“surprising” constant initialization because of definition order

Question

When reading the slides about constexpr the introduction is about \"surprisingly dynamic initialization with consts\". The example is

struct S {
    st         


        

Answer 1

For static initialization one needs, roughly speaking, a constant-expression initializer.

To be a constant-expression, roughly speaking, a variable needs to be of a const type and have a preceding initialization with a constant-expression.

In the first example d's initializer is not a constant-expression, as S::c isn't one (it has no preceding initialization). Hence, d is not statically initialized.

In the second example d's initializer is a constant-expression, and everything is OK.

I'm simplifying matters. In full formal standardese this would be about nine times longer.

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As for constexpr specifier, no object has to be declared constexpr. It is just an additional error-check. (This is about constexpr objects, not constexpr functions).

You may declare S::c constexpr in the second variant if you want some extra error protection (perhaps 5 will start changing its value tomorrow?) Adding constexpr to the first variant cannot possibly help.