Hi.
I looked at the sys_init_module() and found that the ->init callback
for the module is called without the module_mutex held and *after*
the module's symbols are exported. Doesn't this create the race when
loading two modules in parallel? Like this.
Consider the first module to be (without any internal locking for
simplicity)
=== foo.c ===
static struct list_head foo_list;
void foo_register(struct list_head *elem)
{
list_add(elem, &foo_list);
}
EXPORT_SYMBOL(foo_register);
static int foo_init(void)
{
INIT_LIST_HEAD(&foo_list);
}
module_init(foo_init);
===
and the second one
=== bar.c ===
struct bar_struct {
struct list_head list;
...
} bar;
static int bar_init(void)
{
foo_register(&bar.list);
}
module_init(bar_init);
===
If I understand the sys_init_module() right, the following code
flow is possible:
1CPU 2CPU
sys_init_module(/* foo module */) sys_init_module(/* bar module */)
mutex_lock(&module_mutex);
load_module();
/* export the foo_init here */
mutex_unlock(&module_mutex);
mutex_lock(&module_mutex);
load_module();
/* resolve the foo_init here */
mutex_unlock(&mudule_mutex);
mod->init(); /* bar_init */
/*
* OOPS! The foo_list is not ready yet, because the foo_init
* is not called yet.
*/
mod->init(); /* foo_init... too late */
Is this analysis correct?
Thanks,
Pavel
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