Re: pagetables used in interrupt context
Thank you for clearing my doubt. On Mon, Feb 20, 2012 at 8:39 PM, Dave Hylands wrote: > Hi Subin, > > On Mon, Feb 20, 2012 at 6:47 PM, subin gangadharan > wrote: >> Hi All, >> >> Please correct me if I am wrong. In linux each process will have its >> own page tables, so when a interrupt happens processor will switch to >> interrupt context >> and execute the proper handler. So my doubt, if this is the case, >> interrupt hanlder will be using the pagetables of the interrupted >> process or is there a separate page table for this. > > Yep - that's right. Conceptually you can imagine that the kernel page > tables are replicated in each process, so when the interrupt occurs, > the kernel mappings will always be in effect regardless of which task > is running. How this is actually achieved may vary from architecture > to architecture. > > -- > Dave Hylands > Shuswap, BC, Canada > http://www.davehylands.com -- With Regards Subin Gangadharan I am not afraid and I am also not afraid of being afraid. ___ Kernelnewbies mailing list Kernelnewbies@kernelnewbies.org http://lists.kernelnewbies.org/mailman/listinfo/kernelnewbies
Re: cgroup freezer
On 2012年02月22日 01:08, Mulyadi Santosa wrote: > Hi Bill :) > > On Tue, Feb 21, 2012 at 08:23, bill4carson wrote: >> Just one silly question, how could frozen task be moved from node A to >> C ? > > I forgot to mention that I haven't tried the freezer feature in Linux > kernel, but I got impression that is quite similar with so called > checkpointing. > > So assuming that it's indeed checkpointing, what you need to do is > just grab the checkpointed image, copy to another node and execute a > command to unfreeze it. Usually, the image could be similar with ELF > dump core, but it could be something else too. > Thanks for your information, I'll make some time to try what checkpoint does. -- I am a slow learner but I will keep trying to fight for my dreams! --bill ___ Kernelnewbies mailing list Kernelnewbies@kernelnewbies.org http://lists.kernelnewbies.org/mailman/listinfo/kernelnewbies
Re: cgroup freezer
Hi Bill :) On Tue, Feb 21, 2012 at 08:23, bill4carson wrote: > Just one silly question, how could frozen task be moved from node A to > C ? I forgot to mention that I haven't tried the freezer feature in Linux kernel, but I got impression that is quite similar with so called checkpointing. So assuming that it's indeed checkpointing, what you need to do is just grab the checkpointed image, copy to another node and execute a command to unfreeze it. Usually, the image could be similar with ELF dump core, but it could be something else too. -- regards, Mulyadi Santosa Freelance Linux trainer and consultant blog: the-hydra.blogspot.com training: mulyaditraining.blogspot.com ___ Kernelnewbies mailing list Kernelnewbies@kernelnewbies.org http://lists.kernelnewbies.org/mailman/listinfo/kernelnewbies
Re: Semaphore
Hi Santosh, On Tue, Feb 21, 2012 at 5:47 AM, SaNtosh kuLkarni wrote: > A certain kernel process has acquired a semaphore lock and meanwhile during > this period the process gets killed abruptly (and it has not released the > sem lock..//.sem down).and the other processes cannot acquire the lock, > what can be done in this case. Normally, the semaphore would be associated with a driver of some type. If the process opened the driver, then regardless of how the process dies, the file that was opened by the process will eventually be closed by the kernel and the driver's release function will get called. This gives you the opportunity to release the semaphore. The driver would be responsible for tracking the state of the semaphore. Fundamentally the driver is responsible for cleaning up all of the resources that it may be using, and a semaphore is just another resource, like memory. I'm assuming that the semaphore is one which is held across multiple calls into the kernel, otherwise you don't have an issue in the first place, unless there is a bug on the kernel side of things which actually caused the process to terminate. -- Dave Hylands Shuswap, BC, Canada http://www.davehylands.com ___ Kernelnewbies mailing list Kernelnewbies@kernelnewbies.org http://lists.kernelnewbies.org/mailman/listinfo/kernelnewbies
Re: The 2.6.32 Linux kernel patches
On Tue, Feb 21, 2012 at 01:15:48PM +0300, mzee mzee wrote: > Hi, I'm a newbie to the Linux kernel, working on Android msm kernel. Due > to the hack on kernel.org, there was re-organization of if > infrastructure, as a result I couldn't find the 2.6.32 ALL incremental > patches, although I found a few, after long poking around at: > > > http://www.kernel.org/pub/linux/kernel/v2.6/longterm/v2.6.32/incr/ > > > > > > > > The only ones there are from 27-28 to 55-56, where may find the rest? They aren't there, sorry, use the git tree to create them yourself if you really need them. sorry, greg k-h ___ Kernelnewbies mailing list Kernelnewbies@kernelnewbies.org http://lists.kernelnewbies.org/mailman/listinfo/kernelnewbies
