Thanks! That perfectly answered my question.
-x
On 8/21/07, Kevin Hao <[EMAIL PROTECTED]> wrote:
> > I am curious how Linux convert an fd to the pathname? Does it
> > recursively walk back from current dentry to the root?
> Using d_path.
> > Can someone point me to the right place in the kernel
Thanks! That perfectly answered my question.
-x
On 8/21/07, Kevin Hao [EMAIL PROTECTED] wrote:
I am curious how Linux convert an fd to the pathname? Does it
recursively walk back from current dentry to the root?
Using d_path.
Can someone point me to the right place in the kernel where
Hi,
In the Linux proc filesystem, /proc/[pid]/fd is a link to the
actually the actual pathname of the opened file.
I am curious how Linux convert an fd to the pathname? Does it
recursively walk back from current dentry to the root?
Can someone point me to the right place in the kernel where
Hi,
In the Linux proc filesystem, /proc/[pid]/fd is a link to the
actually the actual pathname of the opened file.
I am curious how Linux convert an fd to the pathname? Does it
recursively walk back from current dentry to the root?
Can someone point me to the right place in the kernel where
to prevent a file from containing part
of old data and part of new data.
That's why I am so confused how can these systems provide consistent
snapshotting capability without sacrificing system performance much.
On 7/3/07, Chris Mason <[EMAIL PROTECTED]> wrote:
On Tue, 3 Jul 2007 01:28:57 -040
to prevent a file from containing part
of old data and part of new data.
That's why I am so confused how can these systems provide consistent
snapshotting capability without sacrificing system performance much.
On 7/3/07, Chris Mason [EMAIL PROTECTED] wrote:
On Tue, 3 Jul 2007 01:28:57 -0400
Xin Zhao
TED]> wrote:
It depends how snapshot is being taken. Usually it is Copy On Write. Google
for Copy on Write and you will find the answer for this.
On 7/3/07, Xin Zhao <[EMAIL PROTECTED]> wrote:
>
> Hi,
>
>
> If a file is already opened when snapshot command is issue
Hi,
If a file is already opened when snapshot command is issued, the file
itself could be in an inconsistent state already. Before the file is
closed, maybe part of the file contains old data, the rest contains
new data.
How does a versioning filesystem guarantee that the file snapshot is
in a
Hi,
If a file is already opened when snapshot command is issued, the file
itself could be in an inconsistent state already. Before the file is
closed, maybe part of the file contains old data, the rest contains
new data.
How does a versioning filesystem guarantee that the file snapshot is
in a
:
It depends how snapshot is being taken. Usually it is Copy On Write. Google
for Copy on Write and you will find the answer for this.
On 7/3/07, Xin Zhao [EMAIL PROTECTED] wrote:
Hi,
If a file is already opened when snapshot command is issued, the file
itself could be in an inconsistent
es (BTW ext3cow filesystem
may be interesting for you www.ext3cow.com).
Honza
> On 3/28/07, Dave Kleikamp <[EMAIL PROTECTED]> wrote:
> >On Wed, 2007-03-28 at 02:45 -0400, Xin Zhao wrote:
> >> Hi,
> >>
> >> If a L
).
Honza
On 3/28/07, Dave Kleikamp [EMAIL PROTECTED] wrote:
On Wed, 2007-03-28 at 02:45 -0400, Xin Zhao wrote:
Hi,
If a Linux process opens and reads a file A, then it closes the file.
Will Linux keep the file A's data in cache for a while in case another
process
You are right. If the device is very big, the radix tree could be huge
as well. Maybe the lookup it not that cheap. But the per-device tree
can be optimized too. A simple way I can immediately image is: evenly
split a device into N parts by the sector numbers. For each part, we
maintain a radix
On Wed, 2007-03-28 at 02:45 -0400, Xin Zhao wrote:
> Hi,
>
> If a Linux process opens and reads a file A, then it closes the file.
> Will Linux keep the file A's data in cache for a while in case another
> process opens and reads the same in a short time? I think that is what
&
, 2007-03-28 at 02:45 -0400, Xin Zhao wrote:
Hi,
If a Linux process opens and reads a file A, then it closes the file.
Will Linux keep the file A's data in cache for a while in case another
process opens and reads the same in a short time? I think that is what
I heard before.
Yes.
But after I
You are right. If the device is very big, the radix tree could be huge
as well. Maybe the lookup it not that cheap. But the per-device tree
can be optimized too. A simple way I can immediately image is: evenly
split a device into N parts by the sector numbers. For each part, we
maintain a radix
Hi,
If a Linux process opens and reads a file A, then it closes the file.
Will Linux keep the file A's data in cache for a while in case another
process opens and reads the same in a short time? I think that is what
I heard before.
But after I digged into the kernel code, I am confused.
