*Method)();

    Usage(Method test) { }
};

int main()
{
    Usage<Derived> good(&Derived::norm);

    // "Error: Cannot use void(*)() to initialize Usage<Derived>." on next line
    Usage<Derived> bad(&Derived::func<int>);

    return 0;
}

This code snippet worked just fine on nearly every compiler that I was able to try out; save Sun C++ 5.11 and Sun C++ 5.12.

Should that be a bug? If so, does anyone know if it has been reported to the vendor (currently Oracle)?

Edit:

I'll accept an answer that provide the proper relevant quotations from either the C++03 or C++11 standards documents. Or if you can provide information about a bug report with Oracle.

Answer 1

I just read most of the C++98 standard, chapter 14. There isn't really much said about what type are resulting (specialized) template member, so I assume that it follows the idea that being a template method doesn't make it any less of a method. I f I have a moment, I will see if C++11 says more about it.

From my general idea about C++ I know that your code should pass - and majority of compilers agreeing on that is also a clue, isn't it? :)