官方wp:
进程重影技术：
进程重映像利用了Windows内核中的缓存同步问题，它会导致可执行文件的路径与从该可执行文件创建的映像节区所报告的路径不匹配。通过在一个诱饵路径上加载DLL，然后卸载它，然后从一个新路径加载它，许多Windows API将返回旧路径。这可能可以欺骗安全产品，使其在错误的路径上查找加载的映像。
主要创建方式就是先打开一个新文件，然后把这个文件挂到删除列表上，在关闭文件句柄后文件就会被删除，但是在还没有关闭的时候此时文件还未删除，此时能向文件中写入数据，然后再把这个文件映射到内存上，再关闭文件句柄，此时文件删除，但是内存中还有文件的映像，达到一定的迷惑杀软的目的。
如果是做题的话，直接用IDA附加开启的子进程，然后发现XTEA，解密得到flag
如果是学技术的话，还是要研究一下 “进程重影” 技术思路。
本文两者都会介绍的。
考点：AES\XTEA\进程与子进程\SMC\进程重影
分析
DIE打开，发现是pe64位，无壳
拖进IDA进行分析
习惯操作，先用FindCrypto扫一下，有没有加密
发现AES加密
跟踪main函数，看看它到底要干什么
int __cdecl main_0(int argc, const char **argv, const char **envp){ j___CheckForDebuggerJustMyCode(&unk_1400A80B9);// vs2022调试debug版c++程序会出现该函数，我也不知道啥用 sub_140001FF0(*argv); return 0;}AES核心加密逻辑函数
__int64 __fastcall sub_13F887480(char *a1, char *a2, unsigned int a3, __int64 a4, __int64 a5){ __int64 result; // rax unsigned __int64 i; // [rsp+28h] [rbp+8h] unsigned __int8 v7; // [rsp+44h] [rbp+24h] j___CheckForDebuggerJustMyCode(&unk_13F92800D); v7 = a3 % 0x10; if ( a4 ) { qword_13F91D188 = a4; sub_13F888EA0(); } if ( a5 ) qword_13F91D240 = a5; for ( i = 0i64; i < a3; i += 16i64 ) { sub_13F889C80(a2); j_memmove(a1, a2, 0x10ui64); qword_13F91D180 = (__int64)a1; sub_13F887850(); qword_13F91D240 = (__int64)a1; a2 += 16; a1 += 16; } result = v7; if ( v7 ) { j_memmove(a1, a2, v7); qword_13F91D180 = (__int64)a1; return sub_13F887850(); } return result;}跟进了sub_14000A700函数
signed int __fastcall sub_14000A700(const char *a1){ char *v1; // rdi __int64 i; // rcx DWORD LastError; // eax signed int result; // eax HANDLE CurrentProcess; // rax size_t v6; // rax DWORD dwCreationDisposition; // [rsp+20h] [rbp-40h] char v8; // [rsp+60h] [rbp+0h] BYREF CHAR FileName[48]; // [rsp+68h] [rbp+8h] BYREF __int64 v10[4]; // [rsp+98h] [rbp+38h] BYREF HANDLE hObject; // [rsp+B8h] [rbp+58h] unsigned int v12; // [rsp+D4h] [rbp+74h] char v13[36]; // [rsp+F4h] [rbp+94h] BYREF int v14[12]; // [rsp+118h] [rbp+B8h] BYREF __int64 (__fastcall *v15)(HANDLE, int *, char *, __int64, int); // [rsp+148h] [rbp+E8h] HRSRC hResInfo; // [rsp+168h] [rbp+108h] DWORD v17; // [rsp+184h] [rbp+124h] HGLOBAL hResData; // [rsp+1A8h] [rbp+148h] void *Src; // [rsp+1C8h] [rbp+168h] DWORD NumberOfBytesWritten; // [rsp+1E4h] [rbp+184h] BYREF DWORD nNumberOfBytesToWrite[8]; // [rsp+204h] [rbp+1A4h] BYREF size_t Size; // [rsp+224h] [rbp+1C4h] void *Block; // [rsp+248h] [rbp+1E8h] LPCVOID lpBuffer[4]; // [rsp+268h] [rbp+208h] BYREF __int64 (__fastcall *v25)(__int64 *, __int64, _QWORD, _QWORD, int, int, HANDLE); // [rsp+288h] [rbp+228h] __int64 v26[3]; // [rsp+2A8h] [rbp+248h] BYREF unsigned int v27[9]; // [rsp+2C4h] [rbp+264h] BYREF HANDLE hHandle[4]; // [rsp+2E8h] [rbp+288h] BYREF __int64 (__fastcall *v29)(HANDLE *, __int64, _QWORD, HANDLE, DWORD, __int64, _QWORD, _QWORD); // [rsp+308h] [rbp+2A8h] char v30[76]; // [rsp+328h] [rbp+2C8h] BYREF int v31; // [rsp+374h] [rbp+314h] __int64 (__fastcall *v32)(HANDLE, _QWORD, char *, __int64, _QWORD); // [rsp+398h] [rbp+338h] char *Str; // [rsp+3B8h] [rbp+358h] int v34; // [rsp+3D4h] [rbp+374h] int v35; // [rsp+3F4h] [rbp+394h] int v36; // [rsp+414h] [rbp+3B4h] wchar_t *Dest; // [rsp+438h] [rbp+3D8h] __int64 v38[63]; // [rsp+460h] [rbp+400h] BYREF __int64 v39; // [rsp+658h] [rbp+5F8h] __int64 v40[4]; // [rsp+678h] [rbp+618h] BYREF __int64 (__fastcall *v41)(__int64 *, __int64, _QWORD, HANDLE, __int64, _QWORD, _DWORD, _QWORD, _QWORD, _QWORD, _QWORD); // [rsp+698h] [rbp+638h] int v42; // [rsp+9A4h] [rbp+944h] v1 = &v8; for ( i = 404i64; i; --i ) { *(_DWORD *)v1 = -858993460; v1 += 4; } j___CheckForDebuggerJustMyCode(&unk_1400A80B9); strcpy(FileName, "VNctf2023"); v10[0] = 0x616C7972723073i64; hObject = CreateFileA(FileName, 0xC0010000, 0, 0i64, 2u, 0x80u, 0i64);// 创建一个文件 if ( hObject == (HANDLE)-1i64 ) { // 失败 LastError = GetLastError(); return sub_14000257C("Failed - Error Code %08X\r\n", LastError); } else { v13[0] = 1; // 创建成功 v15 = (__int64 (__fastcall *)(HANDLE, int *, char *, __int64, int))sub_1400016B3("NtSetInformationFile"); v12 = v15(hObject, v14, v13, 1i64, 13); if ( v14[0] >= 0 ) { hResInfo = FindResourceA(0i64, (LPCSTR)0x66, "shell");// 寻找指定类型和名称的资源位置 v17 = SizeofResource(0i64, hResInfo); // 资源大小 hResData = LoadResource(0i64, hResInfo); // 加载资源 Src = LockResource(hResData); NumberOfBytesWritten = 0; nNumberOfBytesToWrite[0] = 0; LODWORD(Size) = v17 - 4; Block = j_j_j__malloc_base(v17 - 4); j_memmove(Block, Src, (unsigned int)Size); j_memmove(nNumberOfBytesToWrite, (char *)Src + (unsigned int)Size, 4ui64); GlobalUnlock(hResData); lpBuffer[0] = 0i64; NumberOfBytesWritten = sub_1400030BC(Block, v10, lpBuffer, (unsigned int)Size); //关键函数，好像对shell资源进行了 AES加密 if ( NumberOfBytesWritten >= nNumberOfBytesToWrite[0] && lpBuffer[0] && WriteFile(hObject, lpBuffer[0], nNumberOfBytesToWrite[0], &NumberOfBytesWritten, 0i64)// 加密后shell 写到资源到文件 && (j_free(Block), // 释放 j_free((void *)lpBuffer[0]), v25 = (__int64 (__fastcall *)(__int64 *, __int64, _QWORD, _QWORD, int, int, HANDLE))sub_1400016B3("NtCreateSection"),// 创建节对象 v12 = v25(v26, 983071i64, 0i64, 0i64, 2, 0x1000000, hObject), (v12 & 0x80000000) == 0) ) // 猜测上面的代码，就是把shell 资源 写到一个叫 vnctf2023的文件里面 { result = sub_14000347C(hObject, v27, nNumberOfBytesToWrite[0]); if ( result ) // 写入成功 { CloseHandle(hObject); hHandle[0] = 0i64; v29 = (__int64 (__fastcall *)(HANDLE *, __int64, _QWORD, HANDLE, DWORD, __int64, _QWORD, _QWORD))sub_1400016B3("NtCreateProcess");// 启动进程 CurrentProcess = GetCurrentProcess(); LOBYTE(dwCreationDisposition) = 1; result = v29(hHandle, 0x1FFFFFi64, 0i64, CurrentProcess, dwCreationDisposition, v26[0], 0i64, 0i64); v12 = result; if ( result >= 0 ) { memset(v30, 0, 0x30ui64); v31 = 0; v32 = (__int64 (__fastcall *)(HANDLE, _QWORD, char *, __int64, _QWORD))sub_1400016B3("NtQueryInformationProcess"); result = v32(hHandle[0], 0i64, v30, 48i64, 0i64); v12 = result; if ( result >= 0 ) { Str = (char *)j_j_j__malloc_base(4ui64); sub_1400039C2(Str, "%x", NumberOfBytesWritten); v34 = j_strlen(FileName); v35 = j_strlen(a1); v36 = j_strlen(Str); j_strcat(FileName, " "); j_strcat(FileName, a1); j_strcat(FileName, " "); j_strcat(FileName, Str); v34 += v35 + v36 + 2; v42 = v34 + 1; v6 = (unsigned int)(2 * (v34 + 1)); if ( !is_mul_ok(2u, v34 + 1) ) v6 = -1i64; Dest = (wchar_t *)j_j_j__malloc_base(v6); sub_140003175(Dest, 0i64, 2 * v34 + 2); j_mbstowcs(Dest, FileName, v34); result = sub_140003DBE(hHandle[0], v30, Dest); if ( result ) { j_free(Dest); memset(v38, 0, 0x1B8ui64); result = sub_140002176(hHandle[0], v30, v38); if ( result ) { v38[59] = v38[2]; v39 = v38[2] + v27[0]; v40[0] = 0i64; v41 = (__int64 (__fastcall *)(__int64 *, __int64, _QWORD, HANDLE, __int64, _QWORD, _DWORD, _QWORD, _QWORD, _QWORD, _QWORD))sub_1400016B3("NtCreateThreadEx"); //启动线程 result = v41(v40, 0x1FFFFFi64, 0i64, hHandle[0], v39, 0i64, 0, 0i64, 0i64, 0i64, 0i64); v12 = result; if ( result >= 0 ) return WaitForSingleObject(hHandle[0], 0xFFFFFFFF); } } } } } } else { return CloseHandle(hObject); } } else { sub_14000257C("Failed - Error Code %08X\r\n", v12); return CloseHandle(hObject); } } return result;}关键函数 sub_140009E20，就是那个对shell资源进行处理的函数
