Will an Implicit conversion function to self ever get called?

Question

Suppose I define an implicit conversion function to myself:

#include <iostream>

class Foo {
 public:
  operator Foo() {
    std::cout << "wha??\n";
    return Foo();
  }
};

void f(Foo f) {}

int main() {
  Foo foo;
  f(foo);
}

Why would I define this? Well I would never write it directly, but it could happen through a template instantiation in a template class I'm writing. If this happens, I hope that defining the conversion function is effectively a no-op and that it is impossible to actually call.

The above program prints nothing (which is great). Is there any circumstance under which this conversion function will actually get called?

Answer 1

It's correct, but will never called, due n3376 12.3.2/1

A conversion function is never used to convert a (possibly cv-qualified) object to the (possibly cv-qualified) same object type (or a reference to it)