Windows?NT/2000/XP下不用驅(qū)動(dòng)的Ring0代碼實(shí)現(xiàn)?????
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????大家知道，Windows?NT/2000為實(shí)現(xiàn)其可靠性，嚴(yán)格將系統(tǒng)劃分為內(nèi)核模式與用戶模式，在i386系統(tǒng)中分別對(duì)應(yīng)CPU的Ring0與Ring3級(jí)別。Ring0下，可以執(zhí)行特權(quán)級(jí)指令，對(duì)任何I/O設(shè)備都有訪問(wèn)權(quán)等等。要實(shí)現(xiàn)從用戶態(tài)進(jìn)入核心態(tài)，即從Ring?3進(jìn)入Ring?0必須借助CPU的某種門機(jī)制，如中斷門、調(diào)用門等。而Windows?NT/2000提供用戶態(tài)執(zhí)行系統(tǒng)服務(wù)(Ring?0例程)的此類機(jī)制即System?Service的int?2eh中斷服務(wù)等，嚴(yán)格的參數(shù)檢查，只能嚴(yán)格的執(zhí)行Windows?NT/2000提供的服務(wù)，而如果想執(zhí)行用戶提供的Ring?0代碼(指運(yùn)行在Ring?0權(quán)限的代碼)，常規(guī)方法似乎只有編寫(xiě)設(shè)備驅(qū)動(dòng)程序。本文將介紹一種在用戶態(tài)不借助任何驅(qū)動(dòng)程序執(zhí)行Ring0代碼的方法。
????Windows?NT/2000將設(shè)備驅(qū)動(dòng)程序調(diào)入內(nèi)核區(qū)域(常見(jiàn)的位于地址0x80000000上)，由DPL為0的GDT項(xiàng)8，即cs為8時(shí)實(shí)現(xiàn)Ring?0權(quán)限。本文通過(guò)在系統(tǒng)中構(gòu)造一個(gè)指向我們的代碼的調(diào)用門(CallGate)，實(shí)現(xiàn)Ring0代碼。基于這個(gè)思路，為實(shí)現(xiàn)這個(gè)目的主要是構(gòu)造自己的CallGate。CallGate由系統(tǒng)中叫Global?Descriptor?Table(GDT)的全局表指定。GDT地址可由i386指令sgdt獲得(sgdt不是特權(quán)級(jí)指令，普通Ring 3程序均可執(zhí)行)。GDT地址在Windows?NT/2000保存于KPCR(Processor?Control?Region)結(jié)構(gòu)中(見(jiàn)《再談Windows?NT/2000環(huán)境切換》)。GDT中的CallGate是如下的格式：
????typedef?struct
????
{?
????????unsigned?short??offset_0_15;
????????unsigned?short??selector;
????????unsigned?char????param_count?:?4;
????????unsigned?char????some_bits???:?4;
????????unsigned?char????type????????:?4;
????????unsigned?char????app_system??:?1;
????????unsigned?char????dpl?????????:?2;
????????unsigned?char????present?????:?1;
????
????????unsigned?short??offset_16_31;
?????}?CALLGATE_DESCRIPTOR;????GDT位于內(nèi)核區(qū)域，一般用戶態(tài)的程序是不可能對(duì)這段內(nèi)存區(qū)域有直接的訪問(wèn)權(quán)。幸運(yùn)的是Windows?NT/2000提供了一個(gè)叫PhysicalMemory的Section內(nèi)核對(duì)象位于\Device的路徑下。顧名思義，通過(guò)這個(gè)Section對(duì)象可以對(duì)物理內(nèi)存進(jìn)行操作。用objdir.exe對(duì)這個(gè)對(duì)象分析如下：
????C:\NTDDK\bin>objdir?/D?\Device
????PhysicalMemory???????????????????
????????Section
????????DACL?-?
???????????Ace[?0]?-?Grant?-?0xf001f?-?NT?AUTHORITY\SYSTEM
?????????????????????????????Inherit:?
?????????????????????????????Access:?0x001F??and??(?D?RCtl?WOwn?WDacl?)
???????????Ace[?1]?-?Grant?-?0x2000d?-?BUILTIN\Administrators
?????????????????????????????Inherit:?
