Why can't I declare a constexpr local, when a const one works?

Question

What the difference between up and up1 here?

Why does const work, but not constexpr?

class vec3 {
    int x, y, z;
public:
    vec3(int x, int y, int z) : x{x}, y{y}, z{z} {}
};

int main()
{
    // Error C2127'up': illegal initialization of 'constexpr'
    // entity with a non-constant expression
    constexpr vec3 up{0, 1, 0};

    const vec3 up1{0, 1, 0};
}

Possible duplicate of [const vs constexpr on variables](https://stackoverflow.com/questions/13346879/const-vs-constexpr-on-variables) — franiis, Aug 09 '18 at 10:40
Please provide [mcve], as text, with a copy-pasted **compiler** error message. What you are trying to show, is **not** an error message from a compiler. — Algirdas Preidžius, Aug 09 '18 at 10:43
Quoting g++: "'vec3' is not an aggregate, does not have a trivial default constructor, and has no constexpr constructor that is not a copy or move constructor" — molbdnilo, Aug 09 '18 at 10:59

geza · Answer 1 · 2018-08-15T13:09:57.490

1

If you want to create a constexpr object, then the corresponding constructor needs to be constexpr as well. So, the constructor needs to be like this:

constexpr vec3(int x, int y, int z) : x{x}, y{y}, z{z} {}

(If the initialization of a constexpr object involves calling a function, then the function needs to be constexpr as well. A constructor is not an exception to this rule.)

edited Aug 15 '18 at 13:09

answered Aug 15 '18 at 12:53

geza

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Why can't I declare a constexpr local, when a const one works?

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