是时候将把戏揭穿了。
请先阅读下文：
https://blog.csdn.net/dog250/article/details/108113329
依照文章中所描述的把戏，我们hack一下两个进程的tcp连接的归属，首先给出hack之前的情景：
[root@localhost ~]# netstat -ntp
Active Internet connections (w/o servers)
Proto Recv-Q Send-Q Local Address Foreign Address State PID/Program name
tcp 0 0 192.168.56.110:22 192.168.56.1:55287 ESTABLISHED 1344/sshd: root@pts
tcp 0 0 192.168.56.110:22 192.168.56.1:55589 ESTABLISHED 4791/sshd: root@pts
然后，在实施hack之后，进程所属的tcp连接被交换了：
[root@localhost ~]# netstat -ntp
Active Internet connections (w/o servers)
Proto Recv-Q Send-Q Local Address Foreign Address State PID/Program name
tcp 0 0 192.168.56.110:22 192.168.56.1:55287 ESTABLISHED 4791/sshd: root@pts
tcp 0 0 192.168.56.110:22 192.168.56.1:55589 ESTABLISHED 1344/sshd: root@pts
如果系统运维或者其它管理员遇到这样的情况，他们到底要瞄准哪个进程来debug呢？
注意⚠️，我们要抛弃事后的预设，事实上，他们根本无法意识到连接被交换了，等到拉了一大堆人一起开完会后很久，他们或许才能意识到事情哪里有点不对劲。
如果有我在，这种事情就不会发生，作为手艺人，我更喜欢直接面对本质，而不是使用工具迂回，在我看来，通过procfs来查找tcp连接和进程之间的关联，就是迂回。
何不直接点儿呢？
来看下面的图：
如果通过那条 不可拆除的线索 来关联tcp_sock和socket，岂不是无需迂回？
OK，下面的代码描述了如何做：
#!/usr/bin/stap -g
// tcpstat(.stp)
%{
#include <net/tcp.h>
#include <linux/fdtable.h>
struct result {
char laddr[16];
char raddr[16];
unsigned short lport;
unsigned short rport;
unsigned long ino;
int pid;
char comm[32];
};
static inline void ip2str(char *to, unsigned int from)
{
int size = snprintf(to, 16, "%pI4", &from);
to[size] = '\0';
}
void traverse(struct sock *sk, unsigned long ino, struct result *ret)
{
struct task_struct *tsk;
int i;
for_each_process(tsk) {
struct file *file;
for (i = 0; i < tsk->files->fdt->max_fds; i++) {
file = tsk->files->fdt->fd[i];
if (file == NULL) {
continue;
}
if (file->f_inode->i_ino == ino) {
char laddr[16], raddr[16];
ip2str(laddr, inet_sk(sk)->inet_rcv_saddr);
ip2str(raddr, inet_sk(sk)->inet_daddr);
memcpy(&ret->laddr[0], laddr, 16);
memcpy(&ret->raddr[0], raddr, 16);
ret->lport = sk->sk_num;
ret->rport = htons(sk->sk_dport);
ret->ino = ino;
ret->pid = tsk->pid;
memcpy(&ret->comm[0], tsk->comm, 32);
}
}
}
}
%}
function dump_tcp_info()
%{
struct task_struct *tsk;
struct inet_hashinfo *hashinfo = &tcp_hashinfo;
struct hlist_nulls_node *node;
struct socket_alloc *sa;
struct sock *sk;
struct result ret;
int i, ino;
for (i = 0; i < INET_LHTABLE_SIZE; i++) {
struct inet_listen_hashbucket *ilb;
ilb = &hashinfo->listening_hash[i];
sk_nulls_for_each(sk, node, &ilb->head) {
unsigned long ino;
sa = (struct socket_alloc *)sk->sk_socket;
ino = sa->vfs_inode.i_ino;
traverse(sk, ino, &ret);
STAP_PRINTF("LISTEN %s:%d inode:%d/[%d] %s\n",
ret.laddr,
ret.lport,
ret.ino,
ret.pid,
ret.comm);
}
}
for (i = 0; i <= hashinfo->ehash_mask; i++) {
struct inet_ehash_bucket *head = &hashinfo->ehash[i];
if (hlist_nulls_empty(&head->chain)) {
continue;
}
sk_nulls_for_each(sk, node, &head->chain) {
unsigned long ino;
sa = (struct socket_alloc *)sk->sk_socket;
ino = sa->vfs_inode.i_ino;
traverse(sk, ino, &ret);
STAP_PRINTF("ESTABLISHED %s:%d %s:%d inode:%d/[%d] %s\n",
ret.laddr,
ret.lport,
ret.raddr,
ret.rport,
ret.ino,
ret.pid,
ret.comm);
}
}
%}
probe begin
{
dump_tcp_info();
exit();
}
来来来，我们运行上面的脚本来试一下能不能把被hack的系统真相找出来：
[root@localhost test]# ./tcpstat
LISTEN 0.0.0.0:22 inode:22852/[1001] sshd
LISTEN 0.0.0.0:22 inode:22850/[1001] sshd
ESTABLISHED 192.168.56.110:22 192.168.56.1:55287 inode:25510/[1344] sshd
ESTABLISHED 192.168.56.110:22 192.168.56.1:55589 inode:28747/[4791] sshd
用这个结果来对比一下被hack后的情形：
[root@localhost ~]# netstat -ntp
Active Internet connections (w/o servers)
Proto Recv-Q Send-Q Local Address Foreign Address State PID/Program name
tcp 0 0 192.168.56.110:22 192.168.56.1:55287 ESTABLISHED 4791/sshd: root@pts
tcp 0 0 192.168.56.110:22 192.168.56.1:55589 ESTABLISHED 1344/sshd: root@pts
哈哈，真相大白！
何为手艺人？不依赖大型工具，不依赖专业知识，小镇上修锅换底而无痕迹者是也。当然，皮鞋也能做。
浙江温州皮鞋湿，下雨进水不会胖。