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[Bug ports/15119] New: unnecessary busy loop in __lll_timedlock_wait on ARM
- From: "katsuki.uwatoko at toshiba dot co.jp" <sourceware-bugzilla at sourceware dot org>
- To: glibc-bugs at sourceware dot org
- Date: Fri, 08 Feb 2013 00:00:53 +0000
- Subject: [Bug ports/15119] New: unnecessary busy loop in __lll_timedlock_wait on ARM
- Auto-submitted: auto-generated
http://sourceware.org/bugzilla/show_bug.cgi?id=15119
Bug #: 15119
Summary: unnecessary busy loop in __lll_timedlock_wait on ARM
Product: glibc
Version: unspecified
Status: NEW
Severity: normal
Priority: P2
Component: ports
AssignedTo: unassigned@sourceware.org
ReportedBy: katsuki.uwatoko@toshiba.co.jp
CC: carlos@redhat.com, roland@gnu.org
Classification: Unclassified
Created attachment 6856
--> http://sourceware.org/bugzilla/attachment.cgi?id=6856
patch for Avoid unneecssary busy loop in __lll_timedlock_wait on ARM.
The following sequence causes unnecessary busy loop.
"A thread" gets the lock. (futex = 1)
"B thread" tries to get the lock, and has not called futex syscall yet. (futex
= 2)
"A thread" releases the lock (futex = 0)
"C thread" gets the lock, and does not call futex syscall because of futex=0
(futex = 1)
"B thread" calls futex syscall, and returns with an error.
Because futex syscall in Linux Kernel checks if the val is changed
from 2, which is the 3rd arg of the syscall at futex_wait_setup(),
but in this case futex is 1.
"B thread" tries to get the lock in userspace but cannot get it
because futex is not 0.
After all the thread keeps calling futex syscall
until "C thread" will release it (futex = 0) or timeout.
Therefore I think that the value should be set 2 in every loop
the same as __lll_lock_wait_private, and attached a patch for this issue.
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