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Argument to function

I am confused by this argument to the function which is defined as

int bind(int s, const struct sockaddr *name, int namelen)

and called as

bind(sd, (struct sockaddr_in *) &addr, length);

I'm unable to interpret what struct sockaddr_in * means here.

Would this work: bind (sd, &addr, length);?

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pushgr8 Avatar asked Nov 11 '12 17:11

pushgr8


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1 Answers

It should be called like this:

bind (sd, (struct sockaddr *) &addr, length);

As for why, this is the C way of doing polymorphism, if you look at the definitions of those structures:

struct sockaddr {
    unsigned short    sa_family;    // address family, AF_xxx
    char              sa_data[14];  // 14 bytes of protocol address
};

// IPv4 AF_INET sockets:
struct sockaddr_in {
    short            sin_family;   // e.g. AF_INET, AF_INET6
    unsigned short   sin_port;     // e.g. htons(3490)
    struct in_addr   sin_addr;     // see struct in_addr, below
    char             sin_zero[8];  // zero this if you want to
};

// IPv6 AF_INET6 sockets:
struct sockaddr_in6 {
    u_int16_t       sin6_family;   // address family, AF_INET6
    u_int16_t       sin6_port;     // port number, Network Byte Order
    u_int32_t       sin6_flowinfo; // IPv6 flow information
    struct in6_addr sin6_addr;     // IPv6 address
    u_int32_t       sin6_scope_id; // Scope ID
};

They all share the same first member, sa_family when you pass a pointer to one of those structures to bind() you first cast it to struct sockaddr * and inside the function, sa_family is used to determine which structure you passed and cast back to the right one, instead of having one function for each structure you have one function that accepts sockaddr*.

Another way to look at it, from an OOP perspective, imagine that sockaddr is the base class for sockaddr_in and sockaddr_in6, and passing a pointer to sockaddr is similar to casting to the base type and calling a generic function. hope this makes it more clear.

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iabdalkader Avatar answered Oct 05 '22 17:10

iabdalkader