[英]What does 'operator sockaddr *()'mean here?
What does 'operator sockaddr *()'mean here? 'operator sockaddr *()' 在这里是什么意思? The class Raw is an inner class, sockaddr is a struct. class Raw 是一个内部 class,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.
.....
.....
This is a user defined conversion function .这是用户定义的转换 function 。 An object of type Raw
can implicitly be converted to a sockaddr
pointer. Raw
类型的 object 可以隐式转换为sockaddr
指针。 For example the following will compile: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`
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