RE: User Level Threads implementation by Linux kernel
Hi Ravi,
Find my inline responses.
From: kernelnewbies-bou...@nl.linux.org
[mailto:kernelnewbies-bou...@nl.linux.org] On Behalf Of ravikumar
Sent: Tuesday, March 02, 2010 4:20 PM
To: Kernel Newbies
Subject: User Level Threads implementation by Linux kernel
Hi all,
Currently I'm going through book 'development of Linux kernel' by Robert
Love.
In the chapter 'Process Management' I read below sentence
"Linux has a unique implementation of threads: It does not differentiate
between threads and processes. To Linux, a thread is just a special kind
of process."
In kernel everything is a just task represented by task_struct
And from some other sources I understand that there is a one-one mapping
between threads created by library API and kernel threads.
Now my questions are :
1. I have Program and I've created two threads say T1 and T2 using
Posix Thread Library. see below
void* function2()
{
while(1)
{
printf("I'm Thread 2\n");
sleep(1);
}
}
void* function1()
{
printf("I'm Thread 1\n");
char buf[1024];
read(1,buf,1023); // Thread 1 will block here
}
int main(int argc,char *argv[])
{
pthread_t pid1,pid2;
pthread_create(&pid1,NULL,function1,NULL);
pthread_create(&pid2,NULL,function2,NULL);
while(1)
{
printf("I'm Main Thread\n");
sleep(1);
}
}
According my understanding there will be three tasks will be created in
the kernel after I ran this program. One is for main process say P and
other two for T1 and T2.
And T1 & T2 share the address space of P. Is my understanding is
correct?
2.Regarding the scheduling.
Upon running the above program , the output is:
I'm Main Thread
I'm Thread 1
I'm Thread 2
I'm Main Thread
I'm Thread 2
I'm Main Thread
I'm Thread 2
I'm Main Thread
I'm Thread 2
I'm Main Thread
I'm Thread 2
Though Thread 1 is blocked Main Thread and Thread 2 are still
running.
That means different user-level threads are scheduled by kernel
separately.Is this correct ?
As per my understanding, yes, for every user thread kernel creates a
separate task_struct. In kernel, scheduling entity is task_struct, so
each user thread will be scheduled independently and will be given its
own time slice for execution. Scheduling will only happen on the
task_struct in run queue (in this case only main thread and thread 2 ,
not thread 1).
If this is correct how time slices will be allocated?
Do please some body clarify my questions and also provide some good
links as references
Thank you all
Regards,
Ravikumar
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Re: User Level Threads implementation by Linux kernel
Sangman Kim wrote:
Hi Ravikumar,
**
And from some other sources I understand that there is a one-one
mapping between threads created by library API and kernel threads.
I think kernel threads here do not seem to indicate conventional
kernel threads. There are kernel thread which handle interrupts, and
there is user-context kernel code on behalf of user threads. I think
what you are mentioning here is the latter, which is not regarded as
kernel "thread," even though they are part of kernel.
Now my questions are :
1. I have Program and I've created two threads say T1 and T2
using Posix Thread Library. see below
*void* function2()
{
while(1)
{
printf("I'm Thread 2\n");
sleep(1);
}
}
void* function1()
{
printf("I'm Thread 1\n");
char buf[1024];
read(0,buf,1023); // Thread 1 will block here
}
int main(int argc,char *argv[])
{
pthread_t pid1,pid2;
pthread_create(&pid1,NULL,function1,NULL);
pthread_create(&pid2,NULL,function2,NULL);
while(1)
{
printf("I'm Main Thread\n");
sleep(1);
}
}
*
According my understanding there will be three tasks will be
created in the kernel after I ran this program. One is for main
process say P and other two for T1 and T2.
And T1 & T2 share the address space of P. Is my understanding is
correct?
Yeah, that's correct. pthread_create() calls create_thread
()
in glibc, and the clone flag contains CLONE_VM flag in calling do_clone().
2.Regarding the scheduling.
*read(0,buf,1023); // Thread 1 will block here
*The integer 1 for fd means reading standard output of itself, which
usually does not make sense.
Even reading from stdin should stall at this point because it is
synchronous read.
Yes your correct 1 is for STDOUT , but I just wanted to block that
thread and just wanted to know how threads will be scheduled.
