To clarify my doubt, it seems to me that it just sits there in memory without it being referenced anymore, is that right?
On Wed, Mar 25, 2015 at 6:38 PM, Leandro M Barbosa <[email protected]> wrote: > Hello all! > > I was reading the __list_splice () function code and I had a doubt. The code > is: > > 274 static inline void __list_splice(const struct list_head *list, > 275 struct list_head *prev, > 276 struct list_head *next) > 277 { > 278 struct list_head *first = list->next; > 279 struct list_head *last = list->prev; > 280 > 281 first->prev = prev; > 282 prev->next = first; > 283 > 284 last->next = next; > 285 next->prev = last; > 286 } > > What happens with the *list head? As I understood, when you call > list_splice (list_a, list_b, list_b->next), the code joins the two > lists together such that the list_a is put before list_b. The code > grabs list->next and list->prev but what about *list itself? > > > -- > Leandro Moreira Barbosa -- Leandro Moreira Barbosa _______________________________________________ Kernelnewbies mailing list [email protected] http://lists.kernelnewbies.org/mailman/listinfo/kernelnewbies
