Why can't I downcast pointer to members in template arguments?

问题

If I make a pointer-to-base-member, I can convert it to a pointer-to-derived-member usually, but not when used within a template like Buzz below, where the first template argument influences the second one. Am I fighting compiler bugs or does the standard really mandate this not work?

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;
}

回答1:

It simply isn't allowed. According to §14.3.2/5:

The following conversions are performed on each expression used as a non-type template-argument. If a non-type template-argument cannot be converted to the type of the corresponding template-parameter then the program is ill-formed.
— for a non-type template-parameter of integral or enumeration type, integral promotions (4.5) and integral conversions (4.7) are applied.
— for a non-type template-parameter of type pointer to object, qualification conversions (4.4) and the array-to-pointer conversion (4.2) are applied. — For a non-type template-parameter of type reference to object, no conversions apply. The type referred to by the reference may be more cv-qualified than the (otherwise identical) type of the template argument. The template-parameter is bound directly to the template-argument, which must be an lvalue.
— For a non-type template-parameter of type pointer to function, only the function-to-pointer conversion (4.3) is applied. If the template-argument represents a set of overloaded functions (or a pointer to such), the matching function is selected from the set (13.4).
— For a non-type template-parameter of type reference to function, no conversions apply. If the template-argument represents a set of overloaded functions, the matching function is selected from the set (13.4).
— For a non-type template-parameter of type pointer to member function, no conversions apply. If the template-argument represents a set of overloaded member functions, the matching member function is selected from the set (13.4).
— For a non-type template-parameter of type pointer to data member, qualification conversions (4.4) are applied.

I've emphasized the conversion regarding pointer to data members. Note that your conversion (§4.11/2) is not listed. In C++0x, it remains the same in this regard.

来源：https://stackoverflow.com/questions/4027154/why-cant-i-downcast-pointer-to-members-in-template-arguments