Why is Overload Resolution favoring unconstrained template function over a more specific one?

Question

I have this minimal expression template library with a multiplication, i.e.

template <typename T, typename U>
struct mul {
    const T &v1;
    const U &v2;
};

template <typename T, typename U>
mul<T, U> operator*(const T &one, const U &two) {
    std::cout << " called: mul<T, U> operator*(const T &one, const T &two)\n";
    return mul<T, U>{one, two};
}

and transpose, i.e.

template <typename T>
struct transpose {
    const T &t;
};

template <typename T>
transpose<T> tran(const T &one) {
    return transpose<T>{one};
}

I will introduce some types A and B, where the latter is a subclass of the former:

template <typename T>
struct A {
    T elem;
};

template <typename T>
struct B : A<T> {
    B(T val) : A<T>{val} {}
};

Then, I can call my expression template library as follows (with an overload for printing to std::cout):

template <typename T, typename U>
std::ostream &operator<<(std::ostream &os, const mul<T, U> &m) {
    os << " unconstrained template \n";
}

int main(int argc, char const *argv[]) {
    B<double> a{2};
    B<double> b{3};
    std::cout << tran(a) * b << "\n";
    return 0;
}

This gives me the output :

called: mul<T, U> operator*(const T &one, const T &two)
unconstrained template 

So far so good. Suppose now that I want a specialized treatment for 'transpose of a variable of type A<T> times a variable of type A<T> for some type T', as I had in my main. To this end, I will introduce

template <typename T>
T operator*(const transpose<A<T>> &one, const A<T> &two) {
    std::cout << " called: T operator*(const A<T> &one, const A<T> &two)\n";
    return one.t.elem * two.elem;
}

I run the same main function as above, and I still get the same output as above (unconstrained template). This is to be expected, since transpose<B<double>> is a completely different type compared to transpose<A<double>>, so overload resolution picks the unconstrained template version of operator*.

(Of course, if I change my variable definitions in main to A instead of B, ADL calls the specialized function and output is called: T operator*(const A<T> &one, const A<T> &two) and 6).

I recently learned about SFINAE, so I expected the following change to the more specific multiplication operator would cause overload resulution to select the specialized function:

template <typename T, typename V>
std::enable_if_t<std::is_base_of<A<T>, V>::value, T> operator*(const transpose<V> &one,
                                                               const V &two) {
    std::cout << " called: std::enable_if_t<std::is_base_of<A<T>, V>::value, T> operator*(const "
                 "transpose<V> &one, const V &two)\n";
    return one.t.elem * two.elem;
}

Even using the SFINAE'd operator* I still get the unconstrained template version. How come? What changes should I make to call the more specialized template function?

Smart

Maybe my understanding of what ADL is is incorrect, but what this has to do with ADL? – Cannell

I thought 'Overload Resolution' and ADL are the same. Apparently, they're not. – Smart

Overloading resolution is choosing between names found incl. those found by ADL. – Wagonlit

Answer 1

The problem is that in the SFINAE overload, T is used in a non-deduced context. You're effectively asking the compiler: "Enable this if there exists a T such that A<T> is a base class of V." Existential quantification is a good indicator that what you're asking for cannot be SFINAEd.

You can see this yourself if you disable the unconstrained template, as I did here. This forces the compiler to spell out why the other function is not admissible.

You can solve this by making T available through your A (and thus B) classes, like this:

template <typename T>
struct A {
    using Type = T;
    T elem;
};


template <typename V>
std::enable_if_t<std::is_base_of<A<typename V::Type>, V>::value, typename V::Type> operator*(const transpose<V> &one,
                                                               const V &two) {
    std::cout << " called: std::enable_if_t<std::is_base_of<A<T>, V>::value, T> operator*(const "
                 "transpose<V> &one, const V &two)\n";
    return one.t.elem * two.elem;
}

[Live example]

Dublin

In fact removing the sfinae, the superfluous template argument then uncommenting the uncontrainted one yield me the correct result. – Arbogast

@GuillaumeRacicot Yes, that works in the reduced example posted. But note that the OP says that the constrained version should only be used when V is derived from A<T> for some A. Removing the sfinae would make it apply to all transpose-nontranspose pairs. – Dublin

Reading this SFINAE it should only allow typename V to be B<T> but given that std::is_base_of<Base, Derived>::value yields true if std::is_same<Base, Derived>::value it also allows A<T>. Shouldn't there be a check to disallow that, too? If I am understand OP's Intention correctly, that is. – Sideslip

@StackDanny I understood the OP's question as "how do I word SFINAE for a special case that involves A<T> or a type derived from A<T>?" That's how I read "Suppose now that I want a specialized treatment for 'transpose of a variable of type A<T> times a variable of type A<T> for some type T', as I had in my main." Note that it was a B<double> in the main, actually. – Dublin

@Angew I just find it weird because if you were to remove struct B from the code the SFINAE (which strictly asks for a std::is_base_of case) will still be called even though there is no inheritance whatsover going on. I just thoughts that's unintentional. I suppose this is to be blamed on the implementation of std::is_base_of. – Sideslip

@Angew You interpreted my intent correctly, although more generally, A<T> would be an abstract class. – Smart

@StackDanny std::is_base_of effectively models IS-A. To be fair, it's much more common in practice to need to know "are you X or one of its descendants" than "are you derived from X, but not X itself." – Dublin