On Tue, 20 Dec 2005, Steven Rostedt wrote:
Ofcourse it is - but it the exact same kind of traversal you do in the
On Tue, 20 Dec 2005, Esben Nielsen wrote:
The same lock taken twice is just a special case of deadlocking. It would
be very hard to check for the general case in the futex code without
"fixing" the rt_mutex. Not that the rt_mutex code is broken - it just
doesn't handle deadlocks very well as it wasn't supposed to. But as the
futex indirectly exposes the rt_mutex to userspace it becomes a problem.
The only _hack_ I can see is to force all robust futex calls to go through
one global lock to prevent the futex deadlocks becomming rt_mutex
deadlocks which again can turn into spin-lock deadlocks.
I instead argue for altering the premisses for the rt_mutex such
they can handle deadlocks without turning them into spin-lock deadlocks
blocking the whole system. Then a futex deadlock will become a rt_mutex
deadlock which can be handled.
For the type of deadlock you are talking about is the following:
P1 -- grabs futex A (no system call)
P2 -- grabs futex B (no system call)
P1 -- tries to grab futex B (system call to block and boost P2)
But holds no other kernel rt_mutex!
P2 -- tries to grab futex A (system call to block and boost P1)
spinning deadlock here,
So, before P2 blocks on P1, can there be a circular check t see if this is
a deadlock. You don't need to worry about other kernel rt_mutexes, you
only need to worry about blocked process.
Is this feasible?
rt_mutex part to resolve the PI boosting. Thus by making the futex code do
this by it's own you essentially just move the complexity in there and
make the total more complex. Notice, that the futex code can't rely on
user-space flag - it can't be trusted for this (opening a local DOS
attack). So it has to rely on the rt_mutex structure to do this - and
therefore also the locks in there to protect against simultanious
unlocking.
-- Steve