Yhyjyk

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.