3.1要简单,爆破dll文件即可!
00454BCC E88BFBFFFF call 0045475C
00454BD1 85C0 test eax, eax
00454BD3 7527 jnz 00454BFC
00454BD5 6A40 push $40
主要是想玩玩线程插入,没加自校验
Var
hMutex: DWORD;
isModify: boolean;
Begin
// 关闭Start.EXE进程
isModify := TheNodeP^.isModify; //Dll文件是否被修改
PostThreadMessage(TheNodeP^.MainThread, WM_QUIT, 0, 0);
// 是否已经插过线程
hMutex := OpenMutex(MUTEX_ALL_ACCESS, FALSE, DllMutex);
If (hMutex = 0) Then
Begin
hMutex := CreateMutex(Nil, TRUE, DllMutex);
// If isModify Then
// MessageBox(0, '文件状态正常 ……', '提示信息', MB_OK + MB_ICONINFORMATION)
// Else
// MessageBox(0, '文件被非法修改!', '提示信息', MB_OK + MB_ICONWARNING);
Application.Initialize;
Application.CreateForm(TForm1, Form1);
Application.Run;
ReleaseMutex(hMutex);
End;
CloseHandle(hMutex);
FreeLibraryAndExitThread(HInstance, 0);
End;
那就将以写好的注释去掉,重新编译上次一份试试!看看能不能再暴破。