Thank you for valuable answers. Could you/somebody please share how
user level thread scheduling will be done in the above example
Sangman
Thanks
Ravikumar
Re: BUG: scheduling while atomic
Hi Vikash... On Tue, Mar 2, 2010 at 1:14 PM, Vikash Kumar wrote: > Hi all, > > I am developing a raid system and I am getting following BUG when I > issue lots of parallel IO. > I am not able to figure out how to go about debugging this dump. It's > evident here that RAID_LAZY_WRITE thread is trying to > schedule while it has locked the CPU but I am not able to trace it > exactly how and why it's happening. > As I don't acquire any lock to issue IO to block layer. > > I enabled spin lock debugging and lock dependency checking so that I > could get detailed debug info. > Here is the whole dump which I got, please provide me tips, ideas on > how to analyze and debug using > this dump. IMHO there are many possibilities here. The best thing to do here is to check the related functions mentioned in the stack frames. Are they really don't schedule while holding any locks on any situations? Are they somekind of callbacks that must satisfy certain kind of rules? Maybe it's like signal handler in user space which must avoid reentrancy etc. Sorry, this tip might sounds blurry for now, but this is the best idea I can offer right now. PS: When you mention "developing RAID system", do you mean "developing RAID handling code"? Or simply implementing RAID 0/1/5/01/10/1+5/5+1/etc disk array? -- regards, Mulyadi Santosa Freelance Linux trainer and consultant blog: the-hydra.blogspot.com training: mulyaditraining.blogspot.com -- To unsubscribe from this list: send an email with "unsubscribe kernelnewbies" to ecar...@nl.linux.org Please read the FAQ at http://kernelnewbies.org/FAQ
Re: Allocating memory with spinlock held
On Thu, Mar 4, 2010 at 2:16 AM, Frederic Weisbecker wrote:
> On Tue, Mar 02, 2010 at 09:32:16AM +0530, Manish Katiyar wrote:
>> Hello all,
>>
>> Is the following code valid ???
>>
>> foo() {
>> ...
>> spin_lock(lock);
>> ptr = kzalloc(size, GFP_KERNEL);
>> spin_unlock(lock);
>> .
>> }
>
>
>
> You can't because GFP_KERNEL involves the fact
> kzalloc might sleep while looking for memory
> somewhere (swapping out pages to get some free
> rooms).
>
> And sleeping while holding a spinlock may
> lead to a deadlock:
>
> Task A holds spinlock in cpu0. Task A sleeps.
> Task B is scheduled in cpu0, tries to take the
> spinlock and then deadlock for ever.
>
> You can use GFP_ATOMIC instead, but it's usually
> not recommended as the GFP_ATOMIC pool is a pretty
> limited resource.
Hi,
Thanks for the clarification. That is what I had suspected too and
wanted to clarify.
>
> The best is to allocate before you take the spinlock.
Yes, that is what I ended coding up.
Thanks again -
Manish
>
>
--
Thanks -
Manish
==
[$\*.^ -- I miss being one of them
==
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Re: User Level Threads implementation by Linux kernel
Hi Ravikumar,
* *
> And from some other sources I understand that there is a one-one mapping
> between threads created by library API and kernel threads.
>
>
I think kernel threads here do not seem to indicate conventional kernel
threads. There are kernel thread which handle interrupts, and there is
user-context kernel code on behalf of user threads. I think what you are
mentioning here is the latter, which is not regarded as kernel "thread,"
even though they are part of kernel.
> Now my questions are :
> 1. I have Program and I've created two threads say T1 and T2 using Posix
> Thread Library. see below
> *void* function2()
> {
>while(1)
>{
> printf("I'm Thread 2\n");
> sleep(1);
>}
> }
> void* function1()
> {
> printf("I'm Thread 1\n");
> char buf[1024];
> read(1,buf,1023); // Thread 1 will block here
> }
> int main(int argc,char *argv[])
> {
>
> pthread_t pid1,pid2;
> pthread_create(&pid1,NULL,function1,NULL);
> pthread_create(&pid2,NULL,function2,NULL);
> while(1)
> {
> printf("I'm Main Thread\n");
> sleep(1);
> }
> }
>
> *
> According my understanding there will be three tasks will be created in the
> kernel after I ran this program. One is for main process say P and other two
> for T1 and T2.
> And T1 & T2 share the address space of P. Is my understanding is correct?
>
>
Yeah, that's correct. pthread_create() calls
create_thread()
in glibc, and the clone flag contains CLONE_VM flag in calling do_clone().
