pcap_lookupnet返回不正确的IP地址 [英] pcap_lookupnet returns incorrect IP address

问题描述

继libpcap的文档例如code产生以下code，应报告（在本例中为eth0）在给定的接口的IP地址[错误检查为简洁起见省略]

Following example code from the libpcap documentation yields the following code which should report the IP address of the given interface (eth0 in this case) [Error checking omitted for brevity]

#include <stdio.h>
#include <pcap.h>
#include <arpa/inet.h>

int main(int argc, char *argv[])
{
  char errbuf[PCAP_ERRBUF_SIZE]; 
  bpf_u_int32 mask;   
  bpf_u_int32 ip;    
  struct in_addr ip_addr;

  /* Find the properties for the device */
  pcap_lookupnet("eth0", &ip, &mask, errbuf);

  ip_addr.s_addr = ip;

  printf("IP Address: %s\n", inet_ntoa(ip_addr));

  return 0;
}

然而，这导致在192.168.1.0，而不是正确的192.168.1.100。不同的子网产量10.0.0.0而不是正确10.0.0.107这使我相信的libpcap是不正确复制最后一个字节不同的机器上运行此。我手动转换由pcap_lookupnet归还，以确保它不是与使用INET_NTOA的（我也试过inet_ntop，具有相同的结果）问题的整数。继这个问题code：
在Linux上接口的获得IP地址
报告正确的IP地址。这是libpcap的一个bug或我做错了什么？

However, this results in 192.168.1.0, rather than the correct 192.168.1.100. Running this on a different machine on a different subnet yields 10.0.0.0 rather than the correct 10.0.0.107 which leads me to believe libpcap is not copying the last octet correctly. I've manually converted the integer returned by pcap_lookupnet to ensure it's not an issue with the use of inet_ntoa (I've also tried inet_ntop, with identical results). Following the code from this question: Get IP address of an interface on linux reports the correct IP address. Is this a bug in libpcap or am I doing something wrong?

推荐答案

您声明，这应报告给定的接口的IP地址是不正确。

Your statement "which should report the IP address of the given interface" is incorrect.

从手册页：

pcap_lookupnet（）被用来确定IPv4网络号码和
  面具
         与网络设备装置相关联。无论NETP和maskp是
         bpf_u_int32指针。

pcap_lookupnet() is used to determine the IPv4 network number and mask associated with the network device device. Both netp and maskp are bpf_u_int32 pointers.

你确定你有分别10.0.0.107或192.168.1.100网络号？听起来相当不寻常的。

are you sure you have a network number of 10.0.0.107 or 192.168.1.100 respectively? Sounds rather unusual.

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