__int64 __fastcall sub_140009E20(__int64 a1, const char *a2, void **a3, unsigned int a4){ char *v4; // rdi __int64 i; // rcx size_t v6; // rax char v8; // [rsp+20h] [rbp+0h] BYREF char v9[44]; // [rsp+28h] [rbp+8h] BYREF unsigned int v10; // [rsp+54h] [rbp+34h] BOOL v11; // [rsp+74h] [rbp+54h] int j; // [rsp+94h] [rbp+74h] size_t Size; // [rsp+168h] [rbp+148h] v4 = &v8; for ( i = 42i64; i; --i ) { *(_DWORD *)v4 = -858993460; v4 += 4; } j___CheckForDebuggerJustMyCode(&unk_1400A80B9, a2, a3); v11 = a4 % 0x10 != 0; v10 = 16 * (v11 + a4 / 0x10); v6 = v10 + 1; if ( v10 == -1 ) v6 = -1i64; *a3 = j_j_j__malloc_base(v6); sub_140003175(*a3, 0i64, v10 + 1); sub_140003175(v9, 0i64, 16i64); if ( j_strlen(a2) <= 0x10 ) Size = j_strlen(a2); else Size = 16i64; j_memmove(v9, a2, Size); for ( j = 0; 16 * j < a4; ++j ) sub_140002DE7(16 * j + a1, v9, (char *)*a3 + 16 * j, 16i64);// AES解密函数 我思考了一下，如果shell进行加密，那么shell还能运行吗？反之，只有解密，才可以正常运行 return v10;}根进观察一下
__int64 __fastcall sub_140007620(const void *a1, __int64 a2, void *a3, unsigned int a4){ j___CheckForDebuggerJustMyCode(&unk_1400A800D, a2, a3); j_memmove(a3, a1, a4); qword_14009D180 = (__int64)a3; qword_14009D188 = a2; sub_140008EA0(); return sub_1400078E0();}发现了一堆函数，一个一个看看
最后发现sub_1400078E0函数的代码，长得特别像AES加密函数。
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从 confuse_us 那题就可以发现，这些代码长得很像.就是AES加密套路
__int64 __fastcall sub_1400078E0(__int64 a1, __int64 a2, __int64 a3){ __int64 v3; // rcx unsigned __int8 i; // [rsp+24h] [rbp+4h] j___CheckForDebuggerJustMyCode(&unk_1400A800D, a2, a3); LOBYTE(v3) = 10; sub_140007760(v3); for ( i = 9; i; --i ) { sub_140008980(); sub_140008DF0(); sub_140007760(i); sub_140007970(); } sub_140008980(); sub_140008DF0(); return sub_140007760(0i64);}跟进sub_140007760 函数，发现就是addroundkey()
__int64 __fastcall sub_140007760(unsigned __int8 a1, __int64 a2, __int64 a3){ __int64 result; // rax unsigned __int8 i; // [rsp+24h] [rbp+4h] unsigned __int8 j; // [rsp+44h] [rbp+24h] j___CheckForDebuggerJustMyCode(&unk_1400A800D, a2, a3); for ( i = 0; ; ++i ) { result = i; if ( i >= 4u ) //这不就是 addroundkey() 函数吗？ break; for ( j = 0; j < 4u; ++j ) *(_BYTE *)(qword_14009D180 + 4i64 * i + j) ^= byte_14009D190[16 * a1 + 4 * i + j]; //对比confuse_us那题，发现没有魔改 ^0x23 } return result;}其余函数也就不用再分析了，可以断定这是AES加密函数。
对shell资源进行AES解密，然后shell程序跑起来
运行该程序，发现我追踪的进程居然挂掉了，又跑起来了另一个进程
可以断定，父进程开子进程
子进程输入flag
我再次运行，发现jiji.exe没了？？？
传递的参数FileNameVNctf2023C:\Users\Le\Desktop\jijiji.exe16000s0rryla附加下子进程
搜索字符串，但是无法找到关键函数
直接dump解密后的exe文件
断点下在此处，然后运行，找到解密后的文件，dump下来
D键转地址
跳转到MZ头部，发现这就是解密后的程序
把它dump下来