When a
Hi,
If a Linux process opens and reads a file A, then it closes the file.
Will Linux keep the file A's data in cache for a while in case another
process opens and reads the same in a short time? I think that is what
I heard before.
But after I digged into the kernel code, I am confused.
When a
I read the code and found that a block buffer is not necessarily freed
even if the corresponding inode is released. Looks like block buffer
can stay around as long as the system still has free memory. Is my
understanding correct?
-x
On 3/9/07, Xin Zhao <[EMAIL PROTECTED]> wrote:
Hi,
Hi,
I am working on a file system that allow multiple files to share data
blocks. That is, a data block can be shared by two or more files. Now
my question is: suppose file A and B share the same data block D. Now
a process open file A and read block D, then this process closes file
A. If
Hi,
I am working on a file system that allow multiple files to share data
blocks. That is, a data block can be shared by two or more files. Now
my question is: suppose file A and B share the same data block D. Now
a process open file A and read block D, then this process closes file
A. If
I read the code and found that a block buffer is not necessarily freed
even if the corresponding inode is released. Looks like block buffer
can stay around as long as the system still has free memory. Is my
understanding correct?
-x
On 3/9/07, Xin Zhao [EMAIL PROTECTED] wrote:
Hi,
I am
EMAIL PROTECTED]> wrote:
On 2007-01-24, Xin Zhao wrote:
> Archived-At: <http://permalink.gmane.org/gmane.linux.kernel/485775>
Hallo.
> I just successfully called do_path_lookup() in my kernel module. I
> just removed the "fastcall" from the declaration of do_path_look
-01-24, Xin Zhao wrote:
Archived-At: http://permalink.gmane.org/gmane.linux.kernel/485775
Hallo.
I just successfully called do_path_lookup() in my kernel module. I
just removed the fastcall from the declaration of do_path_lookup(),
then the problem disappeared. I don't quite understand fastcall
I just successfully called do_path_lookup() in my kernel module. I
just removed the "fastcall" from the declaration of do_path_lookup(),
then the problem disappeared. I don't quite understand "fastcall"
though.
Can someone explain it?
Thanks,
-x
On 1/23/07, Xin Zhao <[
Hi,
I tried to call the following code in a kernel module:
error = do_path_lookup(AT_FDCWD, "/etc/profile.d/glib2.csh",
LOOKUP_PARENT, );
I exported the function do_path_lookup() using
"EXPORT_SYMBOL_GPL(do_path_lookup);"
But do_path_lookup() caused the "general protection fault:
Hi,
I tried to call the following code in a kernel module:
error = do_path_lookup(AT_FDCWD, /etc/profile.d/glib2.csh,
LOOKUP_PARENT, nd);
I exported the function do_path_lookup() using
EXPORT_SYMBOL_GPL(do_path_lookup);
But do_path_lookup() caused the general protection fault:
I just successfully called do_path_lookup() in my kernel module. I
just removed the fastcall from the declaration of do_path_lookup(),
then the problem disappeared. I don't quite understand fastcall
though.
Can someone explain it?
Thanks,
-x
On 1/23/07, Xin Zhao [EMAIL PROTECTED] wrote:
Hi
Sorry if this question is dumb.
SELinux is included in 2.6. But I think it works by putting LSM hooks a lot
of place in Linux and then it can define its own policy enforcement codes.
However, I cannot find hooks in kernel 2.6.9 and 2.6.11. How can
SELinux work with kernel 2.6 to protect system
Sorry if this question is dumb.
SELinux is included in 2.6. But I think it works by putting LSM hooks a lot
of place in Linux and then it can define its own policy enforcement codes.
However, I cannot find hooks in kernel 2.6.9 and 2.6.11. How can
SELinux work with kernel 2.6 to protect system
Hi,
I want to lock down a directory to be read-only, say, /etc, for system
security. Unfortunately, some valid system tools might need to
create/modified files like "/etc/dhclient-eth0.conf". To avoid
disrupting the normal running of those tools, I might have to allow
certain files to be created
Hi,
I want to lock down a directory to be read-only, say, /etc, for system
security. Unfortunately, some valid system tools might need to
create/modified files like /etc/dhclient-eth0.conf. To avoid
disrupting the normal running of those tools, I might have to allow
certain files to be created
Let me clarify with an example:
test.c:
**
open("abc.txt", O_RDONLY);
i ++;
Apparently, after the system call "open" in the kernel, the system
will return to the instruction "i++". My question is: can we find out
the instruction pointer in sys_open? If so, how?
Thanks!
xin
-
To
Let me clarify with an example:
test.c:
**
open(abc.txt, O_RDONLY);
i ++;
Apparently, after the system call open in the kernel, the system
will return to the instruction i++. My question is: can we find out
the instruction pointer in sys_open? If so, how?
