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Redis内存诡异增长如何排查问题

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一、现象

实例名：r-bp1cxxxxxxxxxd04(主从)

问题：一分钟内存上涨了2G，如下图所示：

键值规模：6000万左右

内存一分钟增长2G.png

Redis内存诡异增长如何排查问题

二、Redis内存分析

1. 内存组成

上图中的内存统计的是Redis的info memory命令中的used_memory属性，例如：

redis>infomemory#Memoryused_memory:9195978072used_memory_human:8.56Gused_memory_rss:9358786560used_memory_peak:10190212744used_memory_peak_human:9.49Gused_memory_lua:38912mem_fragmentation_ratio:1.02mem_allocator:jemalloc-3.6.0

每个属性的详细说明

属性名属性说明
used_memoryRedis 分配器分配的内存量，也就是实际存储数据的内存总量
used_memory_human以可读格式返回 Redis 使用的内存总量
used_memory_rss从操作系统的角度，Redis进程占用的总物理内存
used_memory_peak内存分配器分配的最大内存，代表used_memory的历史峰值
used_memory_peak_human以可读的格式显示内存消耗峰值
used_memory_luaLua引擎所消耗的内存
mem_fragmentation_ratioused_memory_rss /used_memory比值，表示内存碎片率
mem_allocatorRedis 所使用的内存分配器。默认: jemalloc

计算公式如下：

used_memory = 自身内存+对象内存+缓冲内存+lua内存used_rss = used_memory + 内存碎片

如下图所示：

Redis内存诡异增长如何排查问题

2. 内存分析

(1) 自身内存：一个空的Redis占用很小，可以忽略不计

(2) kv内存：key对象 + value对象

(3) 缓冲区：客户端缓冲区(普通 + slave伪装 + pubsub)以及aof缓冲区(比较固定，一般没问题)

(4) Lua：Lua引擎所消耗的内存

3. 内存突增常见问题

(1) kv内存：bigkey、大量写入

(2) 客户端缓冲区：一般常见的有普通客户端缓冲区(例如monitor命令)或者pubsub客户端缓冲区

三、问题排查

(1) bigkey ? 经扫描未发现bigkey

Sampled 67234427 keys in the keyspace!
Total key length in bytes is 1574032382 (avg len 23.41)

Biggest string found 'CCARD_DEVICE_CARD_REF_MAP_KEY_016817000004209' has 20862 bytes
Biggest list found 'CCARD_VALID_DEVICE_TRAIN_QUEUE_KEY' has 51 items
Biggest hash found 'CCARD_VALID_DEVICE_TRAIN_MAP_KEY' has 51 fields

67234359 strings with 71767890 bytes (100.00% of keys, avg size 1.07)
67 lists with 151 items (00.00% of keys, avg size 2.25)
0 sets with 0 members (00.00% of keys, avg size 0.00)
1 hashs with 51 fields (00.00% of keys, avg size 51.00)
0 zsets with 0 members (00.00% of keys, avg size 0.00)

(2) 键值个数增加？未发现键值有明显变化

Redis内存诡异增长如何排查问题

(3) 客户端缓冲区

由于内存增上去后，长时间没下落，如果是因为缓冲区问题，会从info clients找到明显问题，执行后发现：

redis> info clients
# Clients
connected_clients:43
client_longest_output_list:0
client_biggest_input_buf:0
blocked_clients:0
admin_clients:6
rejected_vpc_conn_count:0
close_idle_unknown_conn_count:0

