Detecting constexpr with SFINAE

Question

I'm working on upgrading some C++ code to take advantage of the new functionality in C++11. I have a trait class with a few functions returning fundamental types which would most of the time, but not always, return a constant expression. I would like to do different things based on whether the function is constexpr or not. I came up with the following approach:

template<typename Trait>
struct test
{
    template<int Value = Trait::f()>
    static std::true_type do_call(int){ return std::true_type(); }

    static std::false_type do_call(...){ return std::false_type(); }

    static bool call(){ return do_call(0); }
};

struct trait
{
    static int f(){ return 15; }
};

struct ctrait
{
    static constexpr int f(){ return 20; }
};

int main()
{
   std::cout << "regular: " << test<trait>::call() << std::endl;
   std::cout << "constexpr: " << test<ctrait>::call() << std::endl;
}

The extra int/... parameter is there so that if both functions are available after SFINAE, the first one gets chosen by overloading resolution.

Compiling and running this with Clang 3.2 shows:

regular: 0
constexpr: 1

So this appears to work, but I would like to know if the code is legal C++11. Specially since it's my understanding that the rules for SFINAE have changed.

Answer 1

NOTE: I opened a question here about whether OPs code is actually valid. My rewritten example below will work in any case.

---

but I would like to know if the code is legal C++11

It is, although the default template argument may be considered a bit unusual. I personally like the following style better, which is similar to how you (read: I) write a trait to check for a function's existence, just using a non-type template parameter and leaving out the decltype:

#include <type_traits>

namespace detail{
template<int> struct sfinae_true : std::true_type{};
template<class T>
sfinae_true<(T::f(), 0)> check(int);
template<class>
std::false_type check(...);
} // detail::

template<class T>
struct has_constexpr_f : decltype(detail::check<T>(0)){};

Live example.

---

Explanation time~

Your original code works^† because a default template argument's point of instantiation is the point of instantiation of its function template, meaning, in your case, in main, so it can't be substituted earlier than that.

§14.6.4.1 [temp.point] p2

If a function template [...] is called in a way which uses the definition of a default argument of that function template [...], the point of instantiation of the default argument is the point of instantiation of the function template [...].

After that, it's just usual SFINAE rules.

---

† Atleast I think so, it's not entirely clear in the standard.

Answer 2

Prompted by @marshall-clow, I put together a somewhat more-generic version of an type-trait for detecting constexpr. I modelled it on std::invoke_result, but because constexpr depends on the inputs, the template arguments are for the values passed in, rather than the types.

It's somewhat limited, as the template args can only be a limited set of types, and they're all const when they get to the method call. You can easily test a constexpr wrapper method if you need other types, or non-const lvalues for a reference parameter.

So somewhat more of an exercise and demonstration than actually-useful code.

And the use of template<auto F, auto... Args> makes it C++17-only, needing gcc 7 or clang 4. MSVC 14.10.25017 can't compile it.

namespace constexpr_traits {

namespace detail {

// Call the provided method with the provided args.
// This gives us a non-type template parameter for void-returning F.
// This wouldn't be needed if "auto = F(Args...)" was a valid template
// parameter for void-returning F.
template<auto F, auto... Args>
constexpr void* constexpr_caller() {
    F(Args...);
    return nullptr;
}

// Takes a parameter with elipsis conversion, so will never be selected
// when another viable overload is present
template<auto F, auto... Args>
constexpr bool is_constexpr(...) { return false; }

// Fails substitution if constexpr_caller<F, Args...>() can't be
// called in constexpr context
template<auto F, auto... Args, auto = constexpr_caller<F, Args...>()>
constexpr bool is_constexpr(int) { return true; }

}

template<auto F, auto... Args>
struct invoke_constexpr : std::bool_constant<detail::is_constexpr<F, Args...>(0)> {};

}

Live demo with use-cases on wandbox