Initialization by conversion function for direct reference binding - c++

Question

I found out the rules for determining the candidate conversion functions, for direct reference binding, are not described clearly by the standard at least for me.
The rules are found in [over.match.ref]
Under the conditions specified in [dcl.init.ref], a reference can be
bound directly to the result of applying a conversion function to an
initializer expression. Overload resolution is used to select the
conversion function to be invoked. Assuming that “reference to cv1 T”
is the type of the reference being initialized, the candidate
functions are selected as follows:
(1.1) Let R be a set of types including
(1.1.1) “lvalue reference to cv2 T2” (when initializing an lvalue reference or an rvalue reference to function) and
(1.1.2) “cv2 T2” and “rvalue reference to cv2 T2” (when initializing an rvalue reference or an lvalue reference to function) for any T2.
The permissible types for non-explicit conversion functions are the
members of R where “cv1 T” is reference-compatible ([dcl.init.ref])
with “cv2 T2”. For direct-initialization, the permissible types for
explicit conversion functions are the members of R where T2 can be
converted to type T with a (possibly trivial) qualification conversion
([conv.qual]); otherwise there are none.
As far as I understood, I can say that when initializing an "lvalue reference to cv1 T" the conversion function is a candidate if and only if it returns "lvalue reference to cv2 T2", where “cv1 T” is reference-compatible with “cv2 T2”.
And when initializing an "rvalue reference to cv1 T", the conversion function is a candidate if and only if it returns "rvalue reference to cv2 T2" or "cv T2", where “cv1 T” is reference-compatible with “cv2 T2”. Right?
Assuming my understanding is correct, here is a question. Why do we even need a conversion function to convert from cv2 T2 to cv1 T even though “cv1 T” is reference-compatible with “cv2 T2”?
Can you clearly explain the bullet (1.1) with some examples?

Answer 1

Why do we even need a conversion function to convert from cv2 T2 to cv1 T even though “cv1 T” is reference-compatible with “cv2 T2”?
Because only after the conversion is done(using the conversion function), the result is of type cv2 T2. In other words, the result of using/calling the conversion function is cv2 T2 and not whatever expression you have before using the conversion function. This is the whole point of using a conversion function so that we can something that is reference compatible with cv1 T.
Lets look at some examples:
Example 1
struct C {
operator int&();
};
const int& ref= C();
In the above example, the type cv1 T = const int is not reference-compatible with the class-type C. But due to the presence of the conversion function C::operator int&(), a given C can be converted to an lvalue of type cv2 T2 = int. And we know that const int is reference compatible with int. This means that the reference ref can be bound to the result of the conversion function according to your quoted clause.
Example 2
This example 2 does not uses a conversion function but i've added this just to clarify the explanation given at the beginning of my answer.
double val = 4.55;
const int &ref = val;
In the above example, the type cv1 T = const int is not reference-compatible with the type double. And at the end, the reference ref is bound to a materialized temporary(prvalue) of type cv2 T2 = int. This is possible because const int is reference compatible with the type of the materialized temporary int.
The important thing to note here is that a temporary is materialized because originally cv1 T = const int was not reference compatible with double. But after the materialization, cv1 T = const int is reference compatible with the result cv2 T2 = int.