c++ - 运算符必须取 'void'

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 Point 中添加了这个 operator:

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

我在 Debian 上用 g++ 4.9.2 编译，然后收到这个错误:

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

User-defined conversions operators被定义为来自 类型的成员函数，您希望将其转换为不同的类型。签名是(在 Location 类中):

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

另一种可能性是提供 converting constructor对于点:

Point(Location const& location);