* [dpdk-dev] [Bug 260] bugDPDK lock-free hash deletion
@ 2019-04-30 6:51 bugzilla
2019-04-30 6:51 ` bugzilla
0 siblings, 1 reply; 2+ messages in thread
From: bugzilla @ 2019-04-30 6:51 UTC (permalink / raw)
To: dev
https://bugs.dpdk.org/show_bug.cgi?id=260
Bug ID: 260
Summary: bugDPDK lock-free hash deletion
Product: DPDK
Version: 18.11
Hardware: All
OS: All
Status: CONFIRMED
Severity: normal
Priority: Normal
Component: other
Assignee: dev@dpdk.org
Reporter: zhongdahulinfan@163.com
Target Milestone: ---
lock-free版本的哈希表，在创建的时候指定了flag：
RTE_HASH_EXTRA_FLAGS_RW_CONCURRENCY_LF | RTE_HASH_EXTRA_FLAGS_MULTI_WRITER_ADD
以使哈希表支持multi-writer，RTE_HASH_EXTRA_FLAGS_MULTI_WRITER_ADD标记使得每个lcore都可以使用local
cache分配key slot：
struct rte_hash *
rte_hash_create(const struct rte_hash_parameters *params)
{
...
if (params->extra_flag & RTE_HASH_EXTRA_FLAGS_MULTI_WRITER_ADD) {
use_local_cache = 1;
writer_takes_lock = 1;
}
...
/* Store all keys and leave the first entry as a dummy entry for
lookup_bulk */
if (use_local_cache)
/*
* Increase number of slots by total number of indices
* that can be stored in the lcore caches
* except for the first cache
*/
num_key_slots = params->entries + (RTE_MAX_LCORE - 1) *
(LCORE_CACHE_SIZE - 1) + 1;
else
num_key_slots = params->entries + 1;
...
}
这么做的好处是，每个写者可以从本地cache分配key slot，可减少cache miss，提升哈希插入的性能。
这里要先说一下rte_hash的key和data的存储的结构：
| dummy | pdata + key | pdata + key | pdata + key | pdata + key | pdata + key
| pdata + key | pdata + key | pdata + key |...
0 | 1 2 3
4 5 6
7 8 |
| <--------------
bucket
----------> |
struct rte_hash里的成员key_store就是上述数组，存放key的内容和data指针，其中 index 0没被使用，有效数据从index 1
开始。该数组被划分成若干个bucket，每个bucket的大小为8。这么做是有原因的，rte_hash使用cuckoo
哈希实现，引入bucket解决哈希冲突，而非开链。以写为例，在写哈希表的时候，先哈希到primary
bucket，再循环遍历该bucket找到存储位置，如若有空位则插入，否则继续找secondary bucket。
结构体定义如下：
/* Structure that stores key-value pair */
struct rte_hash_key {
union {
uintptr_t idata;
void *pdata;
};
/* Variable key size */
char key[0];
};
/** Bucket structure */
struct rte_hash_bucket {
uint16_t sig_current[RTE_HASH_BUCKET_ENTRIES];
uint32_t key_idx[RTE_HASH_BUCKET_ENTRIES];
uint8_t flag[RTE_HASH_BUCKET_ENTRIES];
void *next;
} __rte_cache_aligned;
然后这些key index，用一个rte_ring来存储：
struct rte_hash *
rte_hash_create(const struct rte_hash_parameters *params)
{
...
/* Populate free slots ring. Entry zero is reserved for key misses. */
for (i = 1; i < num_key_slots; i++)
rte_ring_sp_enqueue(r, (void *)((uintptr_t) i));
...
}
当使用lock-free哈希表时，删除一个表项的时候key和value采用延后回收内存的策略，使得multi-readers访问哈希表不需要加锁，大大减少多核应用程序的性能损耗。我们调用dpdk
API rte_hash_del_key_xxx删除表项时，只将表项从哈希表中摘除，返回一个key的存储位置position，就是上述所说的key
index，然后在所有引用该表项的readers/writers都退出引用后，再根据position回收key和value的存储空间。key的回收调用一下接口：
int __rte_experimental
rte_hash_free_key_with_position(const struct rte_hash *h,
const int32_t position)
{
RETURN_IF_TRUE(((h == NULL) || (position == EMPTY_SLOT)), -EINVAL);
unsigned int lcore_id, n_slots;
struct lcore_cache *cached_free_slots;
const int32_t total_entries = h->num_buckets * RTE_HASH_BUCKET_ENTRIES;
/* Out of bounds */
if (position >= total_entries)
return -EINVAL;
if (h->use_local_cache) {
lcore_id = rte_lcore_id();
cached_free_slots = &h->local_free_slots[lcore_id];
/* Cache full, need to free it. */
if (cached_free_slots->len == LCORE_CACHE_SIZE) {
/* Need to enqueue the free slots in global ring. */
n_slots = rte_ring_mp_enqueue_burst(h->free_slots,
cached_free_slots->objs,
LCORE_CACHE_SIZE, NULL);
cached_free_slots->len -= n_slots;
}
/* Put index of new free slot in cache. */
cached_free_slots->objs[cached_free_slots->len] =
(void *)((uintptr_t)position);
cached_free_slots->len++;
} else {
rte_ring_sp_enqueue(h->free_slots,
(void *)((uintptr_t)position));
}
return 0;
}
仔细看rte_hash的实现，不难发现上述函数有两个地方存在问题：
1、"Out of bounds" 的判断逻辑
这个 "Out of bounds" 的判断逻辑，在哈希表的use_local_cache标识没被置为的时候是成立的，key
index的数量恰好是哈希表entries的数量。但当use_local_cache为真时，它就不正确了。回看一下创建哈希表的函数rte_hash_create，其中key
slots的计算：
/* Store all keys and leave the first entry as a dummy entry for lookup_bulk */
if (use_local_cache)
/*
* Increase number of slots by total number of indices
* that can be stored in the lcore caches
* except for the first cache
*/
num_key_slots = params->entries + (RTE_MAX_LCORE - 1) *
(LCORE_CACHE_SIZE - 1) + 1;
else
num_key_slots = params->entries + 1;
这时候除了分配entries个key内存空间，还要给每个lcore分配LCORE_CACHE_SIZE数量的key空间，那么此时key的数量是会大于哈希表的total
entry的，所以 rte_hash_free_key_with_position的"Out of bounds"判断逻辑有误。
2、position参数问题
position是rte_hash_del_key_xxx()的返回值，为 (key_idx - 1)。前面说过key的index
0未被使用，从1开始有效。那么直接将position enqueue到free_slot队列是不正确的：
rte_ring_sp_enqueue(h->free_slots,
(void *)((uintptr_t)position));
这会导致ring队列里可能存在多个值为0的position，从而损坏ring队列。以ring的size是4为例：
ring初始状态：
|1 | 2 | 3 | 4 |
dequeue完所有key_idx后，返回position为 0,1,2,3，再enqueue回ring队列
一趟dequeue enqueue过后：
| 0 | 1 | 2 | 3 |
dequeue完所有key_idx后，返回position为 0,1,2（其中key_idx 0是无效的，不被使用），再enqueue回ring队列
两趟dequeue enqueue过后：
| 0 | 0 | 1 | 2 |
对比非lock-free的key回收接口，可以发现，remove_entry()是直接将key_indx入队的，所以不存在问题：
static inline void
remove_entry(const struct rte_hash *h, struct rte_hash_bucket *bkt, unsigned i)
{
unsigned lcore_id, n_slots;
struct lcore_cache *cached_free_slots;
if (h->use_local_cache) {
lcore_id = rte_lcore_id();
cached_free_slots = &h->local_free_slots[lcore_id];
/* Cache full, need to free it. */
if (cached_free_slots->len == LCORE_CACHE_SIZE) {
/* Need to enqueue the free slots in global ring. */
n_slots = rte_ring_mp_enqueue_burst(h->free_slots,
cached_free_slots->objs,
LCORE_CACHE_SIZE, NULL);
cached_free_slots->len -= n_slots;
}
/* Put index of new free slot in cache. */
cached_free_slots->objs[cached_free_slots->len] =
(void *)((uintptr_t)bkt->key_idx[i]);
cached_free_slots->len++;
} else {
rte_ring_sp_enqueue(h->free_slots,
(void *)((uintptr_t)bkt->key_idx[i]));
}
}
修复方法
1、修改"Out of bounds" 的判断逻辑
2、position在enqueue回free_slots队列之前加1
--
You are receiving this mail because:
You are the assignee for the bug.
