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Location隐式转换为Point：

Point somePoint;
Location someLocation;

somePoint = someLocation;

所以，我在operator内添加了Point：

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

我在Debian上使用g++ 4.9.2进行编译，并收到此错误：

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

听起来好像编译器希望操作符不接受任何参数，但这似乎并不正确 - 除非我错误地使用了操作符。这个错误背后的真正含义是什么？

Answer 1

User-defined conversions operators被定义为来自的成员函数，您希望将其转换为其他类型。签名是（Location类内）：

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

另一种可能性是为Point提供converting constructor：

Point(Location const& location);

Answer 2

不要使运算符（）过载。你想要做的是创建一个有点意义的自定义构造函数。编译器在执行赋值时调用this或重载赋值运算符。

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

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

这将使其编译

  Point somePoint;
  Location someLocation;

  somePoint = someLocation;
  Location loc = somePoint;

运营商必须采取'无效'

2 个答案: