Question

Consider a type bar which has user-defined conversion operators to references of type bar:

struct bar
{
  operator bar & ();
  operator const bar & () const;
};

When would these conversions be applied? Moreover, what does it imply if these operators were deleted? Is there any interesting use of either feature?

The following program does not appear to apply either conversion:

#include <iostream>

struct bar
{
  operator bar & ()
  {
    std::cout << "operator bar &()" << std::endl;
    return *this;
  }

  operator const bar & () const
  {
    std::cout << "operator const bar &() const" << std::endl;
    return *this;
  }
};

void foo(bar x)
{
}

int main()
{
  bar x;

  bar y = x;         // copy, no conversion

  y = x;             // assignment, no conversion

  foo(x);            // copy, no conversion

  y = (bar&)x;       // no output

  y = (const bar&)x; // no output

  return 0;
}

Answer 1

C++11 §12.3.2

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), to a (possibly cv-qualified) base class of that type (or a reference to it), or to (possibly cv-qualified) void