Clarification of converted constant expression definition

Question

(Following my recent question, and another question.)

[expr.const]/4 says that:

A converted constant expression of type T is an expression, im

Answer 1

First, which expression does "the converted expression" (emphased) refer to?

Implicit conversion introduces a correspondingly initialized temporary:

T e = /* original expression */;

e is the "converted expression". T = int in your case.

then which expression should be constant expression?

e.

Moreover, foo and the conversion operator function that has been implicitly invoked must be constexpr functions as per [expr.const]/(2.2).

Second, what stage of the whole process does "reference binding" refer to?

When T is a reference type, that reference - e in the above example - shall bind directly. No binding necessitated inside the expression or before it is of interest.