Re: [Libguestfs] [PATCH] v2v: virtio-win: include *.dll too
On 29 Oct 2015, at 10:28, Richard W.M. Jones
<rjones(a)redhat.com> wrote:
On Thu, Oct 29, 2015 at 11:05:42AM +1100, Vadim Rozenfeld wrote:
>> 1) tell which devices can be configured
> Not sure that I fully understated your question. but if you are going to
> create some sort of off-line automatic virtio-win drivers update
> utility, then it shouldn't be too complicated.
Roman is talking about what virt-v2v already does.
This is good information:
> First of all you will need
> to obtain the Windows kernel version by reading the following Registry
> key - "HKLM\SOFTWARE\Microsoft\Windows NT\CurrentVersion". Let's say
it
> 6.3, which means that it is Win8.1 or WS2012R2, parsing "BuildLabEx"
> string from the same hive will give you information about the platform
> bitness. Next you need to go through inf files, and find "DriverVer"
> string, like this one . taken from from vioscsi.inf
>
> DriverVer=08/01/2015,62.72.104.10800
>
> This string contains build time and version stamps. The version stamp
> looks as follow "62.72.104.10800"
> where 62 means a target Windows kernel version multiplied by 10. In this
> case it is 6.2 which means Win8 or WS2012
> 72 - the target host platform version multiplied bu 10 (was RHEL7.2)
> 104 just a magic number, but it can be changed, don't make any
> assumption based on this number.
> 10800 our internal build number (build 108) multiplied by 100
> If you found an inf file with the matching minor OS (6 in our case)
> version and matching or less but close minor version number (2 vs 3)
> then you are in the right directory.
Seems like looking at this field in the .inf file is better than our
current approach of matching path names.
You just need to remember that we have binary compatible versions of drivers. For example
virtio-net driver for Windows 2012 (Windows 8) (6.2) will be used for Windows 2012R2
(Windows 8.1) and Windows Server 2016 (Windows 10). So you are getting the lowest
compatible version of the OS from INF file.
Rich.
--
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