C ++ /引用运算符和指针中的Socket connect（）函数说明

Question

I'm new to programming and would like a better understanding of this socket function connect(). I've read many descriptions but I couldn't understand this syntax:

int connect(int sockfd, const struct sockaddr *addr,socklen_t addrlen)

Description: The addr argument is a pointer to a structure specifying the address to which this socket is to be bound.

This is part of my studying code:

  connect(socket_desc ,( struct sockaddr *) &server , sizeof(server)) < 0)

Can anyone explain why there is a structure in parentheses with a pointer operator and the reference operator after it?

Answer 1

The ( struct sockaddr *) &server part of that line can be described in English as:

Take the address of the object server , and treat that memory address (or pointer) as if it were a pointer to a sockaddr object.

The & operator, at least in this context, is taking the memory address of the identifier that follows. An expression that is preceded by (type) is 'cast' to that type. In C, that is the only form of type casting. C++ inherits that functionality, but is discouraged in most C++ code in favor of the other casting operations (static_cast, dynamic_cast, const_cast, reinterpret_cast).

Answer 2

Can anyone explain why there is a structure in parentheses with a pointer operator and the reference operator after it?

The parenthesis are being used for a type cast , and the ampersand is the address-of operator . The code is obtaining the memory address of a struct variable, likely a sockaddr_in or sockaddr_in6 but could be any sockaddr -compatible struct, and then type-casting that memory address to a sockaddr pointer.

Sockets support many different types of addressing schemes, and there is a different sockaddr_... struct type defined for each of them. But the socket API functions that deal with addresses only accept sockaddr* parameters, so type-casting is needed.