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());
}
}