重载运算符

Question

谁能解释一下这个重载运算符是如何调用的？我了解动态转换和条件三元的作用，但我不了解运算符。

头文件：

 // ------------------------------------------------------
Class Base
{
public:
  Base ();
  operator Derived &();
private:
  Base * me;
}
//--------------------------------------------------
Class Derived : public Base
{
public:
Derived ()
}
//----------------------------------------------------
inline Base::operator Derived &() {return *(dynamic_cast<Derived *>(me?me:this));}

源文件：

Base::Base()
{
me = new Derived()
}

Answer 1

运算符operator Derived &()是转换运算符：

12.3.2/1 A member function of a class X having no parameters with a name of the form conversion-function-id: operator conversion-type-id (...)

specifies a conversion from X to the type specified by the conversion-type-id. Such functions are called conversion functions. No return type can be specified.

因此它将 Base 对象转换为对 Derived 的引用。

注意：这种结构看起来非常奇怪和危险：它是一种不小心的沮丧。如果您使用此运算符的对象不是 Derived（例如 "pure" Base 对象或从 Base 派生的另一个 class 但不是来自 Derived) dynamic_cast 会 return 一个 nullptr，这会导致 UB:

8.3.2/5: (...) Note: in particular, a null reference cannot exist in a well-defined program, because the only way to create such a reference would be to bind it to the “object” obtained by indirection through a null pointer, which causes undefined behavior.

重载运算符

Overloading operator

c++

dynamic-cast

operator-overloading