Thank you for your contribution! Sashiko AI review found 2 potential issue(s) to consider: - [High] Iterating the `.initcall.modnames` array locklessly in `initcall_get_modname()` races with `free_initmem()`, causing a Use-After-Free for concurrently loaded modules. - [High] The built-in module blacklisting functionality is completely disabled when CONFIG_MODULES=n. --
commit aca1118f7ac6c8d2b798e7bb31a70a4f01638aee Author: Aaron Tomlin <[email protected]> Subject: module: Extend module_blacklist parameter to built-in modules This patch extends the module_blacklist command-line parameter to apply to built-in modules by checking them during early boot. It introduces a new memory section to map initcalls to module names, allowing the kernel to skip the initcall if the module name is present in the blacklist. > diff --git a/include/linux/module.h b/include/linux/module.h > index 7566815fabbe8..b9b0dc419aeeb 100644 > --- a/include/linux/module.h > +++ b/include/linux/module.h [ ... ] > @@ -756,6 +756,15 @@ void *dereference_module_function_descriptor(struct > module *mod, void *ptr); > int register_module_notifier(struct notifier_block *nb); > int unregister_module_notifier(struct notifier_block *nb); > > +#ifdef CONFIG_MODULES > +extern bool module_is_blacklisted(const char *module_name); > +#else > +static inline bool module_is_blacklisted(const char *module_name) > +{ > + return false; > +} > +#endif [Severity: High] Does this defeat the purpose of blacklisting built-in modules on kernels built without loadable module support? If a kernel is built with CONFIG_MODULES=n, the command-line parsing and state variable in kernel/module/main.c are not compiled. This stub will then unconditionally return false, meaning the module_blacklist parameter will be silently ignored for built-in modules on non-modular kernels. [ ... ] > diff --git a/init/main.c b/init/main.c > index 96f93bb06c490..893b20ddcf658 100644 > --- a/init/main.c > +++ b/init/main.c [ ... ] > @@ -1379,12 +1379,39 @@ static inline void do_trace_initcall_level(const char > *level) > } > #endif /* !TRACEPOINTS_ENABLED */ > > +extern struct initcall_modname __start_initcall_modnames[]; > +extern struct initcall_modname __stop_initcall_modnames[]; > + > +static bool initmem_freed __ro_after_init = false; > + > +static const char *initcall_get_modname(initcall_t fn) > +{ > + struct initcall_modname *p; > + > + if (initmem_freed) > + return NULL; > + > + for (p = __start_initcall_modnames; p < __stop_initcall_modnames; p++) { > + if (p->initcall_fn == fn) > + return p->modname; > + } [Severity: High] Can this race with free_initmem() and cause a use-after-free if a module is loaded concurrently? In kernel_init(), free_initmem() is called just before setting initmem_freed to true: kernel_init() free_initmem(); initmem_freed = true; At this point in the boot sequence, the rootfs is mounted and user-mode helpers are enabled. If a hardware interrupt triggers request_module() concurrently, modprobe could execute and invoke the init_module syscall. When do_init_module() calls do_one_initcall(), it would end up in initcall_get_modname(). Since initmem_freed is still false, it would locklessly scan the .initcall.modnames section while it is actively being freed by free_initmem(). [ ... ] > @@ -1600,6 +1627,7 @@ static int __ref kernel_init(void *unused) > kgdb_free_init_mem(); > exit_boot_config(); > free_initmem(); > + initmem_freed = true; > mark_readonly(); > > /* -- Sashiko AI review ยท https://sashiko.dev/#/patchset/[email protected]?part=1