^ permalink raw reply [flat|nested] 2+ messages in thread
* [dpdk-dev] [Bug 260] bugDPDK lock-free hash deletion
2019-04-30 6:51 [dpdk-dev] [Bug 260] bugDPDK lock-free hash deletion bugzilla
@ 2019-04-30 6:51 ` bugzilla
0 siblings, 0 replies; 2+ messages in thread
From: bugzilla @ 2019-04-30 6:51 UTC (permalink / raw)
To: dev
https://bugs.dpdk.org/show_bug.cgi?id=260
Bug ID: 260
Summary: bugDPDK lock-free hash deletion
Product: DPDK
Version: 18.11
Hardware: All
OS: All
Status: CONFIRMED
Severity: normal
Priority: Normal
Component: other
Assignee: dev@dpdk.org
Reporter: zhongdahulinfan@163.com
Target Milestone: ---
lock-free版本的哈希表，在创建的时候指定了flag：
RTE_HASH_EXTRA_FLAGS_RW_CONCURRENCY_LF | RTE_HASH_EXTRA_FLAGS_MULTI_WRITER_ADD
以使哈希表支持multi-writer，RTE_HASH_EXTRA_FLAGS_MULTI_WRITER_ADD标记使得每个lcore都可以使用local
cache分配key slot：
struct rte_hash *
rte_hash_create(const struct rte_hash_parameters *params)
{
...
if (params->extra_flag & RTE_HASH_EXTRA_FLAGS_MULTI_WRITER_ADD) {
use_local_cache = 1;
writer_takes_lock = 1;
}
...
/* Store all keys and leave the first entry as a dummy entry for
lookup_bulk */
if (use_local_cache)
/*
* Increase number of slots by total number of indices
* that can be stored in the lcore caches
* except for the first cache
*/
num_key_slots = params->entries + (RTE_MAX_LCORE - 1) *
(LCORE_CACHE_SIZE - 1) + 1;
else
num_key_slots = params->entries + 1;
...
}
这么做的好处是，每个写者可以从本地cache分配key slot，可减少cache miss，提升哈希插入的性能。
这里要先说一下rte_hash的key和data的存储的结构：
| dummy | pdata + key | pdata + key | pdata + key | pdata + key | pdata + key
| pdata + key | pdata + key | pdata + key |...
0 | 1 2 3
4 5 6
7 8 |
| <--------------
bucket
----------> |
struct rte_hash里的成员key_store就是上述数组，存放key的内容和data指针，其中 index 0没被使用，有效数据从index 1
开始。该数组被划分成若干个bucket，每个bucket的大小为8。这么做是有原因的，rte_hash使用cuckoo
哈希实现，引入bucket解决哈希冲突，而非开链。以写为例，在写哈希表的时候，先哈希到primary
bucket，再循环遍历该bucket找到存储位置，如若有空位则插入，否则继续找secondary bucket。
结构体定义如下：
/* Structure that stores key-value pair */
struct rte_hash_key {
union {
uintptr_t idata;
void *pdata;
};
/* Variable key size */
char key[0];
};
/** Bucket structure */
struct rte_hash_bucket {
uint16_t sig_current[RTE_HASH_BUCKET_ENTRIES];
uint32_t key_idx[RTE_HASH_BUCKET_ENTRIES];
uint8_t flag[RTE_HASH_BUCKET_ENTRIES];
void *next;
} __rte_cache_aligned;
然后这些key index，用一个rte_ring来存储：
struct rte_hash *
rte_hash_create(const struct rte_hash_parameters *params)
{
...
/* Populate free slots ring. Entry zero is reserved for key misses. */
for (i = 1; i < num_key_slots; i++)
rte_ring_sp_enqueue(r, (void *)((uintptr_t) i));
...