Thanks!
xin
-
To unsubscribe
wrote:
> On Fri, Jul 29, 2005 at 05:00:20PM -0400, Xin Zhao wrote:
> > Thanks for your reply.
> >
> > Below is the code that print the kernel calling trace:
>
> Can I suggest just turning on frame pointers like I suggested?
>
> If you say Y here the resulti
that the show_trace does not scan and guess
the pointers. Instead, it use "previous_esp" to extract the esp and
thus the returning eip. Am I right?
Cheers,
xin
On 7/29/05, bert hubert <[EMAIL PROTECTED]> wrote:
> On Fri, Jul 29, 2005 at 04:27:16PM -0400, Xin Zhao wrote:
> >
I tried to use dump_stack to dump the calling trace in the kernel.
What I did is adding a dump_stack() call in the sys_open function.
Below is the dump out result:
Jul 28 17:33:31 normal kernel: [] sys_open+0xa6/0xb7
Jul 28 17:33:31 normal kernel: [] syscall_call+0x7/0xb
However, if I insert
I know in kernel level we have dump_stack, but is there anyway that we
can dump user stack in the kernel environment? I might want to dump
the calling trace in a system call function.
Thanks in advance!
-x
-
To unsubscribe from this list: send the line "unsubscribe linux-kernel" in
the body of
I know in kernel level we have dump_stack, but is there anyway that we
can dump user stack in the kernel environment? I might want to dump
the calling trace in a system call function.
Thanks in advance!
-x
-
To unsubscribe from this list: send the line unsubscribe linux-kernel in
the body of a
I tried to use dump_stack to dump the calling trace in the kernel.
What I did is adding a dump_stack() call in the sys_open function.
Below is the dump out result:
Jul 28 17:33:31 normal kernel: [c0151601] sys_open+0xa6/0xb7
Jul 28 17:33:31 normal kernel: [c0108e2c] syscall_call+0x7/0xb
. Instead, it use previous_esp to extract the esp and
thus the returning eip. Am I right?
Cheers,
xin
On 7/29/05, bert hubert [EMAIL PROTECTED] wrote:
On Fri, Jul 29, 2005 at 04:27:16PM -0400, Xin Zhao wrote:
I supprisely noticed that the dump_stack results are quite different!
Why did I get
:
On Fri, Jul 29, 2005 at 05:00:20PM -0400, Xin Zhao wrote:
Thanks for your reply.
Below is the code that print the kernel calling trace:
Can I suggest just turning on frame pointers like I suggested?
If you say Y here the resulting kernel image will be slightly larger
and slower
OK. I already read the codes. But an official document is always welcome. :)
On 7/21/05, Xin Zhao <[EMAIL PROTECTED]> wrote:
> I'd like to learn some details of how RPC's FSM works (e.g. the
> changing graph of all states). I know one possibility is to read the
> source code, but
OK. I already read the codes. But an official document is always welcome. :)
On 7/21/05, Xin Zhao [EMAIL PROTECTED] wrote:
I'd like to learn some details of how RPC's FSM works (e.g. the
changing graph of all states). I know one possibility is to read the
source code, but it is a little time
I'd like to learn some details of how RPC's FSM works (e.g. the
changing graph of all states). I know one possibility is to read the
source code, but it is a little time consuming.
Can someone kindly direct me to some documents about the rpc
implementatino? Thanks in advance!
-x
-
To
I'd like to learn some details of how RPC's FSM works (e.g. the
changing graph of all states). I know one possibility is to read the
source code, but it is a little time consuming.
Can someone kindly direct me to some documents about the rpc
implementatino? Thanks in advance!
-x
-
To
Sorry if this question is dumb.
I am trying to modify NFS to add some more features. I changed the
rpc_clnt struct in include/linux/sunrpc/clnt.h to add some more
fields:
vrpc_comm_t *cl_comm;
wait_queue_head_t cl_callwaitq[MAX_PENDING_REQS];
int
Sorry if this question is dumb.
I am trying to modify NFS to add some more features. I changed the
rpc_clnt struct in include/linux/sunrpc/clnt.h to add some more
fields:
vrpc_comm_t *cl_comm;
wait_queue_head_t cl_callwaitq[MAX_PENDING_REQS];
int
nfsd.ko.
How to do this? Thanks!
-x
On 7/5/05, Horst von Brand <[EMAIL PROTECTED]> wrote:
> Xin Zhao <[EMAIL PROTECTED]> wrote:
> > I tried to do "insmod nfsd.ko", but always got the error message
> > "insmod: error inserting 'nfsd.ko': -1 Unknown symbol in mo
I tried to do "insmod nfsd.ko", but always got the error message
"insmod: error inserting 'nfsd.ko': -1 Unknown symbol in module"
Why?
