erlang 虚机crash
现网服务,每次更新一个服务时,另外一个集群所有node 都跟着同时重启一遍,这么调皮,这是闹哪样啊。。
看系统日志:/var/log/messages
Oct 30 15:19:41 localhost kernel: beam.smp[21880]: segfault at 7fa300006d4b ip 00007fa300006d4b sp 00007fa3d0d7c788 error 14 in locale-archive[7fa31616f000+5e91000
beam crash了,好吧开始回忆咱用了哪些c库, 都不应该有问题啊
嗯,打开core dump,再复现一遍。嗯,在线上复现嗷,设计成完全不影响业务的重启还是很有用的。
不一会dump粗来了,挂上gdb 很快找到出错堆栈:
#0 0x00007fa300006d4b in ?? ()
#1 0x00007fa3aa83dd96 in quicksort () from /data0/xxx_0.4.4/lib/hash_ring-0.1.6/priv/hash_ring_drv.so
#2 0x00007fa3aa83d026 in hash_ring_remove_node () from /data0/xxx_0.4.4/lib/hash_ring-0.1.6/priv/hash_ring_drv.so
#3 0x00007fa3aa83c295 in hash_ring_drv_output () from /data0/xxx_0.4.4/lib/hash_ring-0.1.6/priv/hash_ring_drv.so
#4 0x00000000004924ef in call_driver_output (c_p=0x7fa3b0ec8f50, flags=2064, prt=0x7fa3d8d40bc0, from=55559696966931,
list=140341277983642, refp=0x0) at beam/io.c:1768
嗯,定位到出问题位置在于c库依赖hash_ring
顺便说下服务间调用一致性hash的实现:
1. 通过gen_server 定时rpc:call 目标nodes 指定服务运行状态,并动态管理hash_ring 中动态节点。
init([Configs]) -> %hash_ring 需要先启动, link 需要同时重启 link(whereis(hash_ring)), ets:new(?MODULE, [named_table, protected, set, {read_concurrency, true}]), ets:new(?ROUND_ROBIN_ETS, [named_table, public, set, {write_concurrency, true}]), ets:insert(?ROUND_ROBIN_ETS, {inc, 0}), {ok, Routes} = parse_configs(Configs), State = apply_routes(Routes, #state{}), start_check_timer(), {ok, State}.
monitor_route({Svc, Node} = Route) -> % monitor 无法立即返回是否成功,rpc:call 成功后再monitor case catch rpc:call(Node, erlang, whereis, [Svc], 3000) of {'EXIT', Reason} -> {error, Reason}; undefined -> {error, svc_undefined}; Pid when is_pid(Pid) -> Ref = erlang:monitor(process, Route), {ok, Ref}; Error -> {error, Error} end. route_up(Name, Route, Ref, #state{mons=Mons, downs=Downs} = State) -> case lists:member(Route, get_all_routes(Name)) of true -> case lists:member(Route, get_routes(Name)) of false -> ok = hash_ring:add_node({route, Name}, term_to_binary(Route)), add_route(Name, Route); true -> ok end, State#state{mons=dict:store(Ref, {Name, Route}, Mons), downs=Downs -- [{Name, Route}]}; _ -> catch erlang:demonitor