紧急求教，用函数CreateFile打开 COM10以上的虚拟串口为什么会报错！？？？

比如虚拟出来 一对串口 COM3<==>COM10 ，但是在执行
CreateFile(portName,  // Specify port device: default "COM1"
          GENERIC_READ | GENERIC_WRITE,      
                     // Specify    mode that open device.
           0,        // the devide isn't shared.
        NULL,        // the object gets a default security.
    OPEN_EXISTING, // Specify which action to take on file.
    0,            // default.
    NULL);

这个标准函数打开COM 10以上串口时，所不能成功！

请教为什么，大家是怎么打开的！？

 
CreateFile() can be used to get a handle to a serial port. The "Win32
Programmer's Reference" entry for "CreateFile()" mentions that the share
mode must be 0, the create parameter must be OPEN_EXISTING, and the
template must be NULL.
 
CreateFile() is successful when you use "COM1" through "COM9" for the name
of the file; however, the message "INVALID_HANDLE_VALUE" is returned if you
use "COM10" or greater.
 
If the name of the port is .COM10, the correct way to specify the serial
port in a call to CreateFile() is as follows:
 
   CreateFile(
      "\\\\\\\\.\\\\COM10",     // address of name of the communications device
      fdwAccess,          // access (read-write) mode
      0,                  // share mode
      NULL,               // address of security descriptor
      OPEN_EXISTING,      // how to create
      0,                  // file attributes
      NULL                // handle of file with attributes to copy
   );
 
NOTES: This syntax also works for ports COM1 through COM9. Certain boards
will let you choose the port names yourself. This syntax works for those
names as well.




http://support.microsoft.com/?id=115831



[编辑 -  6/30/04 by  KMK]

谢谢！