2
我想了解sin6_scope_id
如何在UNIX C编程中用于IPv6地址。具体来说,我写了这个程序,试图绑定到::1%2
(等接口2环回地址,如果我得到这个权利),即使我回送地址实际上是对接口1当:: 1在接口1上时,为什么我可以将此套接字绑定到:: 1%2?
我预计这将失败。但它成功绑定。为什么?
这里是由ifconfig -a
返回的第一个3个接口:
$ ifconfig -a
lo0: flags=8049<UP,LOOPBACK,RUNNING,MULTICAST> mtu 16384
options=3<RXCSUM,TXCSUM>
inet6 ::1 prefixlen 128
inet 127.0.0.1 netmask 0xff000000
inet6 fe80::1%lo0 prefixlen 64 scopeid 0x1
nd6 options=1<PERFORMNUD>
gif0: flags=8010<POINTOPOINT,MULTICAST> mtu 1280
stf0: flags=0<> mtu 1280
你可以编译这个程序:
cc -Wall -Wextra main.c
这里的注释的源:
#include <unistd.h>
#include <sys/types.h>
#include <sys/socket.h>
#include <netdb.h>
#include <stdio.h>
#include <string.h>
#include <stdlib.h>
int main() {
// Allow any IP address on the interface with scope ID of 2.
const char *hostname = "::1%2";
// Say we want to bind on port 80 (for http).
const char *servname = "1337";
// Store some information about the IP address wanted.
struct addrinfo hints;
// Save addresses in here.
struct addrinfo *addr_list_item = NULL;
// Tell `getaddrinfo` that we want an address for IPv6 TCP.
memset(&hints, 0, sizeof(struct addrinfo));
hints.ai_family = AF_INET6;
hints.ai_socktype = SOCK_STREAM;
hints.ai_protocol = IPPROTO_TCP;
if (getaddrinfo(hostname, servname, &hints, &addr_list_item) != 0) {
printf("Could not read addresses.\n");
exit(1);
}
// Create a socket and bind it to the address we found before. This should fail
// but for some reason I don't understand it doesn't.
if (addr_list_item) {
int sock = socket(AF_INET6, SOCK_STREAM, IPPROTO_TCP);
if (sock != -1) {
if (bind(sock, addr_list_item->ai_addr, addr_list_item->ai_addrlen) != -1) {
printf("Binded succesfully!\n");
} else {
perror(NULL);
}
close(sock);
}
}
// Release memory.
freeaddrinfo(addr_list_item);
return (0);
}
也许它的操作系统特定 - 你在哪个操作系统上运行? – nos 2015-10-07 19:47:18
这是在Linux上,如果我记得正确(抱歉在我的答复延迟)。 – conradkdotcom 2015-11-04 17:06:44