Q: sys_futex() && timespec_valid()

From: Oleg Nesterov
Date: Fri Jun 25 2010 - 15:23:00 EST


Another stupid question about the trivial problem I am going to ask,
just to report the authoritative answer back to bugzilla. The problem
is, personally I am not sure we should/can add the user-visible change
required by glibc maintainers, and I am in no position to suggest them
to fix the user-space code instead.

In short, glibc developers believe that sys_futex(ts) is buggy and
needs the fix to return -ETIMEDOUT instead of -EINVAL in case when
ts->tv_sec < 0 and the timeout is absolute.

Ignoring the possible cleanups/microoptimizations, something like this:

--- x/kernel/futex.c
+++ x/kernel/futex.c
@@ -2625,6 +2625,16 @@ SYSCALL_DEFINE6(futex, u32 __user *, uad
if (copy_from_user(&ts, utime, sizeof(ts)) != 0)
return -EFAULT;
+ // absolute timeout
+ if (cmd != FUTEX_WAIT) {
+ if (ts->tv_nsec >= NSEC_PER_SEC)
+ return -EINVAL;
+ if (ts->tv_sec < 0)
+ return -ETIMEDOUT;
+ }
if (!timespec_valid(&ts))
return -EINVAL;


Otherwise, pthread_rwlock_timedwrlock(ts) hangs spinning in user-space
forever if ts->tv_sec < 0.

To clarify: this depends on libc version and arch.

This happens because pthread_rwlock_timedwrlock(rwlock, ts) on x86_64
roughly does:

for (;;) {
if (fast_path_succeeds(rwlock))
return 0;

if (ts->tv_nsec >= NSEC_PER_SEC)
return EINVAL;

errcode = sys_futex(FUTEX_WAIT_BITSET_PRIVATE, ts);
if (errcode == ETIMEDOUT)

and since the kernel return EINVAL due to !timespec_valid(ts), the
code above loops forever.

(btw, we have same problem with EFAULT, and this is considered as
a caller's problem).

IOW, pthread_rwlock_timedwrlock() assumes that in this case
sys_futex() can return nothing interesting except 0 or ETIMEDOUT.
I guess pthread_rwlock_timedwrlock() is not alone, but I didn't check.

So, the question: do you think we can change sys_futex() to make
glibc happy?

Or, do you think it is user-space who should check tv_sec < 0 if
it wants ETIMEDOUT with the negative timeout ?



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