【破文标题】Windows系统RunAs命令的DIY改造
【破文作者】FishSeeWater
【作者主页】htp://shuihua.hostrocket.com
【作者邮箱】yujianshui99@163.com
【所属组织】BCG
【软件名称】RunAs.EXE
【下载地址】Microsoft的
【破解工具】OllyDbg(cao_cong汉化版) + C32Asm
【保护方式】无
【软件限制】不接受密码参数
【破解难度】简单
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【软件介绍】
Microsoft系统自带的命令，不用我多罗嗦了吧：)
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【DIY声明】
最近由于公司领导要求“计算机资料保密”，我决定采用WIN2000的系统特性加上NTFS权限完成。用Win2003架了台服务器，然后为每位工程师建了User帐户，但是发现好几个CAD软件在User用户下没法运行，但通过右键的“运行方式”以"Administrator"运行就没事儿，调整组权限，整了好几天也没能搞定，微软出于安全考虑 RunAs 命令不能直接跟密码参数。每次都要单独输入密码，这样的话那我岂不要~~*&@#$$@：(。在网上找RunAs命令的程序实现方式，准备写个程序调用CAD来运行，可是用LogonUser()；CreateProcessAsUse()；CreateProcessWithLogonW()这几个函数死活不行，LogonUser()总是返加1314错误，郁闷。无奈之下，还是打RunAs的主意吧。让它支持密码参数不就OK了：)
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【破解分析】
好!开始：
首先预习一下基础知识，控制台程序获取运行参数是用GetCommandLine(VOID)函数
打开OD加载目标程序，参数为 "/user:administrator cmd.exe"
然后下断点BPX GetCommandLine 回车
F9运行，程序断在01001572 CALL DWORD PTR DS:[<&KERNEL32.GetCommandLineW>; |[GetCommandLineW
瞧瞧它是怎样RunAs的：)
F8一步一步走：
[CODE]
01001555 |. 50 PUSH EAX ; /pArgc
01001556 |. 895D>MOV DWORD PTR SS:[EBP-1C],EBX ; |
01001559 |. 895D>MOV DWORD PTR SS:[EBP-20],EBX ; |
0100155C |. 895D>MOV DWORD PTR SS:[EBP-C],EBX ; |
0100155F |. 895D>MOV DWORD PTR SS:[EBP-18],EBX ; |
01001562 |. 895D>MOV DWORD PTR SS:[EBP-38],EBX ; |
01001565 |. 8975>MOV DWORD PTR SS:[EBP-4],ESI ; |
01001568 |. 895D>MOV DWORD PTR SS:[EBP-14],EBX ; |
0100156B |. C745>MOV DWORD PTR SS:[EBP-7C],44 ; |
//得到参数
01001572 |. FF15>CALL DWORD PTR DS:[<&KERNEL32.GetCommandLineW>; |[GetCommandLineW
01001578 |. 50 PUSH EAX ; |CmdLine
//得到参数个数
01001579 |. FF15>CALL DWORD PTR DS:[<&SHELL32.CommandLineToArg>; \CommandLineToArgvW
0100157F |. 8BC8 MOV ECX,EAX
01001581 |. 3BCB CMP ECX,EBX
01001583 |. 894D>MOV DWORD PTR SS:[EBP+8],ECX
01001586 |. 0F85>JNZ runas.0100163E
0100158C |. 895D>MOV DWORD PTR SS:[EBP-4],EBX
.......
.......
.......
