Use sizeof with incomplete type inside std::conditional

Question

Here is a minimal example:

struct incomplete_type;

template<typename T>
struct foo
{
    using type = std::conditional_t<std::is_arithmetic_v<T>,
        std::conditional_t<sizeof(T) < sizeof(void*), int, float>,
        double>;
};

foo<incomplete_type> f; will cause error because it will do sizeof with type, even though incomplete_type is not a arithmetic type(iow, it will not go into sizeof branch logically). live demo

So, I want to defer sizeof:

first attempt(fail)

template<typename T>
auto
foo_aux()
{
    if(sizeof(T) < sizeof(T*))
        return 0;
    else
        return 0.0f;
}

conditional_t<std::is_arithmetic_v<T>, decltype(foo_aux<T>()), double> still trigger the same error.

second attempt(fail)

template<typename T, bool>
struct foo_aux_aux
{
    using type = float;
};
template<typename T>
struct foo_aux_aux<T, true>
{
    using type = int;
};

template<typename T, bool = false>
struct foo_aux : foo_aux_aux<T, sizeof(T) < sizeof(void*)>
{};

conditional_t<std::is_arithmetic_v<T>, typename foo_aux<T>::type, double> still trigger the same error.

third attempt(success)

template<typename T, bool comp>
struct foo_aux_aux
{
    using type = float;
};
template<typename T>
struct foo_aux_aux<T, true>
{
    using type = int;
};

template<typename T, bool isArithmeticType>
struct foo_aux
{
    using type = double;
};

template<typename T>
struct foo_aux<T, true>
{
    using type = typename foo_aux_aux<T, sizeof(T) < sizeof(void*)>::type;
};

Yes, it works as expected, but its really tedious and ugly.

Do you have an elegant way here?

Answer 1

In C++17, you can use if constexpr to do type computation. Just wrap the type into a dummy container and use value computation, then retrieve the type via decltype.

struct foo could be implemented like:

template<class T>
struct type_ {
    using type = T;
};

template<class T>
struct foo {
    auto constexpr static type_impl() {
        if constexpr (std::is_arithmetic<T>{}) {
            if constexpr (sizeof(T) < sizeof(void*)) {
                return type_<int>{};
            } else {
                return type_<float>{};
            }
        } else {
            return type_<double>{};
        }
    }

    using type = typename decltype(type_impl())::type;
};

static_assert(std::is_same<foo<incomplete_type>::type, double>{});

Answer 2

Your second attempt works if you wrap double in type_identity (which is a standard utility in C++20) and move ::type after std::conditional_t<...>:

template<typename T, bool>
struct foo_aux_aux
{
    using type = float;
};
template<typename T>
struct foo_aux_aux<T, true>
{
    using type = int;
};

template<typename T, bool = false>
struct foo_aux : foo_aux_aux<T, sizeof(T) < sizeof(void*)>
{};

template<typename T>
struct type_identity { using type = T; };

typename std::conditional_t<std::is_arithmetic_v<T>, foo_aux<T>, type_identity<double>>::type

Answer 3

Not a great improvement, I suppose, but you can rewrite your third (working) attempt in a little less ugly (IMHO) way using decltype() and declaring (only) some functions.

I mean something as

struct incomplete_type;

constexpr float baz (std::false_type);
constexpr int baz (std::true_type);

template <typename>
constexpr double bar (std::false_type);

template <typename T>
constexpr auto bar (std::true_type)
 -> decltype(baz<std::bool_constant<(sizeof(T) < sizeof(void*))>{});

template<typename T>
struct foo
 { using type = decltype( bar<T>(std::is_arithmetic<T>{}) ); };