static const vs constexpr member fields in C++11/14?

Question

What is the difference between these member variables:

struct my_class {
    static const int i = 0;
    static constexpr int j = 0;
};

If my understanding is correct, I'm able to use both i and j as compile time constants. That is, both std::array<int, my_class::i> and std::array<int,my_class::j> should work.

Possible duplicate of [Difference between \`constexpr\` and \`const\`](http://stackoverflow.com/questions/14116003/difference-between-constexpr-and-const) — Humam Helfawi, Mar 29 '16 at 10:50
After reading that question, i can't still figure out the answer to my question :-) — amin, Mar 29 '16 at 10:52
@HumamHelfawi i want know what difference does it make when the constexpr is static. that is a little bit different. — amin, Mar 29 '16 at 10:54

Columbo · Accepted Answer · 2016-03-29T11:03:48.857

There is no difference for members of integral or enumeration type (as in your example). For all other types, constant expressions require constexpr:

an lvalue-to-rvalue conversion (4.1) unless it is applied to

a non-volatile glvalue of integral or enumeration type that refers to a complete non-volatile const object with a preceding initialization, initialized with a constant expression, or […]

a non-volatile glvalue that refers to a non-volatile object defined with constexpr, or that refers to a non-mutable sub-object of such an object, or […]

static const vs constexpr member fields in C++11/14?

1 Answers1