c++ - 使用运算符 float() 和带有静态成员的 float() const 的隐式转换

Question

我刚刚在尝试重载 operator float() 时遇到了问题和 operator float() const .我以为我可以使用这两个重载来为“做事”和“只读”提供不同的版本……但事实证明，对于包含这些重载的类的静态实例，我不能。

归结起来问题几乎可以简化为:

// Does some math and should convert to float
struct ToFloat
{
  // float conversion here
  operator float()
  {
    cout << "called operator float() of " << s << "\n";
    f += 1.0f;
    return f;
  }

  // just get current value here
  operator float() const
  {
    cout << "called operator float() *const* of " << s << "\n";
    return f;
  }

  float f;
  std::string s;
};

// Uses a static and a normal member of ToFloat
struct Container
{
  // return both instances (with some more math before)
  operator float()
  {
    return s * m;
  }

  // just provide read access
  operator float() const
  {
    return s * m;
  }

  static inline ToFloat s { 1.0f, "static" };
  ToFloat m { 1.0f, "member" };
};

// Uses the container, but must also provide read-only access
struct Use
{
  // Give me operator float() of my container
  float get()
  {
    return c;
  }

  // Give me operator float() const of my container
  float getC() const
  {
    return c;
  }

  Container c {};
};

int main()
{
  Use u {};

  printf("getC() %f \n\n", u.getC());
  printf("get() %f \n\n", u.get());
  printf("getC() %f \n\n", u.getC());
}

产生以下输出...

called operator float() of static
called operator float() *const* of member
getC() 2.000000 

called operator float() of static
called operator float() of member
get() 6.000000 

called operator float() of static
called operator float() *const* of member
getC() 8.000000 

我真的不明白为什么 ToFloat 的静态实例总是使用非 const转换，即使从声明为 const 的函数中调用？这里适用什么规则？

Answer 1

静态数据成员Container::s 只是ToFloat 类型。它总是被直接访问，从不通过对 this 的隐式取消引用。换句话说，容器的 const 运算符实际上是这样的:

operator float() const
{
  return Container::s * this->m;
}

由此可见，Container::s 没有理由仅仅因为 this 是一个常量容器 *。如果您希望它被视为 const，则必须明确限定它:

operator float() const
{
  return std::as_const(s) * m;
}