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[2023]VMP还原day3:模式匹配寻找VIP/VSP和Flow Entry

2023-02-20 huoji 0 条评论

[2022]不使用hypervisor接管windows内核异常

不使用hypervisor的情况下接管内核异常:
how?
让我们检查一下内核处理流程:
触发异常后,在KdTrap中通过 KdpDebugRoutineSelect 来去进行异常处理选择(有调试器走1)
```cpp
__int64 __fastcall KdTrap(int a1, int a2, int a3, int a4, char a5, char a6)
{
  if ( KdpDebugRoutineSelect )
    return KdpTrap_0(a1, a2, a3, a4, a5, a6);
  else
    return KdpStub_0(a1, a2, a3, a4, a5, a6);
}
```
在异常不等于STATUS_BREAKPOINT的情况下,也就是不是int3中断的情况下,走到kdreport
```cpp
char __fastcall KdpTrap_0(
        __int64 trapInfo,
        __int64 commandString,
        __int64 commandType,
        __int64 debuggerContext,
        char response)
{
  char v5; // dl
  _QWORD *contextPointer; // rbx
  unsigned __int64 addressValue; // rax
  __int64 previousAddressValue; // rdi
  int subtractionResult1; // eax
  int subtractionResult2; // eax
  int subtractionResult3; // eax
  int subtractionResult4; // eax
  int printFunctionReturnValue; // eax
  __int64 currentAddressValue; // rax

v5 = 0;
  contextPointer = (_QWORD *)debuggerContext;
  if ( *(_DWORD *)commandType != 0x80000003 )
    return KdpReport(trapInfo, 0, commandType, debuggerContext);
```
只要不是 fast error或者单步异常,走到KdExitDebugger 里面
```cpp
  if ( *a3 == -1073740768
    || (unsigned int)(v6 + 2147483645) <= 1
    || (unsigned int)(v6 - 1073741854) <= 1
    || (NtGlobalFlag & 1) != 0 )
  {
    v9 = a6;
    if ( a6 || (unsigned int)v6 > 0x4000001D && v6 != 0xC0000037 && v6 < 0x40000020 )
      goto LABEL_6;
  }
  else
  {
    v9 = a6;
    if ( a6 )
    {
LABEL_6:
      v10 = ((__int64 (*)(void))KdEnterDebugger)();
      CurrentPrcb = KeGetCurrentPrcb();
      v12 = v10;
      KdpCopyContext(*(_QWORD *)&CurrentPrcb[1].PPPagedLookasideList[3].FreeMisses, *(unsigned int *)(a4 + 48), a4);
      KiSaveProcessorControlState((__int64)&CurrentPrcb->ProcessorState.SpecialRegisters.MsrLStar, v13);
      v14 = *(_QWORD *)&CurrentPrcb[1].PPPagedLookasideList[3].FreeMisses;
      LOBYTE(v15) = v9;
      v16 = *(_DWORD *)(v14 + 48);
      v17 = KdpReportExceptionStateChange((__int64)a3, v14, v15);
      KdpCopyContext(a4, v16, *(_QWORD *)&CurrentPrcb[1].PPPagedLookasideList[3].FreeMisses);
      KiRestoreProcessorControlState(&CurrentPrcb->ProcessorState.SpecialRegisters.MsrLStar);
      LOBYTE(v18) = v12;
      KdExitDebugger_0(v18);
      result = v17;
      KdpControlCPressed_0 = 0;
      return result;
    }
  }
```

