C++ check if statement can be evaluated constexpr

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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());
    }
}

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