//比较参数个数，不等于3就跳走，显示RunAs的帮助(自身的程序名也算一个参数)
0100163E |> \8B45>MOV EAX,DWORD PTR SS:[EBP-14]
01001641 |. 83F8>CMP EAX,5
01001644 |. 74 0>JE SHORT runas.01001650
01001646 |. 83F8>CMP EAX,4
01001649 |. 74 0>JE SHORT runas.01001650
0100164B |. 83F8>CMP EAX,3
0100164E |. 75 7>JNZ SHORT runas.010016CF
01001650 |> 8D50>LEA EDX,DWORD PTR DS:[EAX-2]
//以下是判断并取出第二个参数"administraor" (第一个参数是RunAs.exe)
01001653 |. 8975>MOV DWORD PTR SS:[EBP-8],ESI
01001656 |. 3BD6 CMP EDX,ESI
01001658 |. 7C 7>JL SHORT runas.010016CA
0100165A |. 8D79>LEA EDI,DWORD PTR DS:[ECX+4]
0100165D |> 8B07 MOV EAX,DWORD PTR DS:[EDI]
0100165F |. 66:8>CMP WORD PTR DS:[EAX],2F
01001663 |. 75 6>JNZ SHORT runas.010016CF
01001665 |. 0FB7>MOVZX ECX,WORD PTR DS:[EAX+2]
01001669 |. 83F9>CMP ECX,4E ; Switch (cases 45..75)
0100166C |. 7F 1>JG SHORT runas.0100167E
0100166E |. 0F84>JE runas.01001BA4
01001674 |. 83F9>CMP ECX,45
01001677 |. 75 5>JNZ SHORT runas.010016CF
01001679 |> 8975>MOV DWORD PTR SS:[EBP-1C],ESI ; Cases 45 ('E'),65 ('e') of switch 01001669
0100167C |. EB 4>JMP SHORT runas.010016BF
0100167E |> 83E9>SUB ECX,50
01001681 |. 74 3>JE SHORT runas.010016BC
01001683 |. 83E9>SUB ECX,5
01001686 |. 74 1>JE SHORT runas.0100169F
01001688 |. 83E9>SUB ECX,10
0100168B |.^ 74 E>JE SHORT runas.01001679
0100168D |. 83E9>SUB ECX,9
01001690 |. 0F84>JE runas.01001BA4
01001696 |. 49 DEC ECX
01001697 |. 49 DEC ECX
01001698 |. 74 2>JE SHORT runas.010016BC
0100169A |. 83E9>SUB ECX,5
0100169D |. 75 3>JNZ SHORT runas.010016CF
0100169F |> 83C0>ADD EAX,4 ; Cases 55 ('U'),75 ('u') of switch 01001669
010016A2 |> 66:8>/MOV CX,WORD PTR DS:[EAX]
.......
.......
.......
010016CD |. /75 0>JNZ SHORT runas.010016DC
010016CF |> |E8 8>CALL runas.0100125B ; Default case of switch 01001669
010016D4 |> |895D>MOV DWORD PTR SS:[EBP-4],EBX
010016D7 |. |E9 5>JMP runas.01001B38
//加载提示资源串"键入密码administrator："
010016DC |> \8B3D>MOV EDI,DWORD PTR DS:[<&USER32.LoadStringW>] ; USER32.LoadStringW
010016E2 |. 8D85>LEA EAX,DWORD PTR SS:[EBP-630]
010016E8 |. 68 F>PUSH 1F4 ; /Count = 1F4 (500.)
010016ED |. 50 PUSH EAX ; |Buffer
010016EE |. 68 7>PUSH 1B76 ; |RsrcID = STRING "Enter password for"
010016F3 |. FF35>PUSH DWORD PTR DS:[1003020] ; |hInst = 01000000
010016F9 |. FFD7 CALL EDI ; \LoadStringW
010016FB |. 8B35>MOV ESI,DWORD PTR DS:[<&USER32.CharToOemW>] ; USER32.CharToOemW
01001701 |. 8D85>LEA EAX,DWORD PTR SS:[EBP-8EC]
01001707 |. 50 PUSH EAX ; /pDest
01001708 |. 8D85>LEA EAX,DWORD PTR SS:[EBP-630] ; |
0100170E |. 50 PUSH EAX ; |pSrc
0100170F |. FFD6 CALL ESI ; \CharToOemW
01001711 |. FF75>PUSH DWORD PTR SS:[EBP-C] ; /String