Re: Semaphore
On Tue, Feb 21, 2012 at 3:47 PM, SaNtosh kuLkarni wrote: > A certain kernel process has acquired a semaphore lock and meanwhile during > this period > the process gets killed abruptly (and it has not released the sem > lock..//.sem down). > and the other processes cannot acquire the lock, what can be done in this > case. The only check I see is debug_check_no_locks_held() in kernel/exit.c file, do_exit(). But it seems to be relevant only for 'CONFIG_LOCKDEP'... Does anyone know the correct answer? Thanks, --- KostaZ ___ Kernelnewbies mailing list Kernelnewbies@kernelnewbies.org http://lists.kernelnewbies.org/mailman/listinfo/kernelnewbies
Semaphore
A certain kernel process has acquired a semaphore lock and meanwhile during this period the process gets killed abruptly (and it has not released the sem lock..//.sem down).and the other processes cannot acquire the lock, what can be done in this case. -- *Regards, Santosh Kulkarni* ___ Kernelnewbies mailing list Kernelnewbies@kernelnewbies.org http://lists.kernelnewbies.org/mailman/listinfo/kernelnewbies
Re:Re: How to clean up buffer or cache momory?
I know, but it can not allocate all the cache memory, so you will fail sometimes if the memory has not been reclaimed. 在 2012-02-17 15:13:18，"Wang Chao" 写道： >On 02/16/12 at 09:33pm, 黄小伟 wrote: >> I know this, but I want to reclaim dynamically. Say in a shout time, several >> seconds. >Why do want to do that? Does it have a bad influence on performance? >Kernel will reclaim these cache itself when your machine runs out of mm. >> >> >> >> >> >> 在 2012-02-16 21:21:11，"Manish Katiyar" 写道： >> >2012/2/16 黄小伟 : >> >> Hi >> >> I encounter a problem recentely. My system supports USB strorage and >> >> other network applications. >> >> But after I put a file to usb storage, it will consume some many >> >> momory. >> >> (buffer + cache). >> >> And these buffered memory can not be reclaimed(use free command). >> >> So I want to know how to reclaim the buffer memory in a short time. >> > >> >Google for ./sys/vm/drop_caches >> > >> >-- >> >Thanks - >> >Manish >> > >> >___ >> >Kernelnewbies mailing list >> >Kernelnewbies@kernelnewbies.org >> >http://lists.kernelnewbies.org/mailman/listinfo/kernelnewbies > >> ___ >> Kernelnewbies mailing list >> Kernelnewbies@kernelnewbies.org >> http://lists.kernelnewbies.org/mailman/listinfo/kernelnewbies > > >-- >Thanks, >Chao Wang > >___ >Kernelnewbies mailing list >Kernelnewbies@kernelnewbies.org >http://lists.kernelnewbies.org/mailman/listinfo/kernelnewbies ___ Kernelnewbies mailing list Kernelnewbies@kernelnewbies.org http://lists.kernelnewbies.org/mailman/listinfo/kernelnewbies
Re: How to figure out the byteorder only with one byte number?
On Die, 2012-02-21 at 20:30 +0800, Tao Jiang wrote: [...] > Now I know in the most modern machine there is no difference between BE and LE > at so called 'bit order' level. > Right? One main difference between *byte* order and *bit* order is: What are the means to address individual *bits*? a) Bit shift and masking as in "1 << bit-number": This has a mathematical background and - implicitly - the least-significant bit has - thus - the number 0. I can't even think of an insane reason (let alone a sane one) to break the "shift left is for unsigned numbers equivalent to doubling" property - apart from the fact that it is defined in that way by C - and all other languages I came across. And the same holds for all CPUs/assembler instruction sets b) use a bit-field as in "unsigned char b0:1, b1:1, b2:1, b3:1, b4:1, b5:1, b6:1, b7:1;": It is not defined by any C-standard and is - thus - up to the compiler, if b0 == (1 << 0) or b0 == (1 << 7) or anything else. c) bit-test/st/clr assembler instructions in the architecture: Go read *if* your CPU has such stuff and how it relates to the "bit-shift and mask" method. I would be greatly surprised if it is different (on i386, it is equal since ages BTW) mainly because it makes absolutely no sense. d) There is hardware with bit-addressable memory out there. Go read the manual and the same as c) I doubt that it is different even for really old machines Bernd -- Bernd Petrovitsch Email : be...@petrovitsch.priv.at LUGA : http://www.luga.at ___ Kernelnewbies mailing list Kernelnewbies@kernelnewbies.org http://lists.kernelnewbies.org/mailman/listinfo/kernelnewbies
Re: How to figure out the byteorder only with one byte number?
