Cheat Engine 修改器获取内存数据心路历程， THREADSTACK0 C# .NET VB 获取基址 CE修改器

一、CE 找到地址

根据字符串找到地址。

找到结果，双击此行，会加入下面列表

右键点击选择:对这个地址进行指针扫描。

此处选择10级别指针，其他程序 区别对待。

参考此教程:
CE学习资料

经过多次查找可以找到如下结果

平常程序的 基址 都是以.exe或者 .dll 结尾，这里是 THREADSTACK0，表示是 线程0基地址。

二、获取THREADSTACK0 基址

获取基址：
C++项目程序
threadstack.exe 软件

三、修改C++代码，生成dll，使用的是VS2019 。

增加一个方法，作用是传入一个程序的 PID，和线程序号，程序返回线程的基址（10进制）。

int GetThreadStartAddress_SSY(int Pid, int intThreadNo) {
	std::string pid = std::to_string(Pid);
	DWORD dwProcID;
	std::stringstream stringstream(pid);
	stringstream >> std::dec >> dwProcID;
	if (!dwProcID) {
		std::cerr << pid << " is not a valid process id (PID)" << std::endl;
		return 0;
	}
	HANDLE hProcHandle = NULL;
	hProcHandle = OpenProcess(PROCESS_ALL_ACCESS, FALSE, dwProcID);
	std::vector<DWORD> threadId = threadList(dwProcID);
	int stackNum = 0;
	for (auto it = threadId.begin(); it != threadId.end(); ++it) {
		if (stackNum == intThreadNo)
		{
			HANDLE threadHandle = OpenThread(THREAD_GET_CONTEXT | THREAD_QUERY_INFORMATION, FALSE, *it);
			DWORD threadStartAddress = GetThreadStartAddress(hProcHandle, threadHandle);
			printf("TID: 0x%04x = THREADSTACK%2d BASE ADDRESS: 0x%04x\n", *it, stackNum, threadStartAddress);
			return threadStartAddress;
		}
		stackNum++;
	}
	return 0;
}
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添加可被调用函数:

extern "C" FUNHEAD int  GetThreadStartAddress_SSY(int Pid, int intThreadNo);
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修改项目为:动态库(.dll) ,点击F5生成，得到threadstack.dll。

三、VB.NET程序示范

1.获取线程基址
把 threadstack.dll 放入程序运行目录下。
dll下载地址
我会上传到CSDN，付积分下载（有钱真的可以为所欲为哦），当然也可以按照我的方法编译。

    <DllImport("threadstack.dll", CharSet:=CharSet.Unicode, SetLastError:=True, CallingConvention:=CallingConvention.Cdecl)>
    Private Shared Function GetThreadStartAddress_SSY(ByVal hProcess As ULong, ByVal hThread As ULong) As Integer
    End Function
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    Private Sub Button1_Click(sender As Object, e As EventArgs) Handles Button1.Click
        m_csTargetProcess = Process.GetProcesses().FirstOrDefault(Function(k) k.ProcessName = txt_ExeName.Text)
        If m_csTargetProcess Is Nothing Then
            MsgBox("未检索到程序!")
            Return
        End If
        Dim csBaseAddress As System.IntPtr = IntPtr.Zero '线程的基址
        csBaseAddress = GetThreadStartAddress_SSY(m_csTargetProcess.Id, 0)
   
   '下面还有代码
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2.根据基址+偏移获取最终地址。

'此代码接上面的
'获取最终地址
 csBaseAddress = csBaseAddress - &H384 '基址
 Dim hProcess As IntPtr = OpenProcess(&H1F0FFF, False, m_csTargetProcess.Id) '主窗体句柄
 csBaseAddress = hProcess.FindDMAAddy(csBaseAddress， New Integer() {&H28, &H4, &H8, &H120, &H10, &H14, &HC4, &H10, &H8, &H10})

 Dim Size As Integer = 20 '请求多少数据
        Dim array As Byte() = New Byte(Size - 1) {}
        Dim num As Integer '响应多少个数据
        ReadProcessMemory(hProcess, csBaseAddress, array, Size, num)
        Dim strTarge As String = System.Text.UnicodeEncoding.Unicode.GetString(array)
        MsgBox(strTarge)
        CloseHandle(hProcess)

    End Sub 

<DllImportAttribute("kernel32.dll", EntryPoint:="ReadProcessMemory")>
    Public Function ReadProcessMemory(hProcess As IntPtr,
                                             lpBaseAddress As IntPtr,
                                             <Out()> lpBuffer As Byte(),
                                             dwSize As Integer,
                                             <System.Runtime.InteropServices.OutAttribute()>
                                             ByRef lpNumberOfBytesRead As Integer) As <MarshalAs(UnmanagedType.Bool)> Boolean
    End Function

    <System.Runtime.CompilerServices.Extension>
    Public Function FindDMAAddy(ByVal hProc As IntPtr, ByVal ptr As IntPtr, ByVal offsets() As Integer) As IntPtr
        Dim buffer = New Byte(IntPtr.Size - 1) {}
        For Each i As Integer In offsets
            ReadProcessMemory(hProc, ptr, buffer, buffer.Length, 0)
            If IntPtr.Size = 4 Then
                ptr = IntPtr.Add(New IntPtr(BitConverter.ToInt32(buffer, 0)), i)
            Else
                ptr = IntPtr.Add(New IntPtr(BitConverter.ToInt64(buffer, 0)), i)
            End If
        Next i
        Return ptr
    End Function
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四、总结

第一次使用CE写项目，熬了几个大夜终于搞定了。我们遇到的问题刚好在我们的能力范围内，或许再坚持一下，我们就能把问题解决了。
瑞思拜。