c++ - 运营商必须采取“无效”

Question

假设我有两个类：

// A struct to hold a two-dimensional coordinate.
struct Point
{
    float x;
    float y;
};

// A struct identical to Point, to demonstrate my problem
struct Location
{
    float x;
    float y;
};

我想将 a 隐式转换Location为 a Point：

Point somePoint;
Location someLocation;

somePoint = someLocation;

所以，我operator在里面添加了这个Point：

operator Point(Location &other)
{
    // ...
}

g++ 4.9.2我在 Debian 上编译，并收到此错误：

error: 'Point::operator Point(Location &other)' must take 'void'

听起来编译器希望运算符不接受任何参数，但这似乎不对——除非我错误地使用了运算符。这个错误背后的真正含义是什么？

Answer 1

用户定义的转换运算符被定义为要转换为不同类型的类型的成员函数。签名是（类内）：Location

operator Point() const; // indeed takes void
// possibly operator const& Point() const;

另一种可能性是为提供转换构造函数Point：

Point(Location const& location);

Answer 2

不要重载 operator()。您想要做的是创建一个带点的自定义构造函数。当您执行赋值或重载赋值运算符时，编译器会调用它。

   Location( Point& p){
            x = p.x;
            y = p.y;
   }

   Location& operator= ( Point& p){
            x = p.x;
            y = p.y;
            return *this;
   }

这将使它编译

  Point somePoint;
  Location someLocation;

  somePoint = someLocation;
  Location loc = somePoint;