> 2.Regarding the scheduling.
>
*read(1,buf,1023); // Thread 1 will block here
*The integer 1 for fd means reading standard output of itself, which usually
does not make sense.
Even reading from stdin should stall at this point because it is synchronous
read.
Sangman
Re: Allocating memory with spinlock held
El Wed, Mar 03, 2010 at 11:14:35AM +0530 Manish Katiyar ha dit:
> Hello all,
>
> Is the following code valid ???
>
> foo() {
> ...
> spin_lock(lock);
> ptr = kzalloc(size, GFP_KERNEL);
> spin_unlock(lock);
> .
> }
>
>
> I guess the answer is no, but would like to confirm. Detailed/all
> possible reasons for not doing so are welcome. IIRC, I was getting a
> kernel crash with similar code with error message as "spinlock bug :
> unlocked on wrong CPU" which looked like it was getting locked on one
> cpu and due to memory allocation was getting unlocked from another
> cpu.
preemption is disabled when a spinlock is held. memory allocation can
cause the current process to be scheduled out if the requested amount
of memory isn't available immediately, but preemption is disabled due
to the spinlock.
if you *really* need to allocate memory while holding the spinlock you
must use GFP_ATOMIC instead of GFP_KERNEL, but remember that the
allocation can fail and thus you must check the return value
best regards
--
Matthias Kaehlcke
Embedded Linux Developer
Barcelona
For to be free is not merely to cast off
one's chains, but to live in a way that
respects and enhances the freedom of others
(Nelson Mandela)
.''`.
using free software / Debian GNU/Linux | http://debian.org : :' :
`. `'`
gpg --keyserver pgp.mit.edu --recv-keys 47D8E5D4 `-
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To unsubscribe from this list: send an email with
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Please read the FAQ at http://kernelnewbies.org/FAQ
User Level Threads implementation by Linux kernel
Hi all,
Currently I'm going through book 'development of Linux kernel' by Robert
Love.
In the chapter 'Process Management' I read below sentence
"Linux has a unique implementation of threads: It does not differentiate
between threads and processes. To Linux, a thread is just a special kind
of process."
In kernel everything is a just task represented by *task_struct
*
And from some other sources I understand that there is a one-one mapping
between threads created by library API and kernel threads.
Now my questions are :
1. I have Program and I've created two threads say T1 and T2 using
Posix Thread Library. see below
*void* function2()
{
while(1)
{
printf("I'm Thread 2\n");
sleep(1);
}
}
void* function1()
{
printf("I'm Thread 1\n");
char buf[1024];
read(1,buf,1023); // Thread 1 will block here
}
int main(int argc,char *argv[])
{
pthread_t pid1,pid2;
pthread_create(&pid1,NULL,function1,NULL);
pthread_create(&pid2,NULL,function2,NULL);
while(1)
{
printf("I'm Main Thread\n");
sleep(1);
}
}
*
According my understanding there will be three tasks will be created in
the kernel after I ran this program. One is for main process say P and
other two for T1 and T2.
And T1 & T2 share the address space of P. Is my understanding is correct?
2.Regarding the scheduling.
Upon running the above program , the output is:
*I'm Main Thread
I'm Thread 1
I'm Thread 2
I'm Main Thread
I'm Thread 2
I'm Main Thread
I'm Thread 2
I'm Main Thread
I'm Thread 2
I'm Main Thread
I'm Thread 2
*Though Thread 1 is blocked Main Thread and Thread 2 are still
running.
That means different user-level threads are scheduled by kernel
separately.Is this correct ?
If this is correct how time slices will be allocated?
Do please some body clarify my questions and also provide some good
links as references
Thank you all
Regards,
Ravikumar
Kernel tunnel driver and routing?
