An rough implementation of `std::is_constructible`

Question

As a challenge/fun activity/task, I have implemented my version of std::is_constructible.

Here is the source code:

#include <cbrainx/cbrainx.hpp>
#include <iostream>
template <typename T, typename ...V>
auto aux(long) -> cbx::false_type;
template <typename T, typename ...V>
static auto aux(int) -> cbx::type_switch_t<
 cbx::branch<cbx::false_type,
 cbx::detail::void_t<decltype(::new(cbx::declval<void *>()) T{cbx::declval<V>()...})>>,
 cbx::otherwise<cbx::true_type>>;
template <typename T, typename ...V>
auto is_constructible() -> decltype(aux<T, V...>(0));
struct X {
 X() {
 std::cout << "constructed..." << std::endl;
 }
 private:
 X(int) {
 }
 public:
 X(const X &) = delete;
 X(X &&) {}
 ~X() {
 std::cout << "destroyed..." << std::endl;
 }
 static void xd() {
 X{1};
 std::cout << decltype(is_constructible<X, int>())::value << std::endl;
 }
};
auto main() -> int {
 std::cout << std::boolalpha;
 std::cout << decltype(is_constructible<X, X>())::value << std::endl;
 std::cout << decltype(is_constructible<X, X &>())::value << std::endl;
 std::cout << decltype(is_constructible<X, X &&>())::value << std::endl;
 return 0;
}

IMPORTANT:

• cbx::false_type is equivalent to std::false_type. It has a value of type bool. Likewise cbx::true_type.

• cbx::type_switch_t is equivalent to switch clause but, for types and evaluated statically at compile time.

• cbx::branch is like the if branch. The first argument is the condition and must have a value of type bool. The second argument is the type itself.

• cbx::otherwise is like the else branch.

Things I know

1. It does not work for private constructors. (I am okay with that!)
2. It's a rough implementation.

Things I want to know

1. Is this implementation correct?
2. Am I missing anything?
3. How close it is to the actual std::is_constructible?
4. How can I improve?

C++ Standard: 20

Answer 1

It would help to cleanly separate your is_constructible from the "test harness" code that follows it — either with a //----- comment, or by putting your thing in a namespace, or something.

Your thing is a function template, whereas the standard std::is_constructible is a type-trait (that is, a class template).

Without seeing the definitions of branch, otherwise, etc., it's hard to judge whether your code is even correct, let alone performant.

You name both your template parameter T and your parameter pack V... with singular names. It would be helpful to name the pack with a plural name, e.g. Vs... or Args....

Your unit tests take the form of std::cout << foo. Prefer to use static_assert(foo), so that the compiler will tell you whether your tests passed or failed — you shouldn't have to eyeball the terminal output to tell whether you implemented is_constructible right!

---

You should become familiar with the traditional SFINAE idioms. For a simple type-trait like is_constructible, where all you want to know is whether a particular expression is well-formed, the idiom is

template<class T, class = void>
struct is_fooable : std::false_type {};
template<class T>
struct is_fooable<T, decltype((
 your-expression-here
), void())> : std::true_type {};

Or, since you tagged this question c++20, you could just use a requires-expression:

template<class T>
struct is_fooable : std::bool_constant<requires {
 your-expression-here;
}> {};

In the specific case of is_constructible, the expression you care about is new T(args...). So you might write

template<class, class T, class... Args>
struct is_constructible_impl : std::false_type {};
template<class T, class... Args>
struct is_constructible_impl<decltype((
 new T(std::declval<Args>()...)
), void()), T, Args...> : std::true_type {};
template<class T, class... Args>
struct is_constructible : is_constructible_impl<void, T, Args...> {};

Or:

template<class T, class... Args>
struct is_constructible : std::bool_constant<requires (Args&&... args) {
 new T(static_cast<Args&&>(args)...);
}> {};

(Godbolt.)

---

By writing unit tests and comparing them to the standard std::is_constructible, we soon find a problem: std::is_constructible<int&, int&>::value is true, but our new T(args...) formulation yields false, because you can't new a reference type. Solving this problem is left as an exercise for the reader, because I'm too lazy to look it up right now. :)

• Mansoor Ahmed Memon: Did I do it write? godbolt.org/z/T1Kxrjbch @Quuxplusone
• Mansoor Ahmed Memon: Are there any other edge/special/unhandled cases? @Quuxplusone
• Quuxplusone: @MansoorAhmedMemon: Only one way to find out! Write test cases and compare your answers against std::is_constructible's answers. Try tricky things. godbolt.org/z/T1Kxrjbch seems to mess up on is_constructible<int&, float&>. The expression you chose to SFINAE on, T(x), can do a hidden reinterpret_cast.