执行client中也没有明显的omem大于0的情况

id=80207addr=10.xx.0.4:63920fd=46name=age=624idle=1flags=N db=0sub=0psub=0multi=-1qbuf=0qbuf-free=0obl=0oll=0omem=0events=r cmd=ping read=0write=0type=user
id=80215addr=10.xx.0.23:43489fd=36name=age=591idle=1flags=N db=0sub=0psub=0multi=-1qbuf=0qbuf-free=0obl=0oll=0omem=0events=r cmd=ping read=0write=0type=user
id=80366addr=10.xx.0.8:59785fd=18name=age=84idle=1flags=N db=0sub=0psub=0multi=-1qbuf=0qbuf-free=0obl=0oll=0omem=0events=r cmd=delread=0write=0type=user
id=80356addr=10.xx.0.33:32117fd=13name=age=114idle=0flags=N db=0sub=0psub=0multi=-1qbuf=0qbuf-free=32768obl=0oll=0omem=0events=r cmd=ping read=0write=0type=user
id=80064addr=10.xx.59.4:53446fd=38name=age=1070idle=1070flags=N db=0sub=0psub=0multi=-1qbuf=0qbuf-free=0obl=0oll=0omem=0events=r cmd=NULL read=0write=0type=admin
id=80276addr=10.xx.0.23:48511fd=8name=age=387idle=1flags=N db=0sub=0psub=0multi=-1qbuf=0qbuf-free=0obl=0oll=0omem=0events=r cmd=ping read=0write=0type=user
id=80188addr=10.xx.0.33:16265fd=42name=age=681idle=3flags=N db=0sub=0psub=0multi=-1qbuf=0qbuf-free=0obl=0oll=0omem=0events=r cmd=ping read=0write=0type=user
id=80326addr=10.xx.0.32:59779fd=16name=age=209idle=0flags=N db=0sub=0psub=0multi=-1qbuf=0qbuf-free=32768obl=0oll=0omem=0events=r cmd=ping read=0write=0type=user
id=80065addr=10.xx.59.4:53447fd=45name=age=1070idle=1070flags=N db=0sub=0psub=0multi=-1qbuf=0qbuf-free=0obl=0oll=0omem=0events=r cmd=NULL read=0write=0type=admin
id=79936addr=10.xx.0.22:10607fd=30name=age=1480idle=1flags=N db=0sub=0psub=0multi=-1qbuf=0qbuf-free=32768obl=0oll=0omem=0events=r cmd=ping read=0write=0type=user
id=80174addr=10.xx.0.5:60914fd=6name=age=722idle=2flags=N db=0sub=0psub=0multi=-1qbuf=0qbuf-free=0obl=0oll=0omem=0events=r cmd=ping read=0write=0type=user
id=80300addr=10.xx.0.22:22757fd=48name=age=298idle=1flags=N db=0sub=0psub=0multi=-1qbuf=0qbuf-free=32768obl=0oll=0omem=0events=r cmd=ping read=0write=0type=user
id=80037addr=10.xx.0.5:55189fd=15name=age=1143idle=2flags=N db=0sub=0psub=0multi=-1qbuf=0qbuf-free=0obl=0oll=0omem=0events=r cmd=ping read=0write=0type=user
id=80330addr=10.xx.0.8:48533fd=17name=age=199idle=10flags=N db=0sub=0psub=0multi=-1qbuf=0qbuf-free=0obl=0oll=0omem=0events=r cmd=ping read=0write=0type=user
id=79896addr=10.xx.0.30:26814fd=11name=age=1616idle=1flags=N db=0sub=0psub=0multi=-1qbuf=0qbuf-free=0obl=0oll=0omem=0events=r cmd=ping read=0write=0type=user
id=80299addr=10.xx.0.24:11227fd=44name=age=303idle=3flags=N db=0sub=0psub=0multi=-1qbuf=0qbuf-free=0obl=0oll=0omem=0events=r cmd=ping read=0write=0type=user
id=80086addr=10.xx.0.32:52526fd=40name=age=1002idle=1flags=N db=0sub=0psub=0multi=-1qbuf=0qbuf-free=0obl=0oll=0omem=0events=r cmd=ping read=0write=0type=user
id=80202addr=10.xx.0.33:16658fd=26name=age=636idle=3flags=N db=0sub=0psub=0multi=-1qbuf=0qbuf-free=0obl=0oll=0omem=0events=r cmd=ping read=0write=0type=user
id=80256addr=10.xx.0.24:60496fd=19name=age=448idle=2flags=N db=0sub=0psub=0multi=-1qbuf=0qbuf-free=0obl=0oll=0omem=0events=r cmd=ping read=0write=0type=user
id=79908addr=10.xx.0.29:18975fd=12name=age=1583idle=1flags=N db=0sub=0psub=0multi=-1qbuf=0qbuf-free=0obl=0oll=0omem=0events=r cmd=ping read=0write=0type=user
id=80365addr=10.xx.0.29:46429fd=14name=age=85idle=1flags=N db=0sub=0psub=0multi=-1qbuf=0qbuf-free=32768obl=0oll=0omem=0events=r cmd=ping read=0write=0type=user
id=79869addr=10.xx.27.4:48455fd=35name=age=1700idle=1700flags=N db=0sub=0psub=0multi=-1qbuf=0qbuf-free=0obl=0oll=0omem=0events=r cmd=NULL read=0write=0type=admin
