I am trying to run examples from man fanotify.7 but fanotify_mark always
fail with errno = EFAULT.
fanotify_mark declaration is
SYSCALL_DEFINE5(fanotify_mark, int, fanotify_fd, unsigned int, flags,
__u64, mask, int, dfd,
const char __user *, pathname)
When
fanotify_mark(4, FAN_MARK_ADD | FAN_MARK_ONLYDIR,
FAN_CREATE | FAN_ONDIR, AT_FDCWD, 0xdeadc0de)
is called on kernel side I can see in do_syscall_32_irqs_on that CPU
context is
bx = 0x4 = 4
cx = 0x9 = FAN_MARK_ADD | FAN_MARK_ONLYDIR,
dx = 0x40000100 = FAN_CREATE | FAN_ONDIR
si = 0x0
di = 0xffffff9c = AT_FDCWD
bp = 0xdeadc0de
ax = 0xffffffda
orix_ax = 0x153
I am not sure if it is ok because third argument is uint64_t so if I
understand correctly mask should be divided into two registers (dx and
si).
But in fanotify_mark we get
fanotify_fd = 4 = bx
flags = 0x9 = cx
mask = 0x40000100 = dx
dfd = 0 = si
pathname = 0xffffff9c = di
I believe that correct order is
fanotify_fd = 4 = bx
flags = 0x9 = cx
mask = 0x40000100 = (si << 32) | dx
dfd = 0xffffff9c = di
pathname = 0xdeadc0de = bp
I think that we should call COMPAT version of fanotify_mark here
COMPAT_SYSCALL_DEFINE6(fanotify_mark,
int, fanotify_fd, unsigned int, flags,
__u32, mask0, __u32, mask1, int, dfd,
const char __user *, pathname)
or something wrong is with 64-bits arguments.
I am running Linux 5.9.2 i686 on Pentium III (Coppermine).
For tests I am using Debian sid on qemu with 5.9.2 and default kernel
from repositories.
Everything works fine on 5.5 and 5.4.
--
Paweł Jasiak
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