CVE-2024-50002

CVE-2024-50002 · static_call: Handle module init failure correctly in static_call_del_module()

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In the Linux kernel, the following vulnerability has been resolved: static_call: Handle module init failure correctly in static_call_del_module() Module insertion invokes static_call_add_module() to initialize the static calls in a module. static_call_add_module() invokes __static_call_init(), which allocates a struct static_call_mod to either encapsulate the built-in static call sites of the associated key into it so further modules can be added or to append the module to the module chain. If that allocation fails the function returns with an error code and the module core invokes static_call_del_module() to clean up eventually added static_call_mod entries. This works correctly, when all keys used by the module were converted over to a module chain before the failure. If not then static_call_del_module() causes a #GP as it blindly assumes that key::mods points to a valid struct static_call_mod. The problem is that key::mods is not a individual struct member of struct static_call_key, it's part of a union to save space: union { /* bit 0: 0 = mods, 1 = sites */ unsigned long type; struct static_call_mod *mods; struct static_call_site *sites; }; key::sites is a pointer to the list of built-in usage sites of the static call. The type of the pointer is differentiated by bit 0. A mods pointer has the bit clear, the sites pointer has the bit set. As static_call_del_module() blidly assumes that the pointer is a valid static_call_mod type, it fails to check for this failure case and dereferences the pointer to the list of built-in call sites, which is obviously bogus. Cure it by checking whether the key has a sites or a mods pointer. If it's a sites pointer then the key is not to be touched. As the sites are walked in the same order as in __static_call_init() the site walk can be terminated because all subsequent sites have not been touched by the init code due to the error exit. If it was converted before the allocation fail, then the inner loop which searches for a module match will find nothing. A fail in the second allocation in __static_call_init() is harmless and does not require special treatment. The first allocation succeeded and converted the key to a module chain. That first entry has mod::mod == NULL and mod::next == NULL, so the inner loop of static_call_del_module() will neither find a module match nor a module chain. The next site in the walk was either already converted, but can't match the module, or it will exit the outer loop because it has a static_call_site pointer and not a static_call_mod pointer.

CVSS 5.5 · Mittel · EPSS 0,2% · Korrektur verfügbar

Veröffentlicht

2024-10-21T18:02:40.908Z

Aktualisiert

2026-05-11T20:43:38.540Z

Technologie-Stack: system

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