From https://stackoverflow.com/questions/58479267/initialization-order-of-constant-initialized-variable-with-static-storage-durati @sdkrystian

To my understanding, nothing specifies what value b will have.

const int a = 42;
const int b = a;

Variable a is usable in constant expressions, so it is not ODR-used, so [conv.lval]/(2.2) apply:

2 When an lvalue-to-rvalue conversion is applied to an expression E, and either

• (2.1) E is not potentially evaluated, or
• (2.2) the evaluation of E results in the evaluation of a member E_x of the set of potential results of E, and E_x names a variable x that is not odr-used by E_x

the value contained in the referenced object is not accessed.

I think it means (and should say) the object is not accessed. But this is not the main concern. [conv.lval]/3, which says «The result of the conversion is determined according to the following rules», doesn't seem to cover the case when l-to-r conversion is applied to an evaluated expression denoting an object which is not accessed. I believe the bullet (3.3) saying «Otherwise, the object indicated by the glvalue is read ([defns.access]), and the value contained in the object is the prvalue result» can't apply in case the object is not accessed. Or the implied reading of (3.3) is that the part about access is just ignored for non-accessed objects?

(In case there is doubt that [conv.lval] is applied to a in const int b = a;, it can be replaced with const int b = a + 0;)