在Windows上的Python中打开设备的句柄 [英] Opening a handle to a device in Python on Windows

问题描述

我正在尝试使用Giveio.sys驱动程序，该驱动程序需要先打开文件"，然后才能访问受保护的内存.我正在看WinAVR/AVRdude中使用以下语法的C示例:

I'm trying to use the giveio.sys driver which requires a "file" to be opened before you can access protected memory. I'm looking at a C example from WinAVR/AVRdude that uses the syntax:

 #define DRIVERNAME      "\\\\.\\giveio"
 HANDLE h = CreateFile(DRIVERNAME,
            GENERIC_READ,
            0,
            NULL,
            OPEN_EXISTING,
            FILE_ATTRIBUTE_NORMAL,
            NULL);

但是这似乎在Python中不起作用-对于这两种情况，我都只会收到指定的路径无效"错误

but this does not seem to work in Python - I just get a "The specified path is invalid" error, for both

f = os.open("\\\\.\\giveio", os.O_RDONLY)

和

f = os.open("//./giveio", os.O_RDONLY)

这为什么不做同样的事情?

Why doesn't this do the same thing?

已编辑，希望可以减少思想混乱(感谢Will).我确实通过AVRdude随附的批处理文件验证了设备驱动程序是否正在运行.

Edited to hopefully reduce confusion of ideas (thanks Will). I did verify that the device driver is running via the batch files that come with AVRdude.

进一步编辑以阐明SamB的赏金.

Further edited to clarify SamB's bounty.

推荐答案

解决方案:在python中，您必须使用win32file.CreateFile()而不是open().感谢大家告诉我我想做什么，这有助于我找到答案！

Solution: in python you have to use win32file.CreateFile() instead of open(). Thanks everyone for telling me what I was trying to do, it helped me find the answer!

这篇关于在Windows上的Python中打开设备的句柄的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！