为什么我的程序在打开网络设备时遇到问题

Why is my program having problems opening a Network Device

本文关键字：网络设备 遇到 问题 我的 程序 为什么      更新时间：2023-10-16

我看了一下libtins和示例，我找到了ARP监视器示例。 示例的代码如下所示：

#include <tins/tins.h>
#include <map>
#include <iostream>
#include <functional>
using std::cout;
using std::endl;
using std::map;
using std::bind;
using namespace Tins;
class arp_monitor {
public:
void run(Sniffer& sniffer);
private:
bool callback(const PDU& pdu);
map<IPv4Address, HWAddress<6>> addresses;
};
void arp_monitor::run(Sniffer& sniffer) {
sniffer.sniff_loop(
bind(
&arp_monitor::callback,
this,
std::placeholders::_1
)
);
}
bool arp_monitor::callback(const PDU& pdu) {
// Retrieve the ARP layer
const ARP& arp = pdu.rfind_pdu<ARP>();
// Is it an ARP reply?
if (arp.opcode() == ARP::REPLY) {
// Let's check if there's already an entry for this address
auto iter = addresses.find(arp.sender_ip_addr());
if (iter == addresses.end()) {
// We haven't seen this address. Save it.
addresses.insert({ arp.sender_ip_addr(), arp.sender_hw_addr()});
cout << "[INFO] " << arp.sender_ip_addr() << " is at "
<< arp.sender_hw_addr() << std::endl;
}
else {
// We've seen this address. If it's not the same HW address, inform it
if (arp.sender_hw_addr() != iter->second) {
cout << "[WARNING] " << arp.sender_ip_addr() << " is at " 
<< iter->second << " but also at " << arp.sender_hw_addr() 
<< endl;
}
}
}
return true;
}
int main(int argc, char* argv[]) {
if(argc != 2) {
cout << "Usage: " <<* argv << " <interface>" << endl;
return 1;
}
arp_monitor monitor;
// Sniffer configuration
SnifferConfiguration config;
config.set_promisc_mode(true);
config.set_filter("arp");
try {
// Sniff on the provided interface in promiscuous mode
Sniffer sniffer(argv[1], config);

// Only capture arp packets
monitor.run(sniffer);
}
catch (std::exception& ex) {
std::cerr << "Error: " << ex.what() << std::endl;
}
}

我在这里运行了这段代码：

myprogram.exe eth0

结果是：

错误：打开适配器时出错：系统找不到给定的设备。(二十(

以上词语的定义：

eth0：我的网络设备

libtins：高级、多平台C++网络数据包嗅探和制作库

ARP：地址解析协议

我只在Windows上一个名为"Shadow"的用户上运行它

用户目录的图像

根据 te OP 上传的图像，我只能得出结论，我对他在 Windows 上运行该程序的看法是正确的，所以我只从 libtins 的文档中复制：

为了在Windows上捕获数据包，您可以首先列出所有网络接口。您可以使用 NetworkInterface 类轻松完成此操作：

// First fetch all network interfaces
vector<NetworkInterface> interfaces = NetworkInterface::all();
// Now iterate them
for (const NetworkInterface& iface : interfaces) {
// First print the name (GUID)
cout << "Interface name: " << iface.name();
// Now print the friendly name, a wstring that will contain something like 
// "Local Area Connection 2"
wcout << " (" << iface.friendly_name() << ")" << endl;
}

该小代码片段应提供如下所示的输出：

Interface name: {6527cc7d-c647-4986-ac10-7784dc1f2439} (Local Area Connection 1)
Interface name: {309d733f-79bb-41ef-aaec-8a7b83d2adcf} (Local Area Connection 2)
Interface name: {55ab969f-80df-4d51-8130-291d54a752a3} (Local Area Connection 3)

这可能足以让您识别出哪个是要使用的界面。您还可以求助于获取默认接口，这很可能是您想要使用的接口，或者显示每个接口的 IP 地址，直到您识别它们：

// Get the default interface (where the default gateway route is)
NetworkInterface iface = NetworkInterface::default_interface();
// Print the name and the IP address
cout << "Default interface: " << iface.name() 
<< " (" << iface.addresses().ip_addr() << ")" << endl;
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