bpf: Let free_all() return the number of freed elements.

Let free_all() helper return the number of freed elements.
It's not used in this patch, but helps in debug/development of bpf_mem_alloc.

For example this diff for __free_rcu():
-       free_all(llist_del_all(&c->waiting_for_gp_ttrace), !!c->percpu_size);
+       printk("cpu %d freed %d objs after tasks trace\n", raw_smp_processor_id(),
+       	free_all(llist_del_all(&c->waiting_for_gp_ttrace), !!c->percpu_size));

would show how busy RCU tasks trace is.
In artificial benchmark where one cpu is allocating and different cpu is freeing
the RCU tasks trace won't be able to keep up and the list of objects
would keep growing from thousands to millions and eventually OOMing.
Signed-off-by: Alexei Starovoitov <ast@kernel.org>
Signed-off-by: Daniel Borkmann <daniel@iogearbox.net>
Acked-by: Hou Tao <houtao1@huawei.com>
Link: https://lore.kernel.org/bpf/20230706033447.54696-4-alexei.starovoitov@gmail.com

kernel/bpf/memalloc.c

@@ -223,12 +223,16 @@ static void free_one(void *obj, bool percpu)
 	kfree(obj);
 }
 
-static void free_all(struct llist_node *llnode, bool percpu)
+static int free_all(struct llist_node *llnode, bool percpu)
 {
 	struct llist_node *pos, *t;
+	int cnt = 0;
 
-	llist_for_each_safe(pos, t, llnode)
+	llist_for_each_safe(pos, t, llnode) {
 		free_one(pos, percpu);
+		cnt++;
+	}
+	return cnt;
 }
 
 static void __free_rcu(struct rcu_head *head)