c++ - 'operator sockaddr *()' 在这里是什么意思?
问题描述
'operator sockaddr *()' 在这里是什么意思?Raw 类是一个内部类,sockaddr 是一个结构。
struct sockaddr
{
__SOCKADDR_COMMON (sa_); /* Common data: address family and length. */
char sa_data[14]; /* Address data. */
};
class Address {
public:
//! \brief Wrapper around [sockaddr_storage](@ref man7::socket).
//! \details A `sockaddr_storage` is enough space to store any socket address (IPv4 or IPv6).
class Raw {
public:
sockaddr_storage storage{}; //!< The wrapped struct itself.
operator sockaddr *(); // here
operator const sockaddr *() const;
};
private:
socklen_t _size; //!< Size of the wrapped address.
Raw _address{}; //!< A wrapped [sockaddr_storage](@ref man7::socket) containing the address.
.....
.....
解决方案
这是一个用户定义的转换函数。类型的对象Raw
可以隐式转换为sockaddr
指针。例如以下将编译:
void fun(sockaddr *p); // function that takes a sockaddr pointer
Address::Raw r;
fun(r); // implicit conversion occurs from Raw in order to use function `fun`