Greetings I put a question mark at the end of the subject line because I'm not sure if the problem I have is to do with routing or not... First of all, define the problem. I'm using the tun driver within an application that opens a tunnel and reads encapsulated data from the tunnel, munges it about (as applications do!!) and then sends the reply back down the tunnel. So far, I can open the tunnel and read data but the send data seems to just disappear. Looking at ifconfig I see the send packet count and the send byte counts increment but nothing traced on my wireshark monitor:-( I've so far tried 3 methods of handling the IP/IP encap data, all of which have problems - the tun driver is what I'd prefer. 1. a raw-ip socket opened with a protocol value of 4 (IPIP encap type). This receives and transmits the data OK but it appears that the socket is not an exclusive listener because the encap'ed packets result in ICMP messages being returned to the originator of the tunnel. 2. opening a socket and then doing an SIOCADDTUNNEL ioctl on it to create a tunnel. There are no failures but my application doesn't receive any data so it never sends anything!! ifconfig indicates RX packets and data count but no idea where they go! 3. using the tun device as per the example in tuntap.txt. This is the one that receives but doesn't transmit. I have notices that 'ip link' reports that the encap type is ppp but I haven't found a way of changing that yet. I've spent over 100 hours on this now, I seem to have read the iproute2 and iptables code a dozen times but still no further in understanding what is going on... -- Robin Gilks zl3rob/g8ecj Internet: g8...@gilks.orghttp://www.gilks.org -- To unsubscribe from this list: send an email with "unsubscribe kernelnewbies" to ecar...@nl.linux.org Please read the FAQ at http://kernelnewbies.org/FAQ
Allocating memory with spinlock held
Hello all,
Is the following code valid ???
foo() {
...
spin_lock(lock);
ptr = kzalloc(size, GFP_KERNEL);
spin_unlock(lock);
.
}
I guess the answer is no, but would like to confirm. Detailed/all
possible reasons for not doing so are welcome. IIRC, I was getting a
kernel crash with similar code with error message as "spinlock bug :
unlocked on wrong CPU" which looked like it was getting locked on one
cpu and due to memory allocation was getting unlocked from another
cpu.
--
Thanks -
Manish
==
[$\*.^ -- I miss being one of them
==
--
To unsubscribe from this list: send an email with
"unsubscribe kernelnewbies" to ecar...@nl.linux.org
Please read the FAQ at http://kernelnewbies.org/FAQ
User Level Threads implementation by Linux kernel
Hi all,
Currently I'm going through book 'development of Linux kernel' by Robert
Love.
In the chapter 'Process Management' I read below sentence
"Linux has a unique implementation of threads: It does not differentiate
between threads and processes. To Linux, a thread is just a special kind
of process."
In kernel everything is a just task represented by *task_struct
*
And from some other sources I understand that there is a one-one mapping
between threads created by library API and kernel threads.
Now my questions are :
1. I have Program and I've created two threads say T1 and T2 using
Posix Thread Library. see below
*void* function2()
{
while(1)
{
printf("I'm Thread 2\n");
sleep(1);
}
}
void* function1()
{
printf("I'm Thread 1\n");
char buf[1024];
read(1,buf,1023); // Thread 1 will block here
}
int main(int argc,char *argv[])
{
pthread_t pid1,pid2;
pthread_create(&pid1,NULL,function1,NULL);
pthread_create(&pid2,NULL,function2,NULL);
while(1)
{
printf("I'm Main Thread\n");
sleep(1);
}
}
*
According my understanding there will be three tasks will be created in
the kernel after I ran this program. One is for main process say P and
other two for T1 and T2.
And T1 & T2 share the address space of P. Is my understanding is correct?
2.Regarding the scheduling.
Upon running the above program , the output is:
*I'm Main Thread
I'm Thread 1
I'm Thread 2
I'm Main Thread
I'm Thread 2
I'm Main Thread
I'm Thread 2
I'm Main Thread
I'm Thread 2
I'm Main Thread
I'm Thread 2
*Though Thread 1 is blocked Main Thread and Thread 2 are still
running.
That means different user-level threads are scheduled by kernel
separately.Is this correct ?
If this is correct how time slices will be allocated?
