Re: Boolean algebra error in 2.6.24?

Rainer Weikusat <rweikusat@xxxxxxxxxxx> wrote in
news:87skyc83u4.fsf@xxxxxxxxxxxxxxxxx:

static int check_kill_permission(int sig, struct siginfo *info,
struct task_struct *t)
if (info == SEND_SIG_NOINFO || (!is_si_special(info) &&
SI_FROMUSER(info))) {
[snip]
return error;

int cap_task_kill(struct task_struct *p, struct siginfo *info,
int sig, u32 secid)
{
if (info != SEND_SIG_NOINFO && (is_si_special(info) ||
SI_FROMKERNEL(info)))
return 0;

Defining a ::= 'signal has info', b ::= 'info is special' and
c ::= 'sent from user', the first condition can be written as

!a || (!b && c) [1]

and the second as

a && (b || !c) [2]

[1] can be transformed as follows:

!(a && (b || !c)) [3, !!]

meaning [2] is actually the opposite of [1], IOW when check_ oks the
signal, cap_ will deny it.

Am I missing something here?

I think I agree that the two conditions being checked are exactly
opposite as you say. However, it appears to me that the resulting
action is also opposite. Defining your "(a && (b || !c))" [2] condition
as X, the first case (check_kill_permission) is doing essentially
"if (!X) deny"
while the second case (cap_task_kill) is doing
"if (X) allow"

That seems consistent to me.

GH
.