* Assorted questions
@ 2007-08-09 14:34 Matthew Booth
2007-08-09 21:07 ` Debora Velarde
2007-08-10 14:35 ` Steve Grubb
0 siblings, 2 replies; 3+ messages in thread
From: Matthew Booth @ 2007-08-09 14:34 UTC (permalink / raw)
To: linux-audit
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Questions relate to RHEL4 (unless they don't).
What are the meanings of the following fields from the SYSCALL record:
* items
* fsuid
* fsgid
What are the meanings of the following fields from the PATH record:
* flags
* rdev
How can I programmatically translate an architecture into human, eg
40000003 => 'i686'?
Is there a way of doing a syscall name lookup without having root?
In RHEL5, what's the equivalent of 'auditctl -t'?
Is there any master documentation I've missed? I'm only aware of the man
pages.
Thanks,
Matt
--
Matthew Booth, RHCA, RHCSS
Red Hat, Global Professional Services
M: +44 (0)7977 267231
GPG ID: D33C3490
GPG FPR: 3733 612D 2D05 5458 8A8A 1600 3441 EA19 D33C 3490
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^ permalink raw reply [flat|nested] 3+ messages in thread
* Re: Assorted questions
2007-08-09 14:34 Assorted questions Matthew Booth
@ 2007-08-09 21:07 ` Debora Velarde
2007-08-10 14:35 ` Steve Grubb
1 sibling, 0 replies; 3+ messages in thread
From: Debora Velarde @ 2007-08-09 21:07 UTC (permalink / raw)
To: Matthew Booth; +Cc: linux-audit
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linux-audit-bounces@redhat.com wrote on 08/09/2007 07:34:06 AM:
Hi Matt,
> Questions relate to RHEL4 (unless they don't).
>
> What are the meanings of the following fields from the SYSCALL record:
> * items
the number of path records in the event
> * fsuid
Filesystem User ID
> * fsgid
Filesystem Group ID
>
> What are the meanings of the following fields from the PATH record:
> * flags
file system namei flags
> * rdev
device identifier
>
> How can I programmatically translate an architecture into human, eg
> 40000003 => 'i686'?
When creating a rule with auditctl, you should be able to use either 'b32'
or 'b64' for the architecture.
If you're trying to read the audit log, ausearch has an option "-i" that
interprets numeric items into text. I'm not sure how well it works with
the arch fields, but might be worth a try.
>
> Is there a way of doing a syscall name lookup without having root?
Without root access, I'm not sure. You could probably find the syscall
table for your arch type online.
>
> In RHEL5, what's the equivalent of 'auditctl -t'?
Sorry I've forgotten what -t meant in auditctl.
>
> Is there any master documentation I've missed? I'm only aware of the man
> pages.
http://people.redhat.com/sgrubb/audit/
Hope that helps,
debora
----
Debora Velarde
Linux Security
IBM Linux Technology Center
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* Re: Assorted questions
2007-08-09 14:34 Assorted questions Matthew Booth
2007-08-09 21:07 ` Debora Velarde
@ 2007-08-10 14:35 ` Steve Grubb
1 sibling, 0 replies; 3+ messages in thread
From: Steve Grubb @ 2007-08-10 14:35 UTC (permalink / raw)
To: linux-audit
On Thursday 09 August 2007 10:34:06 Matthew Booth wrote:
> Questions relate to RHEL4 (unless they don't).
>
> How can I programmatically translate an architecture into human, eg
> 40000003 => 'i686'?
>From ausearch source code, ausearch-report.c:
static void print_arch(const char *val)
{
unsigned int ival;
const char *ptr;
errno = 0;
ival = strtoul(val, NULL, 16);
if (errno) {
printf("conversion error(%s) ", val);
return;
}
machine = audit_elf_to_machine(ival);
if (machine < 0) {
printf("unknown elf type(%s) ", val);
return;
}
ptr = audit_machine_to_name(machine);
printf("%s ", ptr);
}
> Is there a way of doing a syscall name lookup without having root?
You do not have to have root to call either of these functions:
extern int audit_name_to_syscall(const char *sc, int machine);
extern const char *audit_syscall_to_name(int sc, int machine);
> In RHEL5, what's the equivalent of 'auditctl -t'?
auditctl -t does not exist anymore.
-Steve
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2007-08-09 21:07 ` Debora Velarde
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