Do please some body clarify my questions and also provide some good
links as references
Thank you all
Regards,
Ravikumar
BUG: scheduling while atomic
Hi all, I am developing a raid system and I am getting following BUG when I issue lots of parallel IO. I am not able to figure out how to go about debugging this dump. It's evident here that RAID_LAZY_WRITE thread is trying to schedule while it has locked the CPU but I am not able to trace it exactly how and why it's happening. As I don't acquire any lock to issue IO to block layer. I enabled spin lock debugging and lock dependency checking so that I could get detailed debug info. Here is the whole dump which I got, please provide me tips, ideas on how to analyze and debug using this dump. BUG: scheduling while atomic: RAID_LAZY_WRITE/102/0x0205 no locks held by RAID_LAZY_WRITE/102. Showing all locks held in the system: = Pid: 102, comm: RAID_LAZY_WRITE CPU: 0Not tainted (2.6.24 #274) PC is at add_preempt_count+0xa0/0xbc LR is at raid_write_stripe+0x388/0x544 pc : []lr : []psr: 8013 sp : cfc63ec4 ip : cfc63ed4 fp : cfc63ed0 r10: cf5ce118 r9 : 0001 r8 : 00b4 r7 : 0001 r6 : cf403cd8 r5 : cfc62000 r4 : 0002 r3 : 0001 r2 : 0001 r1 : cfc62000 r0 : 0001 Flags: Nzcv IRQs on FIQs on Mode SVC_32 ISA ARM Segment kernel Control: 397f Table: 0f558000 DAC: 0017 [] (show_regs+0x0/0x4c) from [] (__schedule_bug+0x54/0x68) r4:cfc9b0a0 [] (__schedule_bug+0x0/0x68) from [] (schedule+0x74/0x2f4) r5:cfc9b0a0 r4:c034d13c [] (schedule+0x0/0x2f4) from [] (io_schedule+0x30/0x54) [] (io_schedule+0x0/0x54) from [] (get_request_wait+0xac/0x140) r4: [] (get_request_wait+0x0/0x140) from [] (__make_request+0x478/0x544) [] (__make_request+0x0/0x544) from [] (generic_make_request+0x2a8/0x2f0) [] (generic_make_request+0x0/0x2f0) from [] (raid_submit_disk_io+0x348/0x37c) [] (raid_submit_disk_io+0x0/0x37c) from [] (perform_raid5_ops+0xdc8/0x1100) [] (perform_raid5_ops+0x0/0x1100) from [] (raid5_RCW_postXOR+0x438/0x4c4) [] (raid5_RCW_postXOR+0x0/0x4c4) from [] (raid5_xor_completed+0x2a4/0x3b0) [] (raid5_xor_completed+0x0/0x3b0) from [] (iop_adma_run_tx_complete_actions+0x64/0xe8) [] (iop_adma_run_tx_complete_actions+0x0/0xe8) from [] (__iop_adma_slot_cleanup+0x2f8/0x3c0) r8:c0352f24 r7: r6: r5: r4:cf4faa7c [] (__iop_adma_slot_cleanup+0x0/0x3c0) from [] (iop_adma_tasklet+0x10/0x14) [] (iop_adma_tasklet+0x0/0x14) from [] (run_timer_softirq+0x17c/0x208) [] (run_timer_softirq+0x0/0x208) from [] (__do_softirq+0x64/0xd0) [] (__do_softirq+0x0/0xd0) from [] (irq_exit+0x48/0x5c) [] (irq_exit+0x0/0x5c) from [] (__exception_text_start+0x48/0x60) [] (__exception_text_start+0x0/0x60) from [] (__irq_svc+0x50/0x7c) Exception stack(0xcfc63e7c to 0xcfc63ec4) 3e60:0001 3e80: cfc62000 0001 0001 0002 cfc62000 cf403cd8 0001 00b4 3ea0: 0001 cf5ce118 cfc63ed0 cfc63ed4 cfc63ec4 c015a83c c004b014 8013 3ec0: r6:0020 r5:cfc63eb0 r4: [] (add_preempt_count+0x0/0xbc) from [] (raid_write_stripe+0x388/0x544) [] (raid_write_stripe+0x0/0x544) from [] (raid_lazy_writer+0x4c4/0x6b8) [] (raid_lazy_writer+0x0/0x6b8) from [] (raid_thread_daemon+0x140/0x15c) [] (raid_thread_daemon+0x0/0x15c) from [] (kthread+0x58/0x90) [] (kthread+0x0/0x90) from [] (do_exit+0x0/0x764) r6: r5: r4: BUG: spinlock cpu recursion on CPU#0, RAID_DAEMON/101 lock: cf4faae0, .magic: dead4ead, .owner: RAID_LAZY_WRITE/102, .owner_cpu: 0 [] (dump_stack+0x0/0x14) from [] (spin_bug+0x90/0xa4) [] (spin_bug+0x0/0xa4) from [] (_raw_spin_lock+0x68/0x15c) r5:c01fb378 r4:cf4faae0 [] (_raw_spin_lock+0x0/0x15c) from [] (_spin_lock_bh+0x4c/0x54) [] (_spin_lock_bh+0x0/0x54) from [] (iop_adma_run_tx_complete_actions+0x38/0xe8) r5: r4:cf4faa7c [] (iop_adma_run_tx_complete_actions+0x0/0xe8) from [] (__iop_adma_slot_cleanup+0x2f8/0x3c0) r8:0001 r7: r6: r5: r4:cf4faa7c [] (__iop_adma_slot_cleanup+0x0/0x3c0) from [] (iop_adma_tasklet+0x10/0x14) [] (iop_adma_tasklet+0x0/0x14) from [] (tasklet_action+0x8c/0xe4) [] (tasklet_action+0x0/0xe4) from [] (__do_softirq+0x64/0xd0) r6:000a r5:0001 r4:c0351f44 [] (__do_softirq+0x0/0xd0) from [] (irq_exit+0x48/0x5c) [] (irq_exit+0x0/0x5c) from [] (__exception_text_start+0x48/0x60) [] (__exception_text_start+0x0/0x60) from [] (__irq_svc+0x50/0x7c) Exception stack(0xcfcb1e64 to 0xcfcb1eac) 1e60: cf652404 0001 0001 cf652404 4013 c0417bec 1e80: 7fff cea81d88 cf6523e0 cfcb1ec0 cfcb1eac cfcb1eac c0298d00 1ea0: c0298d04 6013 r6:0064 r5:cfcb1e98 r4: [] (_spin_unlock_irqrestore+0x0/0x54) from [] (raid_read_stripe+0x204/0x560) r5:0008 r4:cf652404 [] (raid_read_stripe+0x0/0x560) from [] (raid_daemon+0x1d4/0x1110) [] (raid_daemon+0x0/0x1110) from [] (raid_thread_daemon+0x140/0x15c) [] (raid_thread_daemon+0x0/0x15c) from [] (kthread+0x58/0x90) [] (kthread+0x0/0x90) from [] (do_ex
Re: Allocating memory with spinlock held
On Tue, Mar 02, 2010 at 09:32:16AM +0530, Manish Katiyar wrote:
> Hello all,
>
> Is the following code valid ???
>
> foo() {
> ...
> spin_lock(lock);
> ptr = kzalloc(size, GFP_KERNEL);
> spin_unlock(lock);
> .
> }
You can't because GFP_KERNEL involves the fact
kzalloc might sleep while looking for memory
somewhere (swapping out pages to get some free
rooms).
And sleeping while holding a spinlock may
lead to a deadlock:
Task A holds spinlock in cpu0. Task A sleeps.
Task B is scheduled in cpu0, tries to take the
spinlock and then deadlock for ever.
You can use GFP_ATOMIC instead, but it's usually
not recommended as the GFP_ATOMIC pool is a pretty
limited resource.
The best is to allocate before you take the spinlock.
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Allocating memory with spinlock held
Hello all,
Is the following code valid ???
foo() {
...
spin_lock(lock);
ptr = kzalloc(size, GFP_KERNEL);
spin_unlock(lock);
.
}
I guess the answer is no, but would like to confirm. Detailed/all
possible reasons for not doing so are welcome. IIRC, I was getting a
kernel crash with similar code with error message as "spinlock bug :
unlocked on wrong CPU" which looked like it was getting locked on one
cpu and due to memory allocation was getting unlocked from another
cpu.
--
Thanks -
Manish
==
[$\*.^ -- I miss being one of them
==
--
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Re: UTS Version
Hi... On Tue, Mar 2, 2010 at 10:15 PM, Arigead wrote: > I'm getting an error on the UTS Version. I've been searching on the > internet and there is a fix posted on a mailing list for this problem on > a 2.6.18 kernel but the fix, editing a file, does not correspond to the > file I have in my linux source tree for 2.6.33. I am not sure what UTS really means, but I have strong sense it is related to: CONFIG_LOCALVERSION= we usually see in .config file found in top directory of kernel source tree after you do initial make *config. I suggest try to edit that line and change into something like: CONFIG_LOCALVERSION="-custom" and see if it works. -- regards, Mulyadi Santosa Freelance Linux trainer and consultant blog: the-hydra.blogspot.com training: mulyaditraining.blogspot.com -- To unsubscribe from this list: send an email with "unsubscribe kernelnewbies" to ecar...@nl.linux.org Please read the FAQ at http://kernelnewbies.org/FAQ
