gh-115103: Fix unregistering of QSBR state #116480
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If a thread blocks while waiting on the `shared->mutex` lock, the array of QSBR states may be reallocated. The `tstate->qsbr` values before the lock is acquired may not be the same as the value after the lock is acquired.
colesbury
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Mar 7, 2024
In general, when `_PyThreadState_GET()` is non-NULL then the current thread is "attached", but there is a small window during `PyThreadState_DeleteCurrent()` where that's not true: tstate_delete_common is called when the thread is detached, but before current_fast_clear(). This updates _PySemaphore_Wait() to handle that case.
adorilson
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Mar 25, 2024
If a thread blocks while waiting on the `shared->mutex` lock, the array of QSBR states may be reallocated. The `tstate->qsbr` values before the lock is acquired may not be the same as the value after the lock is acquired.
diegorusso
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Apr 17, 2024
If a thread blocks while waiting on the `shared->mutex` lock, the array of QSBR states may be reallocated. The `tstate->qsbr` values before the lock is acquired may not be the same as the value after the lock is acquired.
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If a thread blocks while waiting on the
shared->mutexlock, the array of QSBR states may be reallocated. In other words, the value oftstate->qsbrmay be different before and after thePyMutex_Lock()call, and using the previous value across the lock acquisition is not safe.