KdExitDebugger在处理完一些必要的状态后,走到 KeThawExecution_0(v6);里面
而PoAllProcIntrDisabled_0只要为0 就会调用我们的老朋友 **KeQueryPerformanceCounter**
```cpp
__int64 __fastcall KeThawExecution_0(char a1)
{
  char v2; // di
  __int64 v3; // rcx
  unsigned __int8 v4; // bl
  unsigned __int64 v5; // rcx
  unsigned __int64 v6; // rax
  struct _KPRCB *CurrentPrcb; // rcx
  __int64 result; // rax

v2 = 0;
  if ( (KiFreezeFlag & 8) == 0 )
    v2 = KdPortLocked;
  ((void (__fastcall *)(_QWORD, _QWORD))off_401A88[0])(0i64, 0i64);
  if ( !PoAllProcIntrDisabled_0 )
  {
    MEMORY[0xFFFFF78000000350] = KeQueryPerformanceCounter(0i64).QuadPart;
    KiInterruptTimeErrorAccumulator = 0i64;
  }
```
为什么说 **KeQueryPerformanceCounter**是我们的老朋友呢,因为这个hal层的吊东西已经被日烂了,从infinity hook到infinity hook V2. v3 v4 他已经被日烂了
这次被日的原因和infinity hook v4一模一样,是这一步出了问题:
```cpp
v2 = HalpPerformanceCounter;
  if ( *(_DWORD *)(HalpPerformanceCounter + 0xE4) == 5 )
  {
    v37 = 10000000i64;
    if ( HalpTimerReferencePage )
    {
      if ( (*(_DWORD *)(HalpPerformanceCounter + 224) & 0x10000) != 0 )
        v20 = *(_QWORD *)(HalpPerformanceCounter + 72)
            + (unsigned int)(*(_DWORD *)(HalpPerformanceCounter + 0x50) * HIDWORD(KeGetPcr()[1].LockArray));
      else
        v20 = *(_QWORD *)(HalpPerformanceCounter + 72);
      v21 = (*(__int64 (__fastcall **)(__int64))(HalpPerformanceCounter + 112))(v20);
      v22 = MEMORY[0xFFFFF780000003B8];
      v10.QuadPart = v22 + RtlUnsignedMultiplyHigh(v21, *((_QWORD *)HalpTimerReferencePage + 1));
    }
    else
    {
      do
      {
        v24 = *(_QWORD *)(v2 + 0xD0);
        do
        {
          v25 = *(_QWORD *)(v2 + 0xC8);
          InternalData = HalpTimerGetInternalData(v2);
          v27 = (*(__int64 (__fastcall **)(__int64))(v2 + 0x70))(InternalData);
          _InterlockedOr(v36, 0);
          v28 = *(_QWORD *)(v2 + 200);
        }
        while ( v25 != v28 );
      }
```
注意其中的
```cpp
v25 = *(_QWORD *)(v2 + 0xC8);
          InternalData = HalpTimerGetInternalData(v2);
          v27 = (*(__int64 (__fastcall **)(__int64))(v2 + 0x70))(InternalData);
```
v2+0x70指向的是HalpHvCounterQueryCounter
在最新的infinity hook中也是他出了毛病,没有被PG保护
因此在满足下列条件后,即可进行内核VEH操作:
```cpp
KdpDebugRoutineSelect = 1
PoAllProcIntrDisabled = 0
HalpPerformanceCounter + 0xDC != 0x40

```
**这会PG吗?**
不 他不会.别问咋知道的,某大型软件现在正在用它全局接管内核异常

完整调用栈:
```cpp
00 fffff803`3d12cfa8 fffff803`3d21ecf8     hal!HalpTimerGetInternalData
01 fffff803`3d12cfb0 fffff803`3d4fd046     hal!KeQueryPerformanceCounter+0x168
02 fffff803`3d12cff0 fffff803`3dabee65     nt!KeThawExecution+0x4a
03 fffff803`3d12d020 fffff803`3d4f4d34     nt!KdExitDebugger+0xb1
04 fffff803`3d12d050 fffff803`3dac2419     nt!KdpReport+0xd8
05 fffff803`3d12d090 fffff803`3d2b9b90     nt!KdpTrap+0x14d
06 fffff803`3d12d0e0 fffff803`3d2b97b0     nt!KdTrap+0x2c
07 fffff803`3d12d120 fffff803`3d453602     nt!KiDispatchException+0x1e0
08 fffff803`3d12d7d0 fffff803`3d44ccf0     nt!KiExceptionDispatch+0xc2
09 fffff803`3d12d9b0 fffff803`3d449471     nt!KiBreakpointTrap+0x370
0a fffff803`3d12db48 fffff803`3d4c2a7e     nt!DbgBreakPointWithStatus+0x1
0b fffff803`3d12db50 fffff803`3d4810a4     nt!KdCheckForDebugBreak+0xdbbe2
0c fffff803`3d12db80 fffff803`3d309b8d     nt!KeAccumulateTicks+0x1743e4
0d fffff803`3d12dbe0 fffff803`3d30ac73     nt!KiUpdateRunTime+0x5d
0e fffff803`3d12dc30 fffff803`3d21e883     nt!KeClockInterruptNotify+0x8f3
0f fffff803`3d12df40 fffff803`3d361195     hal!HalpTimerClockInterrupt+0x63
10 fffff803`3d12df70 fffff803`3d4420ea     nt!KiCallInterruptServiceRoutine+0xa5
11 fffff803`3d12dfb0 fffff803`3d4426d7     nt!KiInterruptSubDispatchNoLockNoEtw+0xea
12 fffff803`3d11f540 fffff803`3d238a0f     nt!KiInterruptDispatchNoLockNoEtw+0x37
13 fffff803`3d11f6d8 fffff803`3d413925     hal!HalProcessorIdle+0xf
14 fffff803`3d11f6e0 fffff803`3d30c3d8     nt!PpmIdleDefaultExecute+0x15
15 fffff803`3d11f710 fffff803`3d30bc05     nt!PpmIdleExecuteTransition+0x648
16 fffff803`3d11fb00 fffff803`3d44435c     nt!PoIdle+0x345
17 fffff803`3d11fc60 00000000`00000000     nt!KiIdleLoop+0x2c
```

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2022-12-21 huoji 2 条评论

系统安全二进制安全 C/C++一线开发