def main(): begin = 0x23F8CB49290 # #需对应修改 size = 0x16000 # #需对应修改 list1 = [] for i in range(size): byte_tmp = get_bytes(begin + i,1) list1.append(ord(byte_tmp)) if (i + 1) % 0x1000 == 0: print("All count:{}, collect current:{}, has finish {}".format(hex(size), hex(i + 1), float(i + 1) / size)) print('collect over') file = "C:\\Users\\Le\\Desktop\\WASS.exe" #需对应修改 #print(bytearray(list1)) buf = bytearray(list1) with open(file, 'wb') as fw: fw.write(buf) print('write over')if __name__=='__main__': main()得到dump下来的程序
分析dump下来的程序
根据关键字符串定位关键函数
__int64 __fastcall sub_1400064F0(__int64 a1, __int64 a2){ char *v2; // rdi __int64 i; // rcx const char *v4; // rax DWORD LastError; // eax char v7[32]; // [rsp+0h] [rbp-20h] BYREF char v8; // [rsp+20h] [rbp+0h] BYREF HANDLE hSnapshot; // [rsp+28h] [rbp+8h] PROCESSENTRY32 pe; // [rsp+50h] [rbp+30h] BYREF BOOL v11; // [rsp+194h] [rbp+174h] unsigned int v12; // [rsp+1B4h] [rbp+194h] DWORD CurrentProcessId; // [rsp+1D4h] [rbp+1B4h] DWORD th32ProcessID; // [rsp+1F4h] [rbp+1D4h] char v15[536]; // [rsp+220h] [rbp+200h] BYREF char v16[64]; // [rsp+438h] [rbp+418h] BYREF LPVOID lpAddress; // [rsp+478h] [rbp+458h] DWORD flOldProtect[9]; // [rsp+494h] [rbp+474h] BYREF LPCSTR lpFileName; // [rsp+4B8h] [rbp+498h] char v20[64]; // [rsp+718h] [rbp+6F8h] BYREF char v21[64]; // [rsp+758h] [rbp+738h] BYREF char v22[48]; // [rsp+798h] [rbp+778h] BYREF __int64 v23; // [rsp+7C8h] [rbp+7A8h] __int64 v24; // [rsp+7D0h] [rbp+7B0h] __int64 v25; // [rsp+7D8h] [rbp+7B8h] __int64 v26; // [rsp+7E0h] [rbp+7C0h] __int64 v27; // [rsp+7E8h] [rbp+7C8h] __int64 v28; // [rsp+7F0h] [rbp+7D0h] v2 = &v8; for ( i = 362i64; i; --i ) { *(_DWORD *)v2 = -858993460; v2 += 4; } sub_14000148D(&unk_1400180F5); hSnapshot = CreateToolhelp32Snapshot(2u, 0); v11 = Process32First(hSnapshot, &pe); v12 = -1; CurrentProcessId = GetCurrentProcessId(); while ( v11 ) { if ( CurrentProcessId == pe.th32ProcessID ) { th32ProcessID = pe.th32ProcessID; v12 = sub_140001267(pe.th32ProcessID); } v11 = Process32Next(hSnapshot, &pe); } if ( v12 != -1 ) { sub_140001479(v12, v15, 10i64); v23 = sub_140001217(v20, " > nul"); v24 = v23; v25 = sub_140001217(v21, "taskkill -f /pid "); v26 = v25; v27 = sub_14000143D(v22, v25, v15); v28 = v27; sub_140001069(v16, v27, v24); sub_1400010F0(v22); sub_1400010F0(v21); sub_1400010F0(v20); v4 = (const char *)sub_14000106E(v16); system(v4); sub_1400010F0(v16); } lpAddress = (LPVOID)sub_140001357(sub_1400010C3); VirtualProtect(lpAddress, 0x400ui64, 0x40u, flOldProtect); lpFileName = *(LPCSTR *)(a2 + 8); qword_140014470 = *(_QWORD *)(a2 + 16); if ( !DeleteFileA(lpFileName) ) { LastError = GetLastError(); sub_14000123A("%d", LastError); } sub_1400010FF(sub_1400010C3, qword_140014470); sub_1400010C3(); sub_1400013C5(v7, &unk_1400101C8); return 0i64;}
动调调试走到flag验证的地方
