Control: reassign -1 libc6 Control: found libc6/2.13-38 Control: affects haskell-regex-compat
Hi,
Am Montag, den 11.03.2013, 11:53 -0400 schrieb Joey Hess:
> Joachim Breitner wrote:
> > I can reproduce it from within ghc’s address space using gdb:
> >
> > (gdb) call malloc(32)
> > $7 = 64943120
> > (gdb) call regcomp(64943120, ".", 0)
> > $8 = 0
> > (gdb) call regexec(64943120,"\242",0,0,0)
> > $9 = 1
> > (gdb) call regexec(64943120,"only_ascii",0,0,0)
> > $10 = 0
> >
> > And even from gdb while debugging “sleep”. So the behaviour is already
> > there in regexec, but for some reason it is not triggered from C code,
> > but only via some variants of FFI (GHC’s or gdb’s).
> >
> > I’ll leave it at that, as this is not really related to GHC or Haskell
> > any more.
>
> That's some deep dive!
>
> Sounds like a reassign to glibc is in order?
if you think so...
@glibc maintainers, here is the short story:
This code prints, as expected 0 (for regex matches):
#include <sys/types.h>
#include <regex.h>
#include <stdio.h>
main () {
regex_t r;
regcomp(&r, ".", 0);
char *s = "\242";
int i = regexec(&r, s, 0, NULL, 0);
printf("%d\n", i);
}
But in some circumstances, this does not work as expected. One such
circumstance is Haskell code doing this via the FFI, but also from gdb:
(gdb) call malloc(32)
$1 = 6332464
(gdb) call memset(6332464, 0, 32)
$3 = 6332464
(gdb) call regcomp(6332464, ".", 0)
$4 = 0
(gdb) call regexec(6332464, "\242",0,0,0)
$5 = 1
It fails if there are no ascii characters around:
(gdb) call regexec(6332464, "\242x",0,0,0)
$6 = 0
(gdb) call regexec(6332464, "\242\242",0,0,0)
$7 = 1
(gdb) call regexec(6332464, "only_ascii",0,0,0)
$8 = 0
Greetings,
Joachim
--
Joachim "nomeata" Breitner
Debian Developer
[email protected] | ICQ# 74513189 | GPG-Keyid: 4743206C
JID: [email protected] | http://people.debian.org/~nomeata
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