Hi: I think I'm clear now. What's I originally wanted to ask for is whether 'bit order' equals to 'byte order' And is there a method to find out the 'bit order' then find out the 'byte order' ? Now I know in the most modern machine there is no difference between BE and LE at so called 'bit order' level. Right? Thank you all. 2012/2/21 Sri Ram Vemulpali : > Guys, > > I was late to the party. But this whole discussion throughs me off. > When you say byte order, it applied when the width of data is more > than a byte, lets say our width is 4 bytes, a typical word length. > > Now how is that there will be byte order on a byte width data. Are you > talking about nibble order. > > When you talk byte order -- either little endian or big endian, we are > talking how is our data should be interpreted. Depending on order we > start reading data from left or right a byte at a time. > > So, I am confused on your discussions. Please clarify. > > Thanks, > Sri. > > On Mon, Feb 20, 2012 at 5:32 PM, THAI NGUYEN wrote: >> >> Just as an FYI, way back in the early '90s, Texas Instruments came out with >> a graphics processor (I believe the TMS340x0 praphics processor) that >> actually did do the little-ending and big-endian down to the bit level. >> >> >> >> From: Subramaniam Appadodharana >> To: Tao Jiang >> Cc: Graeme Russ ; Bernd Petrovitsch >> ; Peter Senna Tschudin ; >> kernelnewbies@kernelnewbies.org >> Sent: Monday, February 20, 2012 8:53:10 AM >> Subject: Re: How to figure out the byteorder only with one byte number? >> >> Hi Tao, >> >> >> On Mon, Feb 20, 2012 at 5:25 AM, Tao Jiang wrote: >>> Hi: >>> >>> Thank you all. >>> >>> Take a byte number 0b0001 for example >>> ^ ^ >>> high bit low bit >>> >>> I used to think in a LE machine it will be stored as 0b1000 low bit >>> first >>> >>> ^ ^ >>> >>> low bit high bit >>> >>> and in a BE machine will be 0b0001 high bit first >>> ^ ^ >>> high bit low bit >>> >>> not only the byteorder is different, but inside a byte is also different. >>> >>> But actually they are the same, right? >> yes they are same. In fact it is termed as 'byte' order not 'bit' >> order. Hope this helps. >>> Thank you. >>> >>> >>> >>> 2012/2/20 Graeme Russ : On 02/20/2012 01:24 AM, Bernd Petrovitsch wrote: > On Sun, 2012-02-19 at 20:08 +0800, Tao Jiang wrote: > [...] >> Is there some difference of the storge between BE and LE machine inside >> a byte? > > No. At least TTBOMK there exists no such hardware. Using SHL/SHR would tell you - SHL normally results in a multiply by 2, SHR a divide by 2. If the byte was little endian, the results would be visa-versa But I agree, I doubt there is any such hardware Regards, Graeme >>> >>> ___ >>> Kernelnewbies mailing list >>> Kernelnewbies@kernelnewbies.org >>> http://lists.kernelnewbies.org/mailman/listinfo/kernelnewbies >> >> ___ >> Kernelnewbies mailing list >> Kernelnewbies@kernelnewbies.org >> http://lists.kernelnewbies.org/mailman/listinfo/kernelnewbies >> >> >> >> ___ >> Kernelnewbies mailing list >> Kernelnewbies@kernelnewbies.org >> http://lists.kernelnewbies.org/mailman/listinfo/kernelnewbies >> > > > > -- > Regards, > Sri. ___ Kernelnewbies mailing list Kernelnewbies@kernelnewbies.org http://lists.kernelnewbies.org/mailman/listinfo/kernelnewbies
Fwd: Linux kernel : sysctl registering table under same proc name (path)
Hello there,
I am creating a sysctl kernel module to register 2 hierarchy under same
proc name. For e.g.
my_net
| |
net1 net2
| |
t1t2
So, I used register_sysctl_paths() API for registering the sysctl tables
net1 net2 under path name my_net.
Problem is when unregistering the tables dmesg shows a warning and call
trace.
