标签: hostent

我遇到了以下代码快照:

struct hostent *hp;
 hp = my_gethostbyname(localhost);
    if (hp == NULL) {
       ls_syslog(LOG_ERR, I18N_FUNC_FAIL, fname, "my_gethostbyname()");
       return -1;
    }
    strcpy(localhost, hp->h_name);

    memcpy(&addr, hp->h_addr, hp->h_length);

我对最后一个语句感到困惑,struct hostent的声明是这样的:

struct hostent {
   char *h_name;       /* official name of host */
   char **h_aliases;   /* alias list */
   int h_addrtype;     /* host address type */
   int h_length;       /* length of address */
   char **h_addr_list; /* list of addresses */
};

它没有名为"h_addr"的字段,但代码确实可以编译,任何人都可以告诉我为什么？谢谢.

我有以下代码来获取主机名和IP地址,

#include <stdlib.h>
#include <stdio.h>
#include <netdb.h> /* This is the header file needed for gethostbyname() */
#include <sys/types.h>
#include <sys/socket.h>
#include <netinet/in.h>


int main(int argc, char *argv[])
{
struct hostent *he;

if (argc!=2){
printf("Usage: %s <hostname>\n",argv[0]);
exit(-1);
}

if ((he=gethostbyname(argv[1]))==NULL){
printf("gethostbyname() error\n");
exit(-1);
}

printf("Hostname : %s\n",he->h_name); /* prints the hostname */
printf("IP Address: %s\n",inet_ntoa(*((struct in_addr *)he->h_addr))); /* prints IP address */
}

但是我在编译期间收到警告:

$cc host.c -o host
host.c: In function ‘main’:
host.c:24: warning: format ‘%s’ expects type ‘char …