CVE-2026-46274: io-wq: check that the predecessor is hashed in io_wq_remove_pending()

Vulnerability summary

CVE-2026-46274 is a high (CVSS 7.8) Use After Free vulnerability in linux. EPSS estimates a 0.14% chance of exploitation in the next 30 days, higher than 3.2% of scored vulnerabilities.

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Timeline

1. 08-06-2026
   PublishedVulnerability first disclosed.
2. 05-08-2026
   Last modifiedVulnerability information updated.

Description

In the Linux kernel, the following vulnerability has been resolved: io-wq: check that the predecessor is hashed in io_wq_remove_pending() io_wq_remove_pending() needs to fix up wq->hash_tail[] if the cancelled work was the tail of its hash bucket. When doing this, it checks whether the preceding entry in acct->work_list has the same hash value, but never checks that the predecessor is hashed at all. io_get_work_hash() is simply atomic_read(&work->flags) >> IO_WQ_HASH_SHIFT, and the hash bits are never set for non-hashed work, so it returns 0. Thus, when a hashed bucket-0 work is cancelled while a non-hashed work is its list predecessor, the check spuriously passes and a pointer to the non-hashed io_kiocb is stored in wq->hash_tail[0]. Because non-hashed work is dequeued via the fast path in io_get_next_work(), which never touches hash_tail[], the stale pointer is never cleared. Therefore, after the non-hashed io_kiocb completes and is freed back to req_cachep, wq->hash_tail[0] is a dangling pointer. The io_wq is per-task (tctx->io_wq) and survives ring open/close, so the dangling pointer persists for the lifetime of the task; the next hashed bucket-0 enqueue dereferences it in io_wq_insert_work() and wq_list_add_after() writes through freed memory. Add the missing io_wq_is_hashed() check so a non-hashed predecessor never inherits a hash_tail[] slot.

• CWE-416Use After FreeThe product reuses or references memory after it has been freed. At some point afterward, the memory may be allocated again and saved in another pointer, while the original pointer references a location somewhere within the new allocation. Any operations using the original pointer are no longer valid because the memory "belongs" to the code that operates on the new pointer.