[C++知识库]c++ 使用gethostbyname_r解析域名

前一帖子，实验了如何使用adns库实现解析，用起来还是麻烦，直接使用系统函数，更直接：

#include <stdio.h>
#include <sys/socket.h>
#include <netinet/in.h>
#include <netinet/ip.h>
#include <netinet/ip_icmp.h>
#include <arpa/inet.h>
#include <sys/types.h>
#include <sys/time.h>
#include <unistd.h>
#include <netdb.h>
#include <string.h>
#include <resolv.h>  // dns

///static thread_local 
char t_resolveBuffer[64 * 1024];

void test(const char *url)
{

    char ip[INET_ADDRSTRLEN];
     char *ptr, **pptr;

    struct hostent hostinfo;
    struct hostent *phost = NULL;
    int hostrrno = 0;
    memset(&hostinfo, 0, sizeof(hostinfo));

    //该函数成功返回0，失败返回一个非0的数。
    int ret = gethostbyname_r(url, &hostinfo, t_resolveBuffer, sizeof(t_resolveBuffer), &phost, &hostrrno);

    if (ret == 0 && phost != NULL)
    {
        // assert(phost->h_addrtype == AF_INET && phost->h_length == sizeof(uint32_t));
        // out->addr_.sin_addr = *reinterpret_cast<struct in_addr*>(phost->h_addr);
        printf("official hostname: %s\n", phost->h_name);

        for (pptr = phost->h_aliases; *pptr != NULL; pptr++)
        {
            printf("\talias: %s\n", *pptr);
        }
        switch (phost->h_addrtype)
        {
        case AF_INET:
            pptr = phost->h_addr_list;
            for (; *pptr != NULL; pptr++)
            {
                // inet_ntoa只适用于ipv4地址，而inet_ntop适用ipv4和ipv6地址
                printf("\taddress: %s\n", inet_ntop(phost->h_addrtype, *pptr, ip, sizeof(ip)));
            }
            break;
        default:
            printf("unknow address type.");
            break;
        }
    }
    else
    {
        printf("ret =%d, err=%d\n", ret, hostrrno);
        return;
    }
}

int main()
{

    char host[128];

    res_init();

    while (1)
    {

        scanf("%s", host);
        if (strlen(host) == 4 && 0 == strcmp(host, "exit"))
            break;
        test(host);
    }

    return 0;
}

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