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Is there a method to decide whether something can be constexpr evaluated, and use the result as a constexpr boolean? My simplified use case is as follows:

template <typename base>

class derived

{

    template<size_t size>

    void do_stuff() { (...) }



    void do_stuff(size_t size) { (...) }

public:

    void execute()

    {

        if constexpr(is_constexpr(base::get_data())

        {

            do_stuff<base::get_data()>();

        }

        else

        {

            do_stuff(base::get_data());

        }

    }

}

My target is C++2a.

I found the following reddit thread, but I'm not a big fan of the macros. https://www.reddit.com/r/cpp/comments/7c208c/is_constexpr_a_macro_that_check_if_an_expression/

Not exactly what you asked (I've developer a custom type trait specific for a get_value() static method... maybe it's possible to generalize it but, at the moment, I don't know how) but I suppose you can use SFINAE and make something as follows

#include <iostream>

#include <type_traits>



template <typename T>

constexpr auto icee_helper (int)

   -> decltype( std::integral_constant<decltype(T::get_data()), T::get_data()>{},

                std::true_type{} );



template <typename>

constexpr auto icee_helper (long)

   -> std::false_type;



template <typename T>

using isConstExprEval = decltype(icee_helper<T>(0));



template <typename base>

struct derived

 {

   template <std::size_t I>

   void do_stuff()

    { std::cout << "constexpr case (" << I << ')' << std::endl; }



   void do_stuff (std::size_t i)

    { std::cout << "not constexpr case (" << i << ')' << std::endl; }



   void execute ()

    {

      if constexpr ( isConstExprEval<base>::value )

         do_stuff<base::get_data()>();

      else

         do_stuff(base::get_data());

    }

 };



struct foo

 { static constexpr std::size_t get_data () { return 1u; } };



struct bar

 { static std::size_t get_data () { return 2u; } };



int main ()

 { 

   derived<foo>{}.execute(); // print "constexpr case (1)"

   derived<bar>{}.execute(); // print "not constexpr case (2)"

 }

Here's another solution, which is more generic (applicable to any expression, without defining a separate template each time).

This solution leverages that (1) lambda expressions can be constexpr as of C++17 (2) the type of a captureless lambda is default constructible as of C++20.

The idea is, the overload that returns true is selected when and only when Lambda{}() can appear within a template argument, which effectively requires the lambda invocation to be a constant expression.

template<class Lambda, int=(Lambda{}(), 0)>

constexpr bool is_constexpr(Lambda) { return true; }

constexpr bool is_constexpr(...) { return false; }



template <typename base>

class derived

{

    // ...



    void execute()

    {

        if constexpr(is_constexpr({ base::get_data(); }))

            do_stuff<base::get_data()>();

        else

            do_stuff(base::get_data());

    }

}

template<auto> struct require_constant;

template<class T>

concept has_constexpr_data = requires { typename require_constant<T::get_data()>; };

This is basically what's used by std::ranges::split_view.