我有一个带有方法GetParentProcess(IntPtr handle)的结构,用于通过传递句柄返回父进程。
[StructLayout(LayoutKind.Sequential)]
public struct ParentProcessUtilities
{
internal IntPtr Reserved1;
internal IntPtr PebBaseAddress;
internal IntPtr Reserved2_0;
internal IntPtr Reserved2_1;
internal IntPtr UniqueProcessId;
internal IntPtr InheritedFromUniqueProcessId;
[DllImport("ntdll.dll")]
public static extern int NtQueryInformationProcess(IntPtr processHandle, int processInformationClass, ref ParentProcessUtilities processInformation, int processInformationLength, out int returnLength);
public static Process GetParentProcess()
{
return GetParentProcess(Process.GetCurrentProcess().Handle);
}
public Process GetParentProcess(int id)
{
Process process = Process.GetProcessById(id);
return GetParentProcess(process.Handle);
}
public static Process GetParentProcess(IntPtr handle)
{
ParentProcessUtilities pbi = new ParentProcessUtilities();
int returnLength;
int status = NtQueryInformationProcess(handle, 0, ref pbi, Marshal.SizeOf(pbi), out returnLength);
if (status != 0)
throw new Win32Exception(status);
try
{
return Process.GetProcessById(pbi.InheritedFromUniqueProcessId.ToInt32());
}
catch (ArgumentException)
{
return null;
}
}
}
一切正常,直到我进入进程“ smss.exe”。当我想成为该进程的父进程时,会抛出此异常
Unexpected exception : System.ComponentModel.Win32Exception (0x80004005): Access is denied
at System.Diagnostics.ProcessManager.OpenProcess(Int32 processId, Int32 access, Boolean throwIfExited)
at System.Diagnostics.Process.GetProcessHandle(Int32 access, Boolean throwIfExited)
at System.Diagnostics.Process.OpenProcessHandle(Int32 access)
at System.Diagnostics.Process.get_Handle()
我正在以管理员权限运行应用程序。感谢您的帮助
答案 0 :(得分:0)
public class foo {
private Map<String, Runnable> reflectivelyInitializedMap;
@ExternalApi("makePublic")
private MyObject bar = new MyObject() {
private Runnable makePublic;
}
// I want to generate the following method:
public void makePublic(){
reflectivelyInitializedMap.get("makePublic").run();
}
}
是会话管理器子系统,有关详细信息,请参见此Wikipedia article。这是Windows内核启动的第一个用户模式进程,它执行许多特权操作。因此,您的普通进程将无权管理此进程,因此“访问被拒绝”异常。