# rmmod sysctl.ko
# dmesg
[ cut here ]
WARNING: at kernel/sysctl.c:2270 unregister_sysctl_table+0xb1/0x120()
(Tainted: PW )
Hardware name: VMware Virtual Platform
Modules linked in: sysctl(-)(U) ip6table_filter ip6_tables ebtable_nat
ebtables xt_CHECKSUM iptable_mangle ipt_MASQUERADE iptable_nat nf_nat
nf_conntrack_ipv4 nf_defrag_ipv4 xt_state nf_conntrack ipt_REJECT bridge
autofs4 sunrpc fcoe libfcoe libfc scsi_transport_fc scsi_tgt 8021q garp stp
llc cachefiles fscache ipv6 xt_physdev iptable_filter ip_tables dm_mirror
dm_region_hash dm_log uinput ppdev parport_pc parport e1000 pcnet32 mii
vmware_balloon sg i2c_piix4 i2c_core shpchp ext4 mbcache jbd2 vxspec(P)(U)
vxio(P)(U) vxdmp(P)(U) sd_mod crc_t10dif sr_mod cdrom mptspi mptscsih
mptbase scsi_transport_spi ata_generic pata_acpi ata_piix dm_mod [last
unloaded: sysctl]
Pid: 22463, comm: rmmod Tainted: PW
2.6.32-71.el6.x86_64 #1
Call Trace:
[] warn_slowpath_common+0x87/0xc0
[] warn_slowpath_null+0x1a/0x20
[] unregister_sysctl_table+0xb1/0x120
[] ? stop_machine_destroy+0x4c/0x50
[] ? stop_machine+0x4b/0x60
[] sysctl_module_exit+0x21/0x23 [sysctl]
[] sys_delete_module+0x1a6/0x280
[] ? syscall_trace_enter+0x1d8/0x1e0
[] tracesys+0xd9/0xde
---[ end trace fee2c1c8e41aeb36 ]---
Kindly, help me to resolve this warning.
Following is module.
#include
#include
#include
static struct ctl_table_header * test_sysctl_header1;
static struct ctl_table_header * test_sysctl_header2;
int value1 = 0;
int value2 = 1;
int min = 10;
int max = 20;
static ctl_table test_table1[] = {
{
.ctl_name = CTL_UNNUMBERED,
.procname = "value1",
.data = &value1,
.maxlen = sizeof(int),
.mode = 0644,
.proc_handler
= &proc_dointvec_minmax,
.strategy=
&sysctl_intvec,
.extra1 = &min,
.extra2 = &max
},
{
.ctl_name = CTL_UNNUMBERED,
.procname = "value2",
.data = &value2,
.maxlen = sizeof(int),
.mode = 0644,
.proc_handler
= &proc_dointvec_minmax,
.strategy=
&sysctl_intvec,
.extra1 = &min,
.extra2 = &max
},
{ .ctl_name = 0 }
};
static ctl_table test_table2[] = {
{
.ctl_name = CTL_UNNUMBERED,
.procname = "value1",
.data = &value1,
.maxlen = sizeof(int),
.mode = 0644,
.proc_handler
= &proc_dointvec_minmax,
.strategy=
&sysctl_intvec,
.extra1 = &min,
.extra2 = &max
},
{
.ctl_name = CTL_UNNUMBERED,
.procname = "value2",
.data = &value2,
.maxlen = sizeof(int),
.mode = 0644,
.proc_handler
= &proc_dointvec_minmax,
.strategy=
&sysctl_intvec,
.extra1 = &min,
.extra2 = &max
},
{ .ctl_name = 0 }
};
static ctl_table test_net_table1[] = {
{
.ctl_name = CTL_UNNUMBERED,
.procname = "test1",
.mode = 0555,
.child = test_table1
},
The 2.6.32 Linux kernel patches
Hi, I'm a newbie to the Linux kernel, working on Android msm kernel. Due to the hack on kernel.org, there was re-organization of if infrastructure, as a result I couldn't find the 2.6.32 ALL incremental patches, although I found a few, after long poking around at: > http://www.kernel.org/pub/linux/kernel/v2.6/longterm/v2.6.32/incr/ The only ones there are from 27-28 to 55-56, where may find the rest? Also I'm new to git, does any one have a good tutorial on this with focus on Linux kernel development? Thanks ___ Kernelnewbies mailing list Kernelnewbies@kernelnewbies.org http://lists.kernelnewbies.org/mailman/listinfo/kernelnewbies
Re: Re: does anyone know how to communicate between domUs on xen?
>Check the website of xen, they might have something. One which I knew was >xensocket, but I guess you would need to port it to latest xen kernel as it >was not part of it. There may be other mechanisms also. Thanks! But the truth is I've tried to search material on the official website of xen, but didn't get what i want. ___ Kernelnewbies mailing list Kernelnewbies@kernelnewbies.org http://lists.kernelnewbies.org/mailman/listinfo/kernelnewbies
