CVE-2026-97603: In the Linux kernel, the following vulnerability has been resolved: idpf: disable DIM work before freeing q_vectors idpf never drains the Tx/Rx DIM works before freeing the memory they live in. tx_dim and rx_dim are embe

Testo acquisito dalla fonte

In the Linux kernel, the following vulnerability has been resolved: idpf: disable DIM work before freeing q_vectors idpf never drains the Tx/Rx DIM works before freeing the memory they live in. tx_dim and rx_dim are embedded in struct idpf_q_vector, they are queued from the NAPI poll via net_dim(), and idpf_vport_intr_rel() ends with kfree(rsrc->q_vectors). Nothing in the driver cancels them. idpf_tx_dim_work() and idpf_rx_dim_work() then run on freed memory: idpf_vport_intr_write_itr() writes the ITR register through q_vector->intr_reg.tx_itr / rx_itr, void __iomem pointers loaded out of the freed q_vector. No configuration is needed to get there -- IDPF_ITR_IS_DYNAMIC() is defined as (itr_mode) and idpf_vport_alloc() initialises both modes to IDPF_ITR_DYNAMIC. Draining after idpf_vport_intr_napi_dis_all() is not enough on its own. idpf_net_dim() is called from inside the "if (napi_complete_done(napi, work_done))" branch of the poll, and napi_complete_done() has already cleared NAPIF_STATE_SCHED by then. napi_disable_locked() waits only while (val & (NAPIF_STATE_SCHED | NAPIF_STATE_NPSVC)), so napi_disable() can return while the poll tail is still queueing the work, and a plain cancel_work_sync() would be re-armed behind the drain. Use disable_work_sync(): schedule_work() on a work with a non-zero disable count is dropped by clear_pending_if_disabled() before __queue_work() is reached. Move idpf_init_dim() to idpf_vport_intr_alloc() so the works are initialised on every path that can reach the drain -- the three "goto intr_deinit" sites between idpf_vport_intr_init() and idpf_vport_intr_ena() get there without the enable side having run. Nothing re-enables them: rsrc->q_vectors is freed on every exit from idpf_vport_open() and on every idpf_vport_stop(), so the count d