01001714 |. 8B1D>MOV EBX,DWORD PTR DS:[<&KERNEL32.lstrlenW>] ; |kernel32.lstrlenW
0100171A |. FFD3 CALL EBX ; \lstrlenW
0100171C |. 3D F>CMP EAX,1F4
01001721 |. 76 0>JBE SHORT runas.0100172E
01001723 |. 8B45>MOV EAX,DWORD PTR SS:[EBP-C]
01001726 |. 66:8>AND WORD PTR DS:[EAX+3E6],0
0100172E |> 8D85>LEA EAX,DWORD PTR SS:[EBP-D9C]
01001734 |. 50 PUSH EAX
01001735 |. FF75>PUSH DWORD PTR SS:[EBP-C]
01001738 |. FFD6 CALL ESI
0100173A |. 8D85>LEA EAX,DWORD PTR SS:[EBP-D9C]
01001740 |. 50 PUSH EAX ; /<%s>
01001741 |. 8D85>LEA EAX,DWORD PTR SS:[EBP-8EC] ; |
01001747 |. 50 PUSH EAX ; |<%s>
01001748 |. 68 8>PUSH runas.0100118C ; |format = "%s %s:"
0100174D |. FF15>CALL DWORD PTR DS:[<&MSVCRT.printf>] ; \printf
01001753 |. 83C4>ADD ESP,0C
01001756 |. 8D85>LEA EAX,DWORD PTR SS:[EBP-248]
//这个104是干嘛的？难道是最大长度？请高手指点：)
0100175C |. 68 0>PUSH 104 ; /Arg2 = 00000104
//压入变量地址
01001761 |. 50 PUSH EAX ; |Arg1
//下面是调用ReadConsoleW函数读取键盘输入的密码
01001762 >|. E8 4>CALL runas.010011B0 ; \读入密码
01001767 |. 85C0 TEST EAX,EAX
01001769 |. 74 4>JE SHORT runas.010017AD
0100176B |. 8D85>LEA EAX,DWORD PTR SS:[EBP-630]
........
........
........
0100179B |. 68 3>PUSH runas.01001134 ; |format = "%s"
010017A0 |. FF15>CALL DWORD PTR DS:[<&MSVCRT.printf>] ; \printf
010017A6 |. 59 POP ECX
010017A7 |. 59 POP ECX
010017A8 |. E9 4>JMP runas.010019ED
//处理输入的密码
010017AD |> 8D85>LEA EAX,DWORD PTR SS:[EBP-248]
010017B3 |. 50 PUSH EAX ; /Translation
010017B4 |. 8D85>LEA EAX,DWORD PTR SS:[EBP-248] ; |
010017BA |. 50 PUSH EAX ; |OemString
010017BB |. FF15>CALL DWORD PTR DS:[<&USER32.OemToCharA>] ; \OemToCharA
010017C1 |. 8D85>LEA EAX,DWORD PTR SS:[EBP-248]
//将密码复制后保存
010017C7 |. 50 PUSH EAX ; /<%S>
010017C8 |. 8D85>LEA EAX,DWORD PTR SS:[EBP-1198] ; |
010017CE |. 68 8>PUSH runas.01001184 ; |Format = "%S"
010017D3 |. 50 PUSH EAX ; |s
010017D4 |. FF15>CALL DWORD PTR DS:[<&USER32.wsprintfW>] ; \wsprintfW
010017DA |. 83C4>ADD ESP,0C
010017DD |. 8D85>LEA EAX,DWORD PTR SS:[EBP-E0]
010017E3 |. 6A 3>PUSH 32
010017E5 |. 50 PUSH EAX
010017E6 |. 68 7>PUSH 1B77
010017EB |. FF35>PUSH DWORD PTR DS:[1003020] ; runas.01000000
010017F1 |. FFD7 CALL EDI
010017F3 |. 8B45>MOV EAX,DWORD PTR SS:[EBP-14]
010017F6 |. 8B4D>MOV ECX,DWORD PTR SS:[EBP+8]
//得到命令行参数的第三个参数长度。(DS:[ECX+EAX*4-4]为第三个参数)
010017F9 |. FF74>PUSH DWORD PTR DS:[ECX+EAX*4-4]
010017FD |. FFD3 CALL EBX ; kernel32.lstrlenW
010017FF |. 8BF8 MOV EDI,EAX
01001801 |. 8D85>LEA EAX,DWORD PTR SS:[EBP-E0]
01001807 |. 50 PUSH EAX
01001808 |. FFD3 CALL EBX
0100180A |. FF75>PUSH DWORD PTR SS:[EBP-C]
........