The kernel is 2.6.11.10
The command I used is:
1. insmod lockd.ko ---succeed
2. exportfs -r ---succeed
3. insmod nfsd.ko --- failed
Moreover, I noticed
Please discard the previous message.
I just noticed that something was wrong in the nfsd implementation
(slightly modified by me). After I fixed the problem, I can rmmod nfsd
now.
Sorry for the confusion.
-x
On 7/5/05, Xin Zhao <[EMAIL PROTECTED]> wrote:
> I compile kernel
Please discard the previous message.
I just noticed that something was wrong in the nfsd implementation
(slightly modified by me). After I fixed the problem, I can rmmod nfsd
now.
Sorry for the confusion.
-x
On 7/5/05, Xin Zhao [EMAIL PROTECTED] wrote:
I compile kernel 2.6.11.10 and configure
I tried to do insmod nfsd.ko, but always got the error message
insmod: error inserting 'nfsd.ko': -1 Unknown symbol in module
Why?
The kernel is 2.6.11.10
The command I used is:
1. insmod lockd.ko ---succeed
2. exportfs -r ---succeed
3. insmod nfsd.ko --- failed
Moreover, I noticed that if
nfsd.ko.
How to do this? Thanks!
-x
On 7/5/05, Horst von Brand [EMAIL PROTECTED] wrote:
Xin Zhao [EMAIL PROTECTED] wrote:
I tried to do insmod nfsd.ko, but always got the error message
insmod: error inserting 'nfsd.ko': -1 Unknown symbol in module
Use modprobe(8), it knows about module
Sorry for this dumb question.
I am trying to install Fedora Core 2 on a dell PowerEdge 2850 with
three 73GB SCSI disks on a RAID 4e/DI controller. I set it up as Raid
5. but when I tried to install FC2, it always complaint that no disk
drive can be found.
Can anybody give me some advice on how
We can certainly harden the system, but sometime the vulnerability in
kernel is hard to detect and protect. For example, the brk()
vulnerablitiy found in Linux kernel. All the security mechanisms you
mentioned have to rely on a healthy kernel. Unfortunately, the kernel
itself could be compromised
silly mistakes?
Xin
On 4/17/05, Willy Tarreau <[EMAIL PROTECTED]> wrote:
> On Sun, Apr 17, 2005 at 11:54:34AM -0400, Xin Zhao wrote:
> > Why not simply unset the write bit for all three groups of users?
> > That seems to be enough to prevent file modification.
> >
> >
Why not simply unset the write bit for all three groups of users?
That seems to be enough to prevent file modification.
Immutable seems to only add one more protection level in case of
misconfiguration on standard access right bits. Is that right?
-
To unsubscribe from this list: send the line
Why not simply unset the write bit for all three groups of users?
That seems to be enough to prevent file modification.
Immutable seems to only add one more protection level in case of
misconfiguration on standard access right bits. Is that right?
-
To unsubscribe from this list: send the line
silly mistakes?
Xin
On 4/17/05, Willy Tarreau [EMAIL PROTECTED] wrote:
On Sun, Apr 17, 2005 at 11:54:34AM -0400, Xin Zhao wrote:
Why not simply unset the write bit for all three groups of users?
That seems to be enough to prevent file modification.
Immutable seems to only add one more
We can certainly harden the system, but sometime the vulnerability in
kernel is hard to detect and protect. For example, the brk()
vulnerablitiy found in Linux kernel. All the security mechanisms you
mentioned have to rely on a healthy kernel. Unfortunately, the kernel
itself could be compromised
Sorry for this dumb question.
I am trying to install Fedora Core 2 on a dell PowerEdge 2850 with
three 73GB SCSI disks on a RAID 4e/DI controller. I set it up as Raid
5. but when I tried to install FC2, it always complaint that no disk
drive can be found.
Can anybody give me some advice on how
I have very very fast network and is testing NFS2 over this kind of
network. I noticed that for standard work like read/write a large
file, compile kernels, the performance of NFS2 is good. But if I try
to decompress kernel tar file. The standard ext2 takes 28s while NFS2
takes 81s. Also, if I
I have very very fast network and is testing NFS2 over this kind of
network. I noticed that for standard work like read/write a large
file, compile kernels, the performance of NFS2 is good. But if I try
to decompress kernel tar file. The standard ext2 takes 28s while NFS2
takes 81s. Also, if I
Sorry for the dumb question.
I am trying to develop a new filesystem based on NFS, which runs in a
very fast network environment. I used the source code of NFS2, but
noticed that NFS write is very slow. Even if I changed wsize to 8192,
it still can only reach 1MB/s. I don't know why. Because the
Sorry for the dumb question.
I am trying to develop a new filesystem based on NFS, which runs in a
very fast network environment. I used the source code of NFS2, but
noticed that NFS write is very slow. Even if I changed wsize to 8192,
it still can only reach 1MB/s. I don't know why. Because the
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