C++ operator 的一种不会的用法

自认为对C++比较熟悉，忽然看到一些奇怪的代码（在看网上下载的代码Sockets）：

class SocketAddress
{
public:
	virtual ~SocketAddress() {}

	/** Get a pointer to the address struct. */
	virtual operator struct sockaddr *() = 0;

	/** Get length of address struct. */
	virtual operator socklen_t() = 0;

	/** Compare two addresses. */
	virtual bool operator==(SocketAddress&) = 0;

	/** Set port number.
		\param port Port number in host byte order */
	virtual void SetPort(port_t port) = 0;

	/** Get port number.
		\return Port number in host byte order. */
	virtual port_t GetPort() = 0;

	/** Set socket address.
		\param sa Pointer to either 'struct sockaddr_in' or 'struct sockaddr_in6'. */
	virtual void SetAddress(struct sockaddr *sa) = 0;

	/** Convert address to text. */
	virtual std::string Convert(bool include_port) = 0;

	/** Reverse lookup of address. */
	virtual std::string Reverse() = 0;

	/** Get address family. */
	virtual int GetFamily() = 0;

	/** Address structure is valid. */
	virtual bool IsValid() = 0;

	/** Get a copy of this SocketAddress object. */
	virtual std::auto_ptr<SocketAddress> GetCopy() = 0;
};

奇怪的代码：

/** Get a pointer to the address struct. */
	virtual operator struct sockaddr *() = 0;

	/** Get length of address struct. */
	virtual operator socklen_t() = 0;

最后搜了一下，才知道是对 类型转换 的重载！

socklen_t 是 int 类型 ；

声明：operator XX(); 无返回值，XX就是某个类型；

用法：XX a = (XX)obj; 假设上边SocketAddress非abstract类，例如：SocketAddress sa; int a = (socklen_t)sa; 也就是此时会调用原成员函数operator XX(); 通常返回XX类型值，能够理解成类型转换的重载！

另外：我竟不知道一个自定义类的 构造函数能够用做隐形的类型转换，例如：Class A { A(int i) { val = i};  private int val;} , A a = 5; 解：5 首先经过构造函数A(int)隐形的转换为A类型，而后调用默认的operator=赋值函数，赋值给a；