Re: kernel BUG in new r8188eu

From: Fabio M. De Francesco
Date: Mon Aug 02 2021 - 07:45:12 EST


Hi Greg,

On Monday, August 2, 2021 12:36:18 PM CEST Greg Kroah-Hartman wrote:
> On Sun, Aug 01, 2021 at 03:15:52PM -0500, Larry Finger wrote:
> > On 8/1/21 1:31 AM, Greg Kroah-Hartman wrote:
> > > I am worried that my "remove the wrapper" logic got something wrong
> > > here, so if you could test the revert of that, I would appreciate it.
> > >
> > > I think I need to go buy one of these devices so I can test cleanups
> > > locally...
> >
> > The bad commit was 9ff1cc4ef80e ("staging: r8188eu: remove rtw_buf_free()
> > function").
> >
> > Looking at it, the only difference between the original wrapper and the
new
> > code is that the wrapper zeros the len variable before the kfree() call,
but
> > making that change did not help. Reverting that patch restores the old
> > behavior.
>
> Ah, doh, that was my fault, sorry, that patch was incorrect (odd git id,
> don't know where that came from, it's a different id in my tree.) Let
> me revert this commit and then will redo it correctly.

The commit to revert should be c7e88ecbe328.

>
> thanks for finding this.
>
> > There are lots of these on Ebay from $2.47 up with free shipping!
>
> Any hints on the name for how to figure out which devices are supported
> by this driver?

The following link points to a device that seems to be supported by this
driver. It is just $4.99, but I'm not sure whether or not they ship to Europe
(I suppose you're here):

https://www.amazon.com/gp/product/B00L28AN88/ref=ox_sc_act_title_1

About this item:
* Chipset:RTL8188EU
* Standard: IEEE 802.11n,IEEE 802.11g, IEEE 802.11b
* Supports 64/128 bit WEP, WPA encryption

Otherwise, I think that the following uses the same chip as the above.
However, it costs a little more and, above all else, I'm not really sure if it
works with r8188eu, since they write that the chip is a Realtek RTL8188EUS
(please note that final "EUS"):

https://www.amazon.it/Asus-USB-N10-NANO-N150-compatibile/dp/B081758CWC/
ref=sr_1_1

Thanks,

Fabio


> thanks,
>
> greg k-h