如何在Python ctypes中将“文件”传递给需要“ Buffer”的cpp函数？

Question

我想将'Open_Buffer_Continue'函数添加到MediaInfoDLL.py ctypes包装器中，这里的绑定和MediaInfoDLL.cs C＃绑定已经实现了此功能，因此是可能的。

我如何通过以下内容：

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

到mediainfo cpp Open_Buffer_Continue，它期望：

(   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，但看起来好像ZenLib::int8u*是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]然后将文件读入字符串并将其传递：with open('./file.avi','rb') as f:    data = f.read()MediaInfo_Open_Buffer_Continue(data,len(data))