........
........
01001A0B |. C745>MOV DWORD PTR SS:[EBP-24],400
01001A12 |. EB 0>JMP SHORT runas.01001A17
01001A14 |> 8B5D>MOV EBX,DWORD PTR SS:[EBP-38]
01001A17 |> 8D45>LEA EAX,DWORD PTR SS:[EBP-34]
//调用CreateProcessWithLogonW函数以"administrator"身份创建进程
01001A1A |. 50 PUSH EAX ; /Arg11
01001A1B |. 8D45>LEA EAX,DWORD PTR SS:[EBP-7C] ; |
01001A1E |. 50 PUSH EAX ; |Arg10
01001A1F |. 8B45>MOV EAX,DWORD PTR SS:[EBP-14] ; |
01001A22 |. 53 PUSH EBX ; |Arg9
01001A23 |. FF75>PUSH DWORD PTR SS:[EBP-20] ; |Arg8
01001A26 |. FF75>PUSH DWORD PTR SS:[EBP-24] ; |Arg7
01001A29 |. FF74>PUSH DWORD PTR DS:[EDI+EAX*4-4] ; |Arg6
01001A2D |. 8D85>LEA EAX,DWORD PTR SS:[EBP-1198] ; |
01001A33 |. 6A 0>PUSH 0 ; |Arg5 = 00000000
01001A35 |. FF75>PUSH DWORD PTR SS:[EBP-10] ; |Arg4
01001A38 |. 50 PUSH EAX ; |Arg3
01001A39 |. FF75>PUSH DWORD PTR SS:[EBP-18] ; |Arg2
01001A3C |. FF75>PUSH DWORD PTR SS:[EBP-C] ; |Arg1
01001A3F |. FF15>CALL DWORD PTR DS:[<&ADVAPI32.CreateProcessWi>; \CreateProcessWithLogonW
01001A45 |. 85C0 TEST EAX,EAX
01001A47 |. 0F85>JNZ runas.01001B36
01001A4D |. 33FF XOR EDI,EDI
01001A4F |. C745>MOV DWORD PTR SS:[EBP-10],7B
////////////////////////////////////
[CODE]
过程基本分析完毕，要想达到我们的目的，只要在ReadConsoleW函数之后，将我们的已有密码添上就OK了。
思路：1、先将程序中提示输入密码和读取密码的函数NOP掉!从10016e2处开始一直到1001766
2、我们有现成的密码，也不用他来处理，所以嘛处理过程也NOP!从1007AD处开始到1007D4(哈哈，这下有足够的空间来写我们的代码了)
3、由于我们要用参数型式添入密码，而程序中有检测参数个数的语句，所以在CommandLineToArgvW函数后，将参数个数减1。
好!开工：
首先处理步骤1和2
然后修改1001586处
01001586 JNZ runas.0100163E 为 01001586 JNZ runas.010016F2
在010016EF处添加语句
010016EF JMP SHORT runas.01001767 //正常运行时跳过下面处理参数的两句
在010016F2处添加语句
010016F2 DEC DWORD PTR SS:[EBP-14] //将参数个数减1
010016F5 JMP runas.0100163E //跳回去
处理参数完毕，再处理密码：
修改01001767 处
01001767 TEST EAX,EAX 为 01001767 XOR EAX,EAX //标志位为零
从010017AF处开始添加语句
010017AF MOV EAX,DWORD PTR SS:[EBP-14]
010017B2 INC EAX //将参数个数加1
010017B3 MOV ECX,DWORD PTR SS:[EBP+8]
010017B6 NOP
010017B7 NOP
010017B8 MOV EAX,DWORD PTR DS:[ECX+EAX*4-4]
010017BC PUSH EAX ; /String2
010017BD LEA EAX,DWORD PTR SS:[EBP-1198] ; |
010017C3 NOP ; |
010017C4 PUSH EAX ; |String1
010017C5 CALL DWORD PTR DS:[<&KERNEL32.lstrcpyW>] ; \lstrcpyW
OK!!!处理完成!!!运行试试!真爽：)
再写个外壳调用就OK了：)
收工
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本文章写于2006-1-25 09:04:55 By FishSeeWater