动态调试跟进去
成功到达关键的地方，如果不是这么清晰的话，需要U+C键来调
__int64 sub_140005B40(){ char *v0; // rdi __int64 i; // rcx char v3[32]; // [rsp+0h] [rbp-20h] BYREF char v4; // [rsp+20h] [rbp+0h] BYREF char v5[60]; // [rsp+28h] [rbp+8h] BYREF int v6; // [rsp+64h] [rbp+44h] int v7[11]; // [rsp+88h] [rbp+68h] int j; // [rsp+B4h] [rbp+94h] int k; // [rsp+D4h] [rbp+B4h] int *v10; // [rsp+F8h] [rbp+D8h] unsigned int v11; // [rsp+114h] [rbp+F4h] unsigned int v12; // [rsp+134h] [rbp+114h] unsigned int v13; // [rsp+154h] [rbp+134h] int v14; // [rsp+174h] [rbp+154h] int m; // [rsp+194h] [rbp+174h] int n; // [rsp+1B4h] [rbp+194h] v0 = &v4; for ( i = 110i64; i; --i ) { *(_DWORD *)v0 = -858993460; v0 += 4; } sub_14000148D(&unk_1400180F5); memset(v5, 0, 0x21ui64); v6 = 0; v7[0] = 98; v7[1] = 111; v7[2] = 109; v7[3] = 98; sub_14000123A("your flag:"); sub_1400010BE("%s", v5); for ( j = 0; v5[j]; ++j ) ; if ( j != 32 ) sub_14000123A("bad, ji~ji~ji\n"); for ( k = 0; k < 4; ++k ) { v10 = (int *)&v5[8 * k]; v11 = *v10; v12 = v10[1]; v13 = 0; v14 = -2009038745; for ( m = 0; m > 5) ^ (16 * v12))); v12 += (v7[(v13 >> 11) & 3] + v13) ^ (v11 + ((v11 >> 5) ^ (16 * v11))); v13 += v14; } *v10 = v11; v10[1] = v12; } for ( n = 0; n < 32; ++n ) { if ( v5[n] != byte_140014000[n] ) { v6 = 1; break; } } if ( v6 == 1 ) { sub_14000123A("bad, ji~ji~ji\n"); } else if ( !v6 ) { sub_14000123A("yesssss \n"); } return sub_1400013C5(v3, &unk_140010160);}
XTEA加密
#include #include void decrypt(unsigned int* v, unsigned int* key,unsigned int round) { unsigned int l = v[0], r = v[1], sum = 0, delta = 0x88408067; sum = delta * round; for (size_t i = 0; i < round; i++) { sum -= delta; r -= (((l <> 5)) + l) ^ (sum + key[(sum >> 11) & 3]); l -= (((r <> 5)) + r) ^ (sum + key[sum & 3])^sum; } v[0] = l; v[1] = r;}int main(){ unsigned int v[11] = { 0xADD4F778, 0xA6D7F132, 0x61813290, 0x2D4A40A6, 0x00B05F11, 0xB6D59424, 0x231BBFC6, 0xCD405B31, 0x03020100, 0x00C30504}; unsigned int key[4] = {98,111,109,98}; for(int i=0;i<4;i++){ decrypt(v+i*2,key,33); } for(int i=0;i<8;i++) { printf("%c%c%c%c",*((char*)&v[i]+0),*((char*)&v[i]+1),*((char*)&v[i]+2),*((char*)&v[i]+3)); } return 0;}//2d326e43eb8fea8837737fc0f50f83f2flag{2d326e43eb8fea8837737fc0f50f83f2}
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