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In the Linux kernel, the following vulnerability has been resolved: iommu/vt-d: Fix potential lockup if qi_submit_sync called with 0 count If qi_submit_sync() is invoked with 0 invalidation descriptors (for instance, for DMA draining purposes), we can run into a bug where a submitting thread fails to detect the completion of invalidation_wait. Subsequently, this led to a soft lockup. Currently, there is no impact by this bug on the existing users because no callers are submitting invalidations with 0 descriptors. This fix will enable future users (such as DMA drain) calling qi_submit_sync() with 0 count. Suppose thread T1 invokes qi_submit_sync() with non-zero descriptors, while concurrently, thread T2 calls qi_submit_sync() with zero descriptors. Both threads then enter a while loop, waiting for their respective descriptors to complete. T1 detects its completion (i.e., T1's invalidation_wait status changes to QI_DONE by HW) and proceeds to call reclaim_free_desc() to reclaim all descriptors, potentially including adjacent ones of other threads that are also marked as QI_DONE. During this time, while T2 is waiting to acquire the qi->q_lock, the IOMMU hardware may complete the invalidation for T2, setting its status to QI_DONE. However, if T1's execution of reclaim_free_desc() frees T2's invalidation_wait descriptor and changes its status to QI_FREE, T2 will not observe the QI_DONE status for its invalidation_wait and will indefinitely remain stuck. This soft lockup does not occur when only non-zero descriptors are submitted.In such cases, invalidation descriptors are interspersed among wait descriptors with the status QI_IN_USE, acting as barriers. These barriers prevent the reclaim code from mistakenly freeing descriptors belonging to other submitters. Considered the following example timeline: T1 T2 ======================================== ID1 WD1 while(WD1!=QI_DONE) unlock lock WD1=QI_DONE* WD2 while(WD2!=QI_DONE) unlock lock WD1==QI_DONE? ID1=QI_DONE WD2=DONE* reclaim() ID1=FREE WD1=FREE WD2=FREE unlock soft lockup! T2 never sees QI_DONE in WD2 Where: ID = invalidation descriptor WD = wait descriptor * Written by hardware The root of the problem is that the descriptor status QI_DONE flag is used for two conflicting purposes: 1. signal a descriptor is ready for reclaim (to be freed) 2. signal by the hardware that a wait descriptor is complete The solution (in this patch) is state separation by using QI_FREE flag for #1. Once a thread's invalidation descriptors are complete, their status would be set to QI_FREE. The reclaim_free_desc() function would then only free descriptors marked as QI_FREE instead of those marked as QI_DONE. This change ensures that T2 (from the previous example) will correctly observe the completion of its invalidation_wait (marked as QI_DONE).
Rejected reason: This CVE ID has been rejected or withdrawn by its CVE Numbering Authority.
CVSS 3.1 Base Score 5.5. CVSS Attack Vector: local. CVSS Attack Complexity: low. CVSS Vector: (CVSS:3.1/AV:L/AC:L/PR:L/UI:N/S:U/C:N/I:N/A:H).
In the following Java snippet, methods are defined to get and set a long field in an instance of a class that is shared across multiple threads. Because operations on double and long are nonatomic in Java, concurrent access may cause unexpected behavior. Thus, all operations on long and double fields should be synchronized.
}return someLongValue;someLongValue = l;This code tries to obtain a lock for a file, then writes to it.
fclose($logFile);}//attempt to get logfile lockflock($logfile, LOCK_UN);// unlock logfileprint "Could not obtain lock on logFile.log, message not recorded\n";PHP by default will wait indefinitely until a file lock is released. If an attacker is able to obtain the file lock, this code will pause execution, possibly leading to denial of service for other users. Note that in this case, if an attacker can perform an flock() on the file, they may already have privileges to destroy the log file. However, this still impacts the execution of other programs that depend on flock().
The following function attempts to acquire a lock in order to perform operations on a shared resource.
}
pthread_mutex_unlock(mutex);/* access shared resource */However, the code does not check the value returned by pthread_mutex_lock() for errors. If pthread_mutex_lock() cannot acquire the mutex for any reason the function may introduce a race condition into the program and result in undefined behavior.
In order to avoid data races correctly written programs must check the result of thread synchronization functions and appropriately handle all errors, either by attempting to recover from them or reporting it to higher levels.
}
return pthread_mutex_unlock(mutex);return result;/* access shared resource */It may seem that the following bit of code achieves thread safety while avoiding unnecessary synchronization...
return helper;
}}helper = new Helper();The programmer wants to guarantee that only one Helper() object is ever allocated, but does not want to pay the cost of synchronization every time this code is called.
Suppose that helper is not initialized. Then, thread A sees that helper==null and enters the synchronized block and begins to execute:
helper = new Helper();If a second thread, thread B, takes over in the middle of this call and helper has not finished running the constructor, then thread B may make calls on helper while its fields hold incorrect values.
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