?????????????????????????????Access:?0x000D??and??(?RCtl?) ???從dump出的這個(gè)對(duì)象DACL的Ace可以看出默認(rèn)情況下只有SYSTEM用戶才有對(duì)這個(gè)對(duì)象的讀寫(xiě)權(quán)限，即對(duì)物理內(nèi)存有讀寫(xiě)能力，而Administrator只有讀權(quán)限，普通用戶根本就沒(méi)有權(quán)限。不過(guò)如果我們有Administrator權(quán)限就可以通過(guò)GetSecurityInfo、SetEntriesInAcl與SetSecurityInfo這些API來(lái)修改這個(gè)對(duì)象的ACE。這也是我提供的代碼需要Administrator的原因。實(shí)現(xiàn)的代碼如下：
???VOID?SetPhyscialMemorySectionCanBeWrited(HANDLE?hSection)
????{?
???????PACL?pDacl=NULL;
???????PACL?pNewDacl=NULL;
???????PSECURITY_DESCRIPTOR?pSD=NULL;
???????DWORD?dwRes;
???????EXPLICIT_ACCESS?ea;
???????if(dwRes=GetSecurityInfo(hSection,SE_KERNEL_OBJECT,DACL_SECURITY_INFORMATION,
??????????????????NULL,NULL,&pDacl,NULL,&pSD)!=ERROR_SUCCESS)
??????????{?
?????????????printf(?"GetSecurityInfo?Error?%u\n",?dwRes?);
?????????????goto?CleanUp;
???????????}
???????ZeroMemory(&ea,?sizeof(EXPLICIT_ACCESS));
???????ea.grfAccessPermissions?=?SECTION_MAP_WRITE;
???????ea.grfAccessMode?=?GRANT_ACCESS;
???????ea.grfInheritance=?NO_INHERITANCE;
???????ea.Trustee.TrusteeForm?=?TRUSTEE_IS_NAME;
???????ea.Trustee.TrusteeType?=?TRUSTEE_IS_USER;
???????ea.Trustee.ptstrName?=?"CURRENT_USER";
???????if(dwRes=SetEntriesInAcl(1,&ea,pDacl,&pNewDacl)!=ERROR_SUCCESS)
??????????{?
?????????????printf(?"SetEntriesInAcl?%u\n",?dwRes?);
?????????????goto?CleanUp;
???????????}
???????if(dwRes=SetSecurityInfo(hSection,SE_KERNEL_OBJECT,DACL_SECURITY_INFORMATION,NULL,NULL,pNewDacl,NULL)!=ERROR_SUCCESS)
??????????{?
?????????????printf("SetSecurityInfo?%u\n",dwRes);
?????????????goto?CleanUp;
???????????}
????CleanUp:
???????if(pSD)
??????????LocalFree(pSD);
???????if(pNewDacl)
??????????LocalFree(pSD);
?????}
????這段代碼對(duì)給定HANDLE的對(duì)象增加了如下的ACE:?
????PhysicalMemory???????????????????
????????Section
????????DACL?-?
???????????Ace[?0]?-?Grant?-?0x2?-?WEBCRAZY\Administrator
?????????????????????????????Inherit:?
