Re: [PATCH bpf-next v2 4/5] bpf: lru: Fix unintended eviction when updating lru hash maps

[bot+bpf-ci@kernel.org]

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> diff --git a/kernel/bpf/hashtab.c b/kernel/bpf/hashtab.c
> index d029690246f8..8665eb6b8a7d 100644
> --- a/kernel/bpf/hashtab.c
> +++ b/kernel/bpf/hashtab.c

[ ... ]

> +static int htab_lru_map_update_elem_in_place(struct bpf_htab *htab, void *key, void *value,
> +					     u64 map_flags, struct bucket *b,
> +					     struct hlist_nulls_head *head, u32 hash,
> +					     bool percpu, bool onallcpus)
> +{
> +	struct htab_elem *l_new, *l_old, *l_free;
> +	struct bpf_map *map = &htab->map;
> +	u32 key_size = map->key_size;
> +	struct bpf_lru_node *node;
> +	unsigned long flags;
> +	void *l_val;
> +	int ret;
> +
> +	node = bpf_lru_pop_extra(&htab->lru);
> +	if (!node)
> +		return -ENOENT;
> +
> +	l_new = container_of(node, struct htab_elem, lru_node);
> +	l_new->hash = hash;
> +	memcpy(l_new->key, key, key_size);
> +	if (!percpu) {
> +		l_val = htab_elem_value(l_new, map->key_size);
> +		copy_map_value(map, l_val, value);
> +		bpf_obj_free_fields(map->record, l_val);
> +	}

Should bpf_obj_free_fields() be called here on l_new? The value was just
copied into l_new and l_new will be inserted into the hash table when
l_old is found. Calling bpf_obj_free_fields() on l_new's value frees the
special fields (kptr, bpf_list_head, etc.) that were just copied, then
l_new is inserted with freed fields. This appears to be a use-after-free
for maps with special fields.

Looking at htab_lru_map_update_elem(), the pattern is to copy the value
and NOT call bpf_obj_free_fields() on the new element. The fields are
only freed when an element is removed from the hash table.

> +
> +	ret = htab_lock_bucket(b, &flags);
> +	if (ret)
> +		goto err_lock_bucket;
> +
> +	l_old = lookup_elem_raw(head, hash, key, key_size);
> +
> +	ret = check_flags(htab, l_old, map_flags);
> +	if (ret)
> +		goto err;
> +
> +	if (l_old) {
> +		bpf_lru_node_set_ref(&l_new->lru_node);
> +		if (percpu) {
> +			/* per-cpu hash map can update value in-place.
> +			 * Keep the same logic in __htab_lru_percpu_map_update_elem().
> +			 */
> +			pcpu_copy_value(htab, htab_elem_get_ptr(l_old, key_size),
> +					value, onallcpus);
> +			l_free = l_new;
> +		} else {
> +			hlist_nulls_add_head_rcu(&l_new->hash_node, head);
> +			hlist_nulls_del_rcu(&l_old->hash_node);
> +			l_free = l_old;
> +		}
> +	} else {
> +		ret = -ENOENT;
> +	}
> +
> +err:
> +	htab_unlock_bucket(b, flags);
> +
> +err_lock_bucket:
> +	bpf_lru_push_free(&htab->lru, ret ? node : &l_free->lru_node);
                                              ^^^^^^^^^^^^^^^^^^

When ret == 0 and !percpu, l_free points to l_old which is being removed
from the hash table. The code calls bpf_lru_push_free() directly instead
of htab_lru_push_free(). Looking at htab_lru_push_free():

    static void htab_lru_push_free(struct bpf_htab *htab, struct htab_elem *elem)
    {
        check_and_free_fields(htab, elem);
        bpf_map_dec_elem_count(&htab->map);
        bpf_lru_push_free(&htab->lru, &elem->lru_node);
    }

It calls check_and_free_fields() to free special fields in the element
before pushing to the free list. The code here bypasses that, which would
leak special fields (kptr, etc.) in l_old when it has them.

Should this use htab_lru_push_free() instead of bpf_lru_push_free() for
the non-error path where l_old is being freed?

> +
> +	return ret;
> +}


---
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CI run summary: https://github.com/kernel-patches/bpf/actions/runs/20720201621