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* Re: Fast testing
@ 2009-04-14 14:44 Tomasz Chmielewski
  0 siblings, 0 replies; 16+ messages in thread
From: Tomasz Chmielewski @ 2009-04-14 14:44 UTC (permalink / raw)
  To: LKML

> On 4/14/09, Dragoslav Zaric <dragoslav.zaric.kd@gmail.com> wrote:
>> Hello,
>>
>> I have question regarding fast testing of some change in some kernel
>> source file. Since rebooting is a long process,

Use kexec to "reboot" faster.


> there must be way to
>> test some change faster [of course only if source file is not some boot
>> connected file, but maybe this is also doable].

Compile the things you want to test as modules, if possible.


-- 
Tomasz Chmielewski
http://wpkg.org


^ permalink raw reply	[flat|nested] 16+ messages in thread
* Re: Fast testing
@ 2009-04-15 12:31 devzero
  2009-04-15 12:49 ` Dragoslav Zaric
  0 siblings, 1 reply; 16+ messages in thread
From: devzero @ 2009-04-15 12:31 UTC (permalink / raw)
  To: linux-kernel; +Cc: dragoslav.zaric.kd

if your kernel changes are not driver/hardware specific and you want to test/develop more
generic features, then you should have a look at UserModeLinux. great for testing & debugging.
you donŽt need any bootloader to run UserModeLinux, you can start it from commandline 
just like any other process. Basically, it`s a Linux kernel ported to Linux API (instead of different ARCH)

regards
roland


List:       linux-kernel
Subject:    Re: Fast testing
From:       Dragoslav Zaric <dragoslav.zaric.kd () gmail ! com>
Date:       2009-04-14 16:44:43
Message-ID: 2d05c4580904140944s1dec7dfewf42862ac698bf669 () mail ! gmail ! com
[Download message RAW]

Well Arjan it looks that you have a nice laptop if that is your
reboot time :)

The idea about this is that when you have some idea and you
write code, doing reboot and stuff can make you forgot your idea.
If you write normal program you probably press "build" or type
"gcc" at least 10 times to see what is going on. I think that only
thing necessary is: system to boot one time and then you can
load image in ram memory as many times as you want. So I am
not sure if this is ok, but at least I will try.

But I am not saying that if this works, the changes are going to
crush or freeze system more often than produce any kind of result :)

For example, we can boot from GRUB command line, so this would
be something like falling back to GRUB command line [to that point]
and load [any] image in ram memory.

thanks,

Dragoslav Zaric

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^ permalink raw reply	[flat|nested] 16+ messages in thread
* Re: Fast testing
@ 2009-04-15 12:19 Tomasz Chmielewski
  2009-04-15 12:45 ` Tilman Schmidt
  0 siblings, 1 reply; 16+ messages in thread
From: Tomasz Chmielewski @ 2009-04-15 12:19 UTC (permalink / raw)
  To: LKML

> Wow. Nice machine. I wish I had one like that. And how long until
> the GUI is usable again and you can actually continue working?
> 
> For comparison: my 3.4 GHz Pentium D takes 30 secs for kernel start
> (ie. until it begins looking for the filesystems),

Including all BIOS routines?


>  2 mins until the graphical login screen appears, and
> 5 mins from logging in to the GUI being completely present.

I have an old 366 MHz / 192 MB machine which boots to GUI much faster.

Something's wrong with your setup (unless you start really lots of 
things or have some other, "uncommon" things).


-- 
Tomasz Chmielewski
http://wpkg.org



^ permalink raw reply	[flat|nested] 16+ messages in thread
* Fast testing
@ 2009-04-14  9:55 Dragoslav Zaric
  2009-04-14 13:36 ` Alan Jenkins
                   ` (2 more replies)
  0 siblings, 3 replies; 16+ messages in thread
From: Dragoslav Zaric @ 2009-04-14  9:55 UTC (permalink / raw)
  To: LKML, Cyrill Gorcunov

Hello,

I have question regarding fast testing of some change in some kernel
source file. Since rebooting is a long process, there must be way to
test some change faster [of course only if source file is not some boot
connected file, but maybe this is also doable].

This is question:

When system is running, file that drives the system is "vmlinux" executable
and it is located in "/" or "/boot" folder. Let's say you change some kernel
source file and run "make" [this is fast since "make" should see that just few
files are changed]. After "make install" you should now have new "vmlinux" file.
Since state of cpu registers and memory locations determine current
system state, the change in binary system driver ["vmlinux"] should not
freeze system ? So if I copy new vmlinux over old vmlinux file can I see
change immediately ? I think it should continue to drive system without
problem because vmlinux is binary algorithm file, it does not contain any
data regarding current system state ? Also this assumes that your change does
not affect some large things, like memory segmentation tables and paging,
let's say it is more algorithmic and not structure change, because it is obvious
if you have structural change, current memory and register states
probably won't map
into new structure without system freeze.

Thanks,

Dragoslav Zaric,

[Programmer, M.Sc Astrophysics]

^ permalink raw reply	[flat|nested] 16+ messages in thread

end of thread, other threads:[~2009-04-16 15:19 UTC | newest]

Thread overview: 16+ messages (download: mbox.gz follow: Atom feed
-- links below jump to the message on this page --
2009-04-14 14:44 Fast testing Tomasz Chmielewski
  -- strict thread matches above, loose matches on Subject: below --
2009-04-15 12:31 devzero
2009-04-15 12:49 ` Dragoslav Zaric
2009-04-15 12:19 Tomasz Chmielewski
2009-04-15 12:45 ` Tilman Schmidt
2009-04-16  0:47   ` Robert Hancock
2009-04-16 15:19   ` Valdis.Kletnieks
2009-04-14  9:55 Dragoslav Zaric
2009-04-14 13:36 ` Alan Jenkins
2009-04-14 14:10   ` Dragoslav Zaric
2009-04-14 14:24   ` Dragoslav Zaric
2009-04-14 14:57     ` Cyrill Gorcunov
2009-04-14 13:53 ` Andi Kleen
2009-04-14 15:05 ` Arjan van de Ven
2009-04-14 16:44   ` Dragoslav Zaric
2009-04-15 11:48   ` Tilman Schmidt

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