getaddrinfo在本地主机的IPv6之前对IPv4进行排序?

问题描述

我为getaddrinfo编写了一个非常简单的测试程序:

I wrote a very simple test program for getaddrinfo:

#include <sys/types.h>
#include <sys/socket.h>
#include <netdb.h>
#include <stdio.h>
#include <string.h>

int main() {
    struct addrinfo hints;
    struct addrinfo *res, *rp;
    char hoststr[64], servstr[8];

    memset(&hints, 0, sizeof(hints));
    hints.ai_family = AF_UNSPEC;
    hints.ai_socktype = SOCK_STREAM;
    hints.ai_flags = AI_PASSIVE;

    getaddrinfo(NULL, "9998", &hints, &res);
    for (rp = res; rp != NULL; rp = rp->ai_next) {
        getnameinfo(rp->ai_addr, rp->ai_addrlen, hoststr, sizeof(hoststr),
                servstr, sizeof(servstr), NI_NUMERICHOST | NI_NUMERICSERV);
        printf("%s:%s\n", hoststr, servstr);
    }
}

当我编译并运行该程序时，它会在IPv6地址之前给出IPv4地址:

When I compile and run this program, it's giving the IPv4 address before the IPv6 address:

# gcc -o getaddrinfo getaddrinfo.c
# ./getaddrinfo 
0.0.0.0:9998
:::9998

据我从其他消息来源得知，IPv6地址应该比IPv4更可取.我正在使用默认的/etc/gai.conf，这意味着IPv6应该比IPv4更受青睐.那么为什么getaddrinfo会以这种方式排序?

As far as I can tell from other sources, IPv6 addresses are supposed to be preferred over IPv4. I'm using the default /etc/gai.conf which implies that IPv6 should be preferred over IPv4. So why is getaddrinfo sorting this way?

# ip addr show lo
1: lo: <LOOPBACK,UP,LOWER_UP> mtu 65536 qdisc noqueue state UNKNOWN group default 
    link/loopback 00:00:00:00:00:00 brd 00:00:00:00:00:00
    inet 127.0.0.1/8 scope host lo
       valid_lft forever preferred_lft forever
    inet6 ::1/128 scope host 
       valid_lft forever preferred_lft forever

推荐答案

从手册页开始:

您描述的行为是针对使用getaddrinfo()查找名称以准备建立传出连接的行为.因为您要进行相反的操作，所以查找本地地址以绑定服务，因此显示地址的顺序无关紧要.

The behavior you're describing is for when you use getaddrinfo() to look up a name in preparation to make an outgoing connection. Because you're doing the reverse, looking up a local address to bind for a service, the order the addresses are presented is irrelevant.

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