id=80334addr=10.xx.0.23:50012fd=39name=age=189idle=1flags=N db=0sub=0psub=0multi=-1qbuf=0qbuf-free=32768obl=0oll=0omem=0events=r cmd=ping read=0write=0type=user
id=80041addr=10.xx.0.32:51107fd=33name=age=1132idle=3flags=N db=0sub=0psub=0multi=-1qbuf=0qbuf-free=0obl=0oll=0omem=0events=r cmd=ping read=0write=0type=user
id=79992addr=10.xx.0.22:12068fd=28name=age=1289idle=1flags=N db=0sub=0psub=0multi=-1qbuf=0qbuf-free=32768obl=0oll=0omem=0events=r cmd=ping read=0write=0type=user
id=80251addr=10.xx.0.30:44213fd=23name=age=468idle=1flags=N db=0sub=0psub=0multi=-1qbuf=0qbuf-free=32768obl=0oll=0omem=0events=r cmd=ping read=0write=0type=user
id=80006addr=10.xx.0.2:45895fd=31name=age=1242idle=1flags=N db=0sub=0psub=0multi=-1qbuf=0qbuf-free=0obl=0oll=0omem=0events=r cmd=ping read=0write=0type=user
id=80321addr=10.xx.0.30:48048fd=5name=age=224idle=3flags=N db=0sub=0psub=0multi=-1qbuf=0qbuf-free=0obl=0oll=0omem=0events=r cmd=ping read=0write=0type=user
id=80381addr=10.xx.0.8:13360fd=22name=age=24idle=1flags=N db=0sub=0psub=0multi=-1qbuf=0qbuf-free=32768obl=0oll=0omem=0events=r cmd=delread=0write=0type=user
id=80200addr=10.xx.0.24:59183fd=24name=age=640idle=0flags=N db=0sub=0psub=0multi=-1qbuf=0qbuf-free=32768obl=0oll=0omem=0events=r cmd=ping read=0write=0type=user
id=80113addr=10.xx.0.2:52492fd=21name=age=915idle=1flags=N db=0sub=0psub=0multi=-1qbuf=0qbuf-free=32768obl=0oll=0omem=0events=r cmd=ping read=0write=0type=user
id=174addr=11.216.117.242:53027fd=9name=age=281390idle=0flags=S db=0sub=0psub=0multi=-1qbuf=0qbuf-free=32768obl=0oll=0omem=0events=r cmd=replconf read=0write=0type=admin
id=79991addr=10.xx.0.4:48412fd=25name=age=1296idle=0flags=N db=0sub=0psub=0multi=-1qbuf=0qbuf-free=32768obl=0oll=0omem=0events=r cmd=ping read=0write=0type=user
id=80301addr=127.0.0.1:47869fd=49name=age=291idle=261flags=N db=0sub=0psub=0multi=-1qbuf=0qbuf-free=0obl=0oll=0omem=0events=r cmd=strlen read=0write=0type=admin
id=80047addr=10.xx.59.4:53184fd=41name=age=1114idle=1114flags=N db=0sub=0psub=0multi=-1qbuf=0qbuf-free=0obl=0oll=0omem=0events=r cmd=NULL read=0write=0type=admin
id=80236addr=10.xx.0.5:62546fd=47name=age=516idle=1flags=N db=0sub=0psub=0multi=-1qbuf=0qbuf-free=32768obl=0oll=0omem=0events=r cmd=ping read=0write=0type=user
id=80364addr=10.xx.0.4:18794fd=7name=age=85idle=1flags=N db=0sub=0psub=0multi=-1qbuf=0qbuf-free=32768obl=0oll=0omem=0events=r cmd=ping read=0write=0type=user
id=80175addr=10.xx.0.4:62245fd=29name=age=718idle=1flags=N db=0sub=0psub=0multi=-1qbuf=0qbuf-free=32768obl=0oll=0omem=0events=r cmd=ping read=0write=0type=user
id=80336addr=10.xx.0.29:45701fd=50name=age=180idle=1flags=N db=0sub=0psub=0multi=-1qbuf=0qbuf-free=32768obl=0oll=0omem=0events=r cmd=ping read=0write=0type=user
id=80050addr=10.xx.59.4:53188fd=43name=age=1114idle=1114flags=N db=0sub=0psub=0multi=-1qbuf=0qbuf-free=0obl=0oll=0omem=0events=r cmd=NULL read=0write=0type=admin
id=79765addr=10.xx.0.2:33832fd=37name=age=2027idle=177flags=N db=0sub=0psub=0multi=-1qbuf=0qbuf-free=0obl=0oll=0omem=0events=r cmd=info read=0write=0type=user
id=80170addr=10.xx.0.2:57853fd=20name=age=728idle=24flags=N db=0sub=0psub=0multi=-1qbuf=0qbuf-free=0obl=0oll=0omem=0events=r cmd=ping read=0write=0type=user
id=80390addr=127.0.0.1:49449fd=27name=age=0idle=0flags=N db=0sub=0psub=0multi=-1qbuf=0qbuf-free=32768obl=0oll=0omem=0events=r cmd=client read=0write=0type=admin

