为什么我不能在模板参数中向下转换指向成员的指针?

如果我创建一个指向基成员的指针,我通常可以将其转换为指向派生成员的指针,但在下面的buzz之类的模板中使用时,则不会,其中第一个模板参数会影响第二个模板参数。我是在与编译器bug作斗争,还是标准确实要求它不起作用?

struct Foo
{
    int x;
};

struct Bar : public Foo
{
};

template<class T, int T::* z>
struct Buzz
{
};

static int Bar::* const workaround = &Foo::x;

int main()
{
    // This works. Downcasting of pointer to members in general is fine.
    int Bar::* y = &Foo::x;

    // But this doesn't, at least in G++ 4.2 or Sun C++ 5.9. Why not?
    // Error: could not convert template argument '&Foo::x' to 'int Bar::*'
    Buzz<Bar, &Foo::x> test;

    // Sun C++ 5.9 accepts this but G++ doesn't because '&' can't appear in
    // a constant expression
    Buzz<Bar, static_cast<int Bar::*>(&Foo::x)> test;

    // Sun C++ 5.9 accepts this as well, but G++ complains "workaround cannot
    // appear in a constant expression"
    Buzz<Bar, workaround> test;

    return 0;
}