Python HANDLE 的 ctypes

Question

我有一个 DLL 函数，其原型如下：

short Connect(HANDLE* pHandle, UINT* pTimerID, LPCTSTR strDestination, LPCTSTR strServiceName, int nProtocol, int nType)

功能说明如下：

pHandle:  socket handle.[out]
pTimerID: Timer ID using sending alive packet to a camera.[out]
strDestination: camera network IP.[in]
strServiceName: camera network port.[in]
nProtocol: address family(AF_INET).[in]
nType: Socket type(SOCK_STREAM or SCOK_DGRAM).[in]

我尝试从 python 调用它 DLL，如下所示：

DLL = windll.LoadLibrary('myDll.dll')
self.UserID = 0
self.keepAliveID = 0
DLL.Connect.restype = ctypes.c_short
DLL.Connect.argtypes = [c_void_p, c_ulong, c_wchar_p, c_wchar_p, c_int, c_int]
try:
    err = DLL.Connect(self.UserID,self.keepAliveID,self.ip,self.port,2,1)
    if err == -1:
        print "connect ErrorCode = ", err
except:
    print "connect Exception = ", sys.exc_info()

在运行程序之后我得到了这个异常：

connect Exception =  (<type 'exceptions.WindowsError'>, WindowsError('exception: access violation writing 0x00000000',), <traceback object at 0x0262C300>)

我的错误在哪里？
C++ DLL中HANDLE的数据类型是什么，如何使用ctypes进行转换？

第二个问题是当我把argtypes改成如下：

DLL.Connect.argtypes = [c_void_p, POINTER(c_ulong), c_wchar_p, c_wchar_p, c_int, c_int]

然后我得到另一个异常：

connect Exception =  (<class 'ctypes.ArgumentError'>, ArgumentError("argument 2: <type 'exceptions.TypeError'>: expected LP_c_ulong instance instead of int",), <traceback object at 0x0261A328>)

请帮忙！

Answer 1

尝试像这样重新定义函数调用：

from ctypes import *
from ctypes.wintypes import *

dll = WinDLL('myDll.dll')
connect = dll.Connect
connect.restype = c_short
connect.argtypes = [POINTER(HANDLE), POINTER(UINT), LPCWSTR, LPCWSTR, c_int, c_int)

然后这样调用，使用byref通过引用传递：

self.user_id = HANDLE()
self.keep_alive_id = UINT()
err = connect(byref(self.user_id), byref(self.keep_alive_id), self.ip, self.port, 2, 1)

当然未经测试，但这至少应该让你更接近。

Python HANDLE 的 ctypes

Python ctypes for HANDLE

python

dll

ctypes