四、揪出元凶

常用的几招都用了，还是不行，同事@径远帮忙一起分析，怀疑是不是因为Redis的kv哈希表做了 rehash。

1. Redis的kv存储结构

如下图所示，Redis的所有kv保存在dict中，其中ht对应两个哈希表ht[0]和ht[1]，平时一个空闲，一个用于存储数据，只有当需要rehash时，ht[1]才会用到。

Redis内存诡异增长如何排查问题

2. Redis的字典rehash

为了保证哈希表的负载，当哈希表的元素个数等于哈希表槽数时候，会进行rehash扩容。扩容后h[1]的容量等于第一个大于等于ht[0].size*2的2n，例如hash表的初始化容量是4，那么下一次扩容就是8，以此类推。

3. 测试

(1) 测试方法

先批量写入到rehash阈值附近，然后在逐条去写，观察内存变化

// 为每个键设置1天过期时间
int expireTime = 60 * 60 * 24;
// rehash阈值 - 50为了方便观察rehash内存变化
int rehashThreshold = (int) Math.pow(2, 25) - 50;

// 1.批量写入：pipeline批量写入，由于是本机测试，这里用10000，实际生产不要这么用
Pipeline pipeline = jedis.pipelined();
pipeline = jedis.pipelined();
for (int i = 0; i < rehashThreshold; i++) {
  pipeline.setex(String.valueOf(i), expireTime, String.valueOf(i));
  if (i % 10000 == 0) {
    pipeline.sync();
  }
}
pipeline.sync();

// 2.等待写增量
TimeUnit.SECONDS.sleep(5);
for (int i = rehashThreshold; i < rehashThreshold + 200; i++) {
  jedis.setex(String.valueOf(i), expireTime, String.valueOf(i));
  TimeUnit.SECONDS.sleep(1);
}

(2) 开始测试

(a) 当阈值=215=32768，从下面可以看出到key的个数为32769时，内存涨了一些，但是还不明显。

keys       mem      clients blocked requests            connections32766      4.69M    3       0       32797 (+2)          4
32767      4.69M    3       0       32799 (+2)          4
32768      4.69M    3       0       32801 (+2)          4
32769      5.44M    3       0       32803 (+2)          4

(b) 当阈值=220=1048576，从下面可以看出到key的个数为1048577时，内存涨了32M。因为rehash会扩容，所以新的哈希表中的槽位变为了221 * 2（因为每个key都设置了过期时间，expires表），指针为8个字节，221 ? 2 ? 8 = 225 = 32MB。

keys       mem      clients blocked requests            connections1048574    128.69M  3       0       3364129 (+2)        16
1048575    128.69M  3       0       3364131 (+2)        16
1048576    128.69M  3       0       3364133 (+2)        16
1048577    160.69M  3       0       3364135 (+2)        16
1048578    160.69M  3       0       3364137 (+2)        16

(c) 当阈值=226=67108864，从下面可以看出到key的个数为67108865时，内存涨了2GB。因为rehash会扩容，所以新的哈希表中的槽位变为了227 * 2（因为每个key都设置了过期时间，expires表），指针为8个字节，227 ? 2 ? 8 = 231 = 2GB。

keys       mem      clients blocked requests            connections67108862   9.70G    3       0       70473683 (+2)       18
67108863   9.70G    3       0       70473685 (+2)       18
67108864   9.70G    3       0       70473687 (+2)       18
67108865   11.70G   3       0       70473689 (+2)       18
67108866   11.70G   3       0       70473691 (+2)       18
67108867   11.70G   3       0       70473693 (+2)       18

回过来看r-bp1c15fd9b142d04的key和内存变化图，可以发现上面的规则是正确的：

 Redis内存诡异增长如何排查问题

4. 后续观察

17点时，rehash结束，内存降了增加的2G的一半。

Redis内存诡异增长如何排查问题

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