ip4數據包組裝為ip6-libnet

Question

我編寫了一個C程序，該程序從鏈路層開始創建一個TCP / IPv4數據包。 該程序的目標是使用開放套接字（原始端口或僅是流套接字）接收單個數據包。 該數據包包括一個以太網頭，一個tcp頭和ipv4：

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <libnet.h>
#include <time.h>

uint8_t *randmac();    //creates my random source-mac, shady I know but I got 
                       //tired of typing in the same input over and over
void gateway();        //gets my standart gateway i.e. the router (as far as 
                       //I understand the ethernet layer is for the router 
                       //and tcp and ip for the rest of the net.)



int main() {
uint32_t src, dst, seq, ack;
char payload[1024], destip[16];
uint16_t sp = 2001, dp = 2000;
libnet_ptag_t tcp;
libnet_ptag_t ip4;
libnet_ptag_t eth;
char errbuf[LIBNET_ERRBUF_SIZE];
int i, bytes;
uint8_t *dstmc;     //destination mac address (router)
uint8_t *smc;       //source mac address
int  len;
libnet_t * l;

l = libnet_init(LIBNET_LINK, NULL, errbuf);
if(l == NULL)

//getting the destination ip-address, works all right...

//payload is fine too

smc = malloc(6);
gatemac = malloc(6);

libnet_seed_prand(l);
src = libnet_get_prand(LIBNET_PRu32);
seq = 0;
ack = libnet_get_prand(LIBNET_PRu32);
dst = libnet_name2addr4(l, destip, LIBNET_DONT_RESOLVE);
smc = randmac();
gateway();

//getting the rest of the required addresses etc, works as well...




tcp = libnet_build_tcp(sp, dp, seq, ack, TH_SYN, 0, 0, 0, LIBNET_TCP_H + 
sizeof(payload),(uint8_t *) payload, sizeof(payload), l, 0);
if(tcp == -1)
printf("(1)unable because: %s\n", libnet_geterror(l));
perror("libnet_build_tcp()");

ip4 = libnet_build_ipv4(LIBNET_TCP_H + LIBNET_IPV4_H + sizeof(payload), 
IPPROTO_TCP, 1, 1024, 60, 0, 0, src, dst, NULL, 0, l, 0);   //or (uint8_t) 
if(ip4 == -1)
printf("(2)unable because: %s\n", libnet_geterror(l));
perror("libnet_build_ipv4()");

eth = libnet_build_ethernet(gatemac, smc, ETHERTYPE_IP, NULL, 0, l, 0);
if(eth == -1)
printf("(3) unable because: %s\n", libnet_geterror(l));
perror("libnet_build_ethernet()");


bytes = libnet_write(l);

//error handling

}

程序本身可以正常工作，沒有錯誤，數據包被寫入等等。 但有趣的是：我無法通過任何原始或ipv4套接字接收它們。 因此，我打開了tcpdump，以查看數據包是否甚至都已發送，結果如下：

 13:43:24.003324 IP 74.253.145.81 > 192.168.88.130: hopopt
 13:43:27.007860 IP 74.253.145.81 > 192.168.88.130: hopopt

 //Random source address, I also sent multiple packets to eliminate the 
 //chances of missing them.
 //There is also no doubt that these are truly my programmed packets due to 
 //their number and random addresses.

我進行了一些研究，發現“ hopopt”代表ipv6使用的逐跳擴展頭。 為什么當我清楚地使用libnet_build_ipv4（）函數時會發生這種情況。 仍然有一種方法可以正確接收這些單個數據包，或者只是使其變為普通的ipv4？

Answer 1

看起來，好像您沒有對libnet_build_ipv4使用正確的參數libnet_build_ipv4 ：

ip4 = libnet_build_ipv4(LIBNET_TCP_H + LIBNET_IPV4_H + sizeof(payload), 

IPPROTO_TCP，1、1024、60、0、0，src，dst，NULL，0，l，0）;

正確的順序是：

libnet_ptag_t libnet_build_ipv4 (u_int16_t len,
u_int8_t tos, u_int16_t id, u_int16_t frag,
u_int8_t ttl, u_int8_t prot, u_int16_t sum,
u_int32_t src, u_int32_t dst, u_int8_t * payload,
u_int32_t payload_s, libnet_t * l, libnet_ptag_t ptag);

因此，您的第六個參數（prot）為0 ，與您在tcpdump中觀察到的逐跳選項（hopopt）完全對應。

您可能想要類似的東西：

ip4 = libnet_build_ipv4(LIBNET_TCP_H + LIBNET_IPV4_H + sizeof(payload), //len
                        0,           // tos
                        1234,        // some id
                        0,           // No fragment
                        0x40,        // Standard TTL (64)
                        IPPROTO_TCP, // Next protocol goes here
                        0,           // Checksum, auto-filled?
                        src,         // IP source address
                        dst,         // IP destination address
                        NULL,        // payload
                        0,           // payload length
                        l,           // libnet handle
                        0            // ptag
);

如果您固定參數的順序，它應該可以正常工作！ 另請參見libnet 教程