Using constexpr as a rvalue

Question

Is there a way to use constexpr without the need to create an extra variable? Take for example the code below, which maps a type to a global object. The why of the mapping should not be relevant to the question.

#include <iostream>

struct S { int val; };

S one{1};
S two{2};

template<typename>
constexpr S& get() = delete;

template<>
constexpr S& get<int>() { return one; };

template<>
constexpr S& get<double>() { return two; };

auto main() -> int {

    constexpr auto& type = get<int>();

    std::cout << type.val << std::endl;
    std::cout << get<double>().val << std::endl;
    return 0; 
}

godbolt

The call to get<double>(); does not get treated as a constexpr, and it's executed at runtime (with, optimizations the call is inlined and it's equivalent to constexpr). The call to constexpr auto& type = get<int>(); does make use of constexpr do requires another variable. Just for usability, is there something to this extent std::cout << (constexpr get<int>()) << std::endl; that could be used?

"with, optimizations the call is inlined and it's equivalent to constexpr" why is that not ok? — 463035818_is_not_an_ai, Jan 23 '23 at 15:05
there are ways to enforce compile time evaluation also in c++14. You could write a class template with `val` as `static` member, but thats possibly a step back compared to what you have — 463035818_is_not_an_ai, Jan 23 '23 at 15:08
@463035818_is_not_a_number It is, I just wanted more consistency if that makes sense. — Lucas Mülling, Jan 23 '23 at 15:11
`constexpr` only implies that `get` can be evaluated at compile time. It does not imply that it is always evaluated at compile time. If you assume the latter then there is inconsistency. Whether it is called at compile time in contexts where it can be called at runtime, the compiler still decides as it does without `constexpr` — 463035818_is_not_an_ai, Jan 23 '23 at 15:20

Using constexpr as a rvalue

0 Answers0