徹底改掉進程名
寫了個改進程名的東西，跟大家分享！技術(shù)含量不高，大牛飄過。
先總結(jié)一下，一個進程的名字有可能從以下部位獲取（參考小偉同學的《偽造進程初探》一文）：
一、EPROCESS中：
??? 1、EPROCESS-->ImageFileName（很常用，冰刃獲取進程名的地方）
??? 2、EPROCESS-->SeAuditProcessCreationInfo->ImageFileName（任務(wù)管理器獲取進程名的地方，NtQueryInformationProcess就是從這里獲取進程名的）
??? 3、EPROCESS->SectionObject->Segment->ControlArea->FileObject->FileName（RKU獲取進程名的方法）
??? 4、VAD（記錄用戶空間內(nèi)存分配情況的數(shù)據(jù)結(jié)構(gòu)，里面當然有進程的exe模塊）
二、PEB中：
??? 1、PEB-->ProcessParameters-->ImagePathName
??? 2、PEB-->ProcessParameters-->CommandLine
??? 3、PEB-->ProcessParameters-->WindowTitle（這個地方比較奇怪，如果雙擊的是exe的快捷方式，則記錄的是快捷方式的路徑，還是一并改掉的好）
??? 4、PEB-->LDR-->InLoadOrderModuleList->第一個結(jié)構(gòu)->FullDllName
??? 5、PEB-->LDR-->InLoadOrderModuleList->第一個結(jié)構(gòu)->BaseDllName
??? 6、PEB-->LDR-->InMemoryOrderModuleList->第一個結(jié)構(gòu)->FullDllName（此處的BaseDllName貌似為NULL，就不管它了）
??? （PEB-->LDR-->InInitializationOrderModuleList這個表里貌似沒有exe模塊，也不管它了）
把這些地方都改掉即可徹底改掉進程名（如果不夠徹底，謝謝補充！）。
示例代碼如下（示例代碼中以winmine.exe做測試）： //Fypher
//http://hi.baidu.com/nmn714
?
VOID ChangeName(ULONG pProcess){
??? ULONG peb,ProcessParameters,ldr;
??? ULONG InLoadOrderModuleList;
??? ULONG InMemoryOrderModuleList;
??? ULONG tmp;
?
??? KAPC_STATE kapc;
??? PUCHAR str;
??? PWCHAR wstr;
???
??? //get PEB
??? peb=*(PULONG)(pProcess + 0x1b0);
???
??? KeStackAttachProcess((PEPROCESS)pProcess,&kapc);
??? __try{
??????? ProcessParameters = *(PULONG)(peb + 0x010);
??????? //ImagePathName
??????? FindAndChangeUni(ProcessParameters+0x038);
??????? //CommandLine
??????? FindAndChangeUni(ProcessParameters+0x040);
??????? //WindowTitle
??????? FindAndChangeUni(ProcessParameters+0x070);
?
??????? //Ldr
??????? ldr = *(PULONG)(peb + 0x00c);
??????? InLoadOrderModuleList = *(PULONG)(ldr+0x00c);
??????? //InLoadOrderModuleList->FullDllName
??????? FindAndChangeUni(InLoadOrderModuleList+0x024);
??????? //InLoadOrderModuleList->BaseDllName
??????? FindAndChangeUni(InLoadOrderModuleList+0x02c);
??????? InMemoryOrderModuleList = *(PULONG)(ldr+0x014);
??????? //InMemoryOrderModuleList->FullDllName
??????? FindAndChangeUni(InMemoryOrderModuleList+0x024);
??? }__except(1){
??????? KdPrint(("exception occured!"));
??? }
??? KeUnstackDetachProcess (&kapc);
?
??? //EPROCESS-->ImageFileName
??? FindAndChangeA(pProcess+0x174,16);
??? //EPROCESS-->SeAuditProcessCreationInfo->ImageFileName
??? FindAndChangeUni(*(PULONG)(pProcess + 0x1F4));
??? //EPROCESS->SectionObject->Segment->ControlArea->FileObject->FileName
??? //should use MmIsAddressValid to verify
??? tmp=*(PULONG)(pProcess+0x138);
??? tmp=*(PULONG)(tmp+0x14);
??? tmp=*(PULONG)tmp;
??? tmp=*(PULONG)(tmp+0x024);
??? FindAndChangeUni(tmp+0x030);
???
??? //VAD
??? //should use MmIsAddressValid to verify
??? tmp=*(PULONG)(pProcess+0x11c);
??? tmp=*(PULONG)(tmp+0x10);
??? tmp=*(PULONG)(tmp+0x018);
??? tmp=*(PULONG)(tmp+0x024);
??? FindAndChangeUni(tmp+0x030);
}
復制代碼其中，F(xiàn)indAndChangeUni和FindAndChangeA的作用是在一個字符串（UNICODE_STRING或CHAR）中定位“winmine.exe”并改成"winxxoo.exe"。代碼如下: //Fypher
//http://hi.baidu.com/nmn714
?
VOID FindAndChangeUni(ULONG strAddr){
??? PUNICODE_STRING uniStr = (PUNICODE_STRING)strAddr;
??? ULONG len = uniStr->Length / 2;
??? ULONG maxLen = uniStr->MaximumLength / 2;
??? PWCHAR str = uniStr->Buffer;
??? ULONG i=0;
?
??? if(!str || len<11|| maxLen<11 )
??????? return;
?
??? for(i=0;i<= len - 11;++i){
??????? if(!_wcsnicmp(str+i,L"winmine.exe",11))
??????????? break;
??? }
?
??? if(i>len - 11)
??????? return;
???
??? _asm{
??????? cli
??????? mov eax, cr0
??????? and eax, not 0x10000
??????? mov cr0, eax
??? }
??? //str可能是PEB中的，故try之
??? __try{
??????? str[i+3]=L'x';
??????? str[i+4]=L'x';
??????? str[i+5]=L'o';
??????? str[i+6]=L'o';
??? }__except(1){
??? }???
??? _asm{
??????? mov eax, cr0
??????? or eax,0x10000
??????? mov cr0,eax
??????? sti
??? }
}
?
VOID FindAndChangeA(ULONG strAddr,ULONG len){
??? PUCHAR str = (PUCHAR)strAddr;
??? ULONG i=0;
?
??? if(!str || len<11 )
??????? return;
?
??? for(i=0;i<= len - 11;++i){
??????? if(!_strnicmp(str+i,"winmine.exe",11))
??????????? break;
??? }
?
??? if(i>len - 11)
??????? return;
???
??? _asm{
??????? cli
??????? mov eax, cr0
??????? and eax, not 0x10000
??????? mov cr0, eax
??? }
??? //str可能是PEB中的，故try之
??? __try{
??????? str[i+3]='x';
??????? str[i+4]='x';
??????? str[i+5]='o';
??????? str[i+6]='o';
??? }__except(1){
??? }???
???
??? _asm{
??????? mov eax, cr0
??????? or eax,0x10000
??????? mov cr0,eax
??????? sti
??? }
}
復制代碼截圖效果：
參考：《偽造進程初探》——小偉同學（膜拜一下）
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