?????????????????????????????Access:?0x0002????//SECTION_MAP_WRITE
???這樣我們?cè)谟蠥dministrator權(quán)限的條件下就有了對(duì)物理內(nèi)存的讀寫(xiě)能力。但若要修改GDT表實(shí)現(xiàn)Ring?0代碼。我們將面臨著另一個(gè)難題，因?yàn)閟gdt指令獲得的GDT地址是虛擬地址(線性地址)，我們只有知道GDT表的物理地址后才能通過(guò)\Device\PhysicalMemory對(duì)象修改GDT表，這就牽涉到了線性地址轉(zhuǎn)化成物理地址的問(wèn)題。我們先來(lái)看一看Windows?NT/2000是如何實(shí)現(xiàn)這個(gè)的：
????kd>?u?nt!MmGetPhysicalAddress?l?30
????ntoskrnl!MmGetPhysicalAddress:
????801374e0?56???????????????push????esi
????801374e1?8b742408?????????mov?????esi,[esp+0x8]
????801374e5?33d2?????????????xor?????edx,edx
????801374e7?81fe00000080?????cmp?????esi,0x80000000
????801374ed?722c?????????????jb????ntoskrnl!MmGetPhysicalAddress+0x2b?(8013751b)
????801374ef?81fe000000a0?????cmp?????esi,0xa0000000
????801374f5?7324?????????????jnb???ntoskrnl!MmGetPhysicalAddress+0x2b?(8013751b)
????801374f7?39153ce71780?????cmp?????[ntoskrnl!MmKseg2Frame?(8017e73c)],edx
????801374fd?741c?????????????jz????ntoskrnl!MmGetPhysicalAddress+0x2b?(8013751b)
????801374ff?8bc6?????????????mov?????eax,esi
????80137501?c1e80c???????????shr?????eax,0xc
????80137504?25ffff0100???????and?????eax,0x1ffff
????80137509?6a0c?????????????push????0xc
????8013750b?59???????????????pop?????ecx
????8013750c?e8d3a7fcff???????call????ntoskrnl!_allshl?(80101ce4)
????80137511?81e6ff0f0000?????and?????esi,0xfff
????80137517?03c6?????????????add?????eax,esi
????80137519?eb17?????????????jmp???ntoskrnl!MmGetPhysicalAddress+0x57?(80137532)
????8013751b?8bc6?????????????mov?????eax,esi
????8013751d?c1e80a???????????shr?????eax,0xa
????80137520?25fcff3f00???????and?????eax,0x3ffffc
????80137525?2d00000040???????sub?????eax,0x40000000
????8013752a?8b00?????????????mov?????eax,[eax]
????8013752c?a801?????????????test????al,0x1
????8013752e?7506?????????????jnz???ntoskrnl!MmGetPhysicalAddress+0x44?(80137536)
????80137530?33c0?????????????xor?????eax,eax
????80137532?5e???????????????pop?????esi
????80137533?c20400???????????ret?????0x4
????從這段匯編代碼可看出如果線性地址在0x80000000與0xa0000000范圍內(nèi)，只是簡(jiǎn)單的進(jìn)行移位操作(位于801374ff-80137519指令間)，并未查頁(yè)表。我想Microsoft這樣安排肯定是出于執(zhí)行效率的考慮。這也為我們指明了一線曙光，因?yàn)镚DT表在Windows?NT/2000中一般情況下均位于這個(gè)區(qū)域(我不知道/3GB開(kāi)關(guān)的Windows?NT/2000是不是這種情況)。
????經(jīng)過(guò)這樣的分析，我們就可以只通過(guò)用戶態(tài)程序修改GDT表了。而增加一個(gè)CallGate就不是我可以介紹的了，找本Intel手冊(cè)自己看一看了。具體實(shí)現(xiàn)代碼如下：
????typedef?struct?gdtr?{?
????????short?Limit;
????????short?BaseLow;
????????short?BaseHigh;
?????}?Gdtr_t,?*PGdtr_t;
????ULONG?MiniMmGetPhysicalAddress(ULONG?virtualaddress)
????{?
????????if(virtualaddress<0x80000000||virtualaddress>=0xA0000000)
???????????return?0;
????????return?virtualaddress&0x1FFFF000;
?????}
????BOOL?ExecRing0Proc(ULONG?Entry,ULONG?seglen)
????{?
???????Gdtr_t?gdt;
???????__asm?sgdt?gdt;
?????
???????ULONG?mapAddr=MiniMmGetPhysicalAddress(gdt.BaseHigh<<16U|gdt.BaseLow);
???????if(!mapAddr)?return?0;
???????HANDLE???hSection=NULL;
???????NTSTATUS?status;
???????OBJECT_ATTRIBUTES????????objectAttributes;
???????UNICODE_STRING?objName;
???????CALLGATE_DESCRIPTOR?*cg;
???????status?=?STATUS_SUCCESS;
???
???????RtlInitUnicodeString(&objName,L"\\Device\\PhysicalMemory");
???????InitializeObjectAttributes(&objectAttributes,
??????????????????????????????????&objName,
??????????????????????????????????OBJ_CASE_INSENSITIVE?|?OBJ_KERNEL_HANDLE,
??????????????????????????????????NULL,
?????????????????????????????????(PSECURITY_DESCRIPTOR)?NULL);
???????status?=?ZwOpenSection(&hSection,SECTION_MAP_READ|SECTION_MAP_WRITE,&objectAttributes);
???????if(status?==?STATUS_ACCESS_DENIED){?
??????????status?=?ZwOpenSection(&hSection,READ_CONTROL|WRITE_DAC,&objectAttributes);
??????????SetPhyscialMemorySectionCanBeWrited(hSection);
??????????ZwClose(hSection);
??????????status?=ZwOpenSection(&hSection,SECTION_MAP_WRITE|SECTION_MAP_WRITE,&objectAttributes);
????????}
???????if(status?!=?STATUS_SUCCESS)
?????????{?
????????????printf("Error?Open?PhysicalMemory?Section?Object,Status:%08X\n",status);
????????????return?0;
??????????}
??????
???????PVOID?BaseAddress;
???????BaseAddress=MapViewOfFile(hSection,
?????????????????????FILE_MAP_READ|FILE_MAP_WRITE,
?????????????????????0,
?????????????????????mapAddr,????//low?part
?????????????????????(gdt.Limit+1));
???????if(!BaseAddress)
??????????{?
?????????????printf("Error?MapViewOfFile:");
?????????????PrintWin32Error(GetLastError());
?????????????return?0;
???????????}
???????BOOL?setcg=FALSE;
???????for(cg=(CALLGATE_DESCRIPTOR?*)((ULONG)BaseAddress+(gdt.Limit&0xFFF8));(ULONG)cg>(ULONG)BaseAddress;cg--)
???????????if(cg->type?==?0){?
?????????????cg->offset_0_15?=?LOWORD(Entry);
?????????????cg->selector?=?8;
?????????????cg->param_count?=?0;
?????????????cg->some_bits?=?0;
?????????????cg->type?=?0xC;??????????//?386?call?gate
?????????????cg->app_system?=?0;??????//?A?system?descriptor
?????????????cg->dpl?=?3;?????????????//?Ring?3?code?can?call
?????????????cg->present?=?1;
?????????????cg->offset_16_31?=?HIWORD(Entry);
?????????????setcg=TRUE;
?????????????break;
???????????}
???????if(!setcg){?
????????????ZwClose(hSection);
????????????return?0;
????????}
???????short?farcall[3];
???????farcall[2]=((short)((ULONG)cg-(ULONG)BaseAddress))|3;??//Ring?3?callgate;
???????if(!VirtualLock((PVOID)Entry,seglen))
??????????{?
?????????????printf("Error?VirtualLock:");
?????????????PrintWin32Error(GetLastError());
?????????????return?0;
???????????}
???????SetThreadPriority(GetCurrentThread(),THREAD_PRIORITY_TIME_CRITICAL);
???????Sleep(0);
???????_asm?call?fword?ptr?[farcall]
???????SetThreadPriority(GetCurrentThread(),THREAD_PRIORITY_NORMAL);
???????VirtualUnlock((PVOID)Entry,seglen);
???????//Clear?callgate
???????*(ULONG?*)cg=0;
???????*((ULONG?*)cg+1)=0;
???????ZwClose(hSection);
???????return?TRUE;
?????}
????我在提供的代碼中演示了對(duì)Control?Register與I/O端口的操作。CIH病毒在Windows?9X中就是因?yàn)楂@得Ring?0權(quán)限才有了一定的危害，但Windows?NT/2000畢竟不是Windows?9X，她已經(jīng)有了比較多的安全審核機(jī)制，本文提供的代碼也要求具有Administrator權(quán)限，但如果系統(tǒng)存在某種漏洞，如緩沖區(qū)溢出等等，還是有可能獲得這種權(quán)限的，所以我不對(duì)本文提供的方法負(fù)有任何的責(zé)任，所有討論只是一個(gè)技術(shù)熱愛(ài)者在討論技術(shù)而已。謝謝！?
????參考資料：
??????1.Intel Corp<<Intel Architecture Software Developer's Manual,Volume 3>>?