
python调用dll怎么返回多个值
1个回答
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多个返回值需要用list集合来解析。。
举例参考一下:
import ctypes
# Load DLL into memory.
hllDll = ctypes.WinDLL ("c:\\PComm\\ehlapi32.dll")
# Set up prototype and parameters for the desired function call.
# HLLAPI
hllApiProto = ctypes.WINFUNCTYPE (
ctypes.c_int, # Return type.
ctypes.c_void_p, # Parameters 1 ...
ctypes.c_void_p,
ctypes.c_void_p,
ctypes.c_void_p) # ... thru 4.
hllApiParams = (1, "p1", 0), (1, "p2", 0), (1, "p3",0), (1, "p4",0),
# Actually map the call ("HLLAPI(...)") to a Python name.
hllApi = hllApiProto (("HLLAPI", hllDll), hllApiParams)
# This is how you can actually call the DLL function.
# Set up the variables and call the Python name with them.
p1 = ctypes.c_int (1)
p2 = ctypes.c_char_p (sessionVar)
p3 = ctypes.c_int (1)
p4 = ctypes.c_int (0)
hllApi (ctypes.byref (p1), p2, ctypes.byref (p3), ctypes.byref (p4))
举例参考一下:
import ctypes
# Load DLL into memory.
hllDll = ctypes.WinDLL ("c:\\PComm\\ehlapi32.dll")
# Set up prototype and parameters for the desired function call.
# HLLAPI
hllApiProto = ctypes.WINFUNCTYPE (
ctypes.c_int, # Return type.
ctypes.c_void_p, # Parameters 1 ...
ctypes.c_void_p,
ctypes.c_void_p,
ctypes.c_void_p) # ... thru 4.
hllApiParams = (1, "p1", 0), (1, "p2", 0), (1, "p3",0), (1, "p4",0),
# Actually map the call ("HLLAPI(...)") to a Python name.
hllApi = hllApiProto (("HLLAPI", hllDll), hllApiParams)
# This is how you can actually call the DLL function.
# Set up the variables and call the Python name with them.
p1 = ctypes.c_int (1)
p2 = ctypes.c_char_p (sessionVar)
p3 = ctypes.c_int (1)
p4 = ctypes.c_int (0)
hllApi (ctypes.byref (p1), p2, ctypes.byref (p3), ctypes.byref (p4))
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