}
当使用lock-free哈希表时，删除一个表项的时候key和value采用延后回收内存的策略，使得multi-readers访问哈希表不需要加锁，大大减少多核应用程序的性能损耗。我们调用dpdk
API rte_hash_del_key_xxx删除表项时，只将表项从哈希表中摘除，返回一个key的存储位置position，就是上述所说的key
index，然后在所有引用该表项的readers/writers都退出引用后，再根据position回收key和value的存储空间。key的回收调用一下接口：
int __rte_experimental
rte_hash_free_key_with_position(const struct rte_hash *h,
const int32_t position)
{
RETURN_IF_TRUE(((h == NULL) || (position == EMPTY_SLOT)), -EINVAL);
unsigned int lcore_id, n_slots;
struct lcore_cache *cached_free_slots;
const int32_t total_entries = h->num_buckets * RTE_HASH_BUCKET_ENTRIES;
/* Out of bounds */
if (position >= total_entries)
return -EINVAL;
if (h->use_local_cache) {
lcore_id = rte_lcore_id();
cached_free_slots = &h->local_free_slots[lcore_id];
/* Cache full, need to free it. */
if (cached_free_slots->len == LCORE_CACHE_SIZE) {
/* Need to enqueue the free slots in global ring. */
n_slots = rte_ring_mp_enqueue_burst(h->free_slots,
cached_free_slots->objs,
LCORE_CACHE_SIZE, NULL);
cached_free_slots->len -= n_slots;
}
/* Put index of new free slot in cache. */
cached_free_slots->objs[cached_free_slots->len] =
(void *)((uintptr_t)position);
cached_free_slots->len++;
} else {
rte_ring_sp_enqueue(h->free_slots,
(void *)((uintptr_t)position));
}
return 0;
}
仔细看rte_hash的实现，不难发现上述函数有两个地方存在问题：
1、"Out of bounds" 的判断逻辑
这个 "Out of bounds" 的判断逻辑，在哈希表的use_local_cache标识没被置为的时候是成立的，key
index的数量恰好是哈希表entries的数量。但当use_local_cache为真时，它就不正确了。回看一下创建哈希表的函数rte_hash_create，其中key
slots的计算：
/* Store all keys and leave the first entry as a dummy entry for lookup_bulk */
if (use_local_cache)
/*
* Increase number of slots by total number of indices
* that can be stored in the lcore caches
* except for the first cache
*/
num_key_slots = params->entries + (RTE_MAX_LCORE - 1) *
(LCORE_CACHE_SIZE - 1) + 1;
else
num_key_slots = params->entries + 1;
这时候除了分配entries个key内存空间，还要给每个lcore分配LCORE_CACHE_SIZE数量的key空间，那么此时key的数量是会大于哈希表的total
entry的，所以 rte_hash_free_key_with_position的"Out of bounds"判断逻辑有误。
2、position参数问题
position是rte_hash_del_key_xxx()的返回值，为 (key_idx - 1)。前面说过key的index
0未被使用，从1开始有效。那么直接将position enqueue到free_slot队列是不正确的：
rte_ring_sp_enqueue(h->free_slots,
(void *)((uintptr_t)position));
这会导致ring队列里可能存在多个值为0的position，从而损坏ring队列。以ring的size是4为例：
ring初始状态：
|1 | 2 | 3 | 4 |
dequeue完所有key_idx后，返回position为 0,1,2,3，再enqueue回ring队列
一趟dequeue enqueue过后：
| 0 | 1 | 2 | 3 |
dequeue完所有key_idx后，返回position为 0,1,2（其中key_idx 0是无效的，不被使用），再enqueue回ring队列
两趟dequeue enqueue过后：
| 0 | 0 | 1 | 2 |
对比非lock-free的key回收接口，可以发现，remove_entry()是直接将key_indx入队的，所以不存在问题：
static inline void
remove_entry(const struct rte_hash *h, struct rte_hash_bucket *bkt, unsigned i)
{
unsigned lcore_id, n_slots;
struct lcore_cache *cached_free_slots;
if (h->use_local_cache) {
lcore_id = rte_lcore_id();
cached_free_slots = &h->local_free_slots[lcore_id];
/* Cache full, need to free it. */
if (cached_free_slots->len == LCORE_CACHE_SIZE) {
/* Need to enqueue the free slots in global ring. */
n_slots = rte_ring_mp_enqueue_burst(h->free_slots,
cached_free_slots->objs,
LCORE_CACHE_SIZE, NULL);
cached_free_slots->len -= n_slots;
}
/* Put index of new free slot in cache. */
cached_free_slots->objs[cached_free_slots->len] =
(void *)((uintptr_t)bkt->key_idx[i]);
cached_free_slots->len++;
} else {
rte_ring_sp_enqueue(h->free_slots,
(void *)((uintptr_t)bkt->key_idx[i]));
}
}
修复方法
1、修改"Out of bounds" 的判断逻辑
2、position在enqueue回free_slots队列之前加1
--
You are receiving this mail because:
You are the assignee for the bug.
^ permalink raw reply [flat|nested] 2+ messages in thread
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