Does copying an empty object involve accessing it

Question

Inspired from this question.

struct E {};
E e;
E f(e);  // Accesses e?

To access is to

read or modify the value of an object<

Answer 1

I think it does not access the object, though a valid object is required to be present.

E f(e);

This calls E's implicitly defined constructor E::E(const E&). Obviously the body of this constructor is empty (because there is nothing to do). So if anything happens, it must happen during argument passing, i.e. during the initialization of const E& from e.

It is self-evident that this initialization does not modify e. Now, to read the value of e, a lvalue-to-rvalue conversion must take place. However, the standard actually says that this conversion does not take place during direct reference binding¹. That is to say, no read is performed.

However, the standard does require that a reference must be initialized to refer to a valid object or function² (though this is subject to CWG 453), so things like E f(*reinterpret_cast(nullptr)); will be ill-formed.

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^{1. This is done by not normatively requiring such conversion, and further strengthened by the non-normative note in [dcl.init.ref].}

^{2. [dcl.ref].}