c++ - 在什么情况下会调用类型自身的转换运算符？

Question

考虑一个 bar 类型，它具有用户定义的转换运算符到 bar 类型的引用:

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

何时应用这些转换？此外，如果这些运算符被删除，这意味着什么？这两个功能有什么有趣的用途吗？

以下程序似乎没有应用任何一种转换:

#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