5 ms·

How so? constexpr doesn't guarantee that an expression is evaluated at compile time, but it does guarantee that it could be.

• jcelerier 4y ago
  > but it does guarantee that it could be. in which sense do you mean this? consider for instance constexpr int foo(int x) { if(x < 0) throw "error"; return x; } I would say that "guarantee that it could be" means that as long as the function definition itself "builds" / is validated by the compiler, you can use it in a constexpr context yet this is not the case here: constexpr int a = foo(123); // works fine constexpr int b = foo(-123); // compile error int c = foo(-123); // works fine so having "constexpr" in the API does not mean that your code will always build ; as soon as you have constexpr the entire implementation of the function is part of the API, thus making the keyword moot like gpderetta says
  • cbsmith 4y ago
    Yup. Turns out, I was wrong.
• gpderetta 4y ago
  That's what everybody originally believes (me included)! constexpr is_right_answer(int x){ if (x==42) return true; else { std::cout <<"try again \n"; return false; } A function can be constexpr as long as at least one possible invocation can be compile time evaluated. In fact compilers can't generally prove that a function can't be constexpr. I believe consteval has the semantics you want but I've yet to study it in details. For more details see: https://www.foonathan.net/2020/10/constexpr-platform/ https://www.foonathan.net/2020/10/constexpr-platform/ .
  • cbsmith 4y ago
    I stand corrected.