python - 如何将“文件”传递给需要 Python ctypes 中“缓冲区”的 cpp 函数？

Question

我想将“Open_Buffer_Continue”函数添加到 MediaInfoDLL.py ctypes 包装器中，绑定在这里，MediaInfoDLL.cs C# 绑定已经实现了这个函数，所以它是可能的。

我如何通过以下内容：

file = open('./file.avi', 'rb')

到 mediainfo cpp Open_Buffer_Continue期望：

(   const ZenLib::int8u *   Buffer,
    size_t                  Buffer_Size 
) 

?

这是我到目前为止所拥有的：

MediaInfo_Open_Buffer_Init = MediaInfoDLL_Handler.MediaInfo_Open_Buffer_Init
MediaInfo_Open_Buffer_Init.argtype = [c_size_t, c_size_t]
MediaInfo_Open_Buffer_Init.restype = None

MediaInfo_Open_Buffer_Continue = MediaInfoDLL_Handler.MediaInfo_Open_Buffer_Continue
MediaInfo_Open_Buffer_Continue.argtype = [c_size_t, c_size_t]  # NOT SURE HERE var 1
MediaInfo_Open_Buffer_Continue.restype = c_size_t

MediaInfo_Open_Buffer_Finalize = MediaInfoDLL_Handler.MediaInfo_Open_Buffer_Finalize
MediaInfo_Open_Buffer_Finalize.argtype = [c_void_p]
MediaInfo_Open_Buffer_Finalize.restype = None

Answer 1

我对此一无所知mediainfo，但它看起来像是ZenLib::int8u*一个指向字节的指针，所以最好使用的是：

MediaInfo_Open_Buffer_Continue.argtype = [c_char_p, c_size_t]

然后将您的文件读入一个字符串并传递它：

with open('./file.avi','rb') as f:
    data = f.read()
MediaInfo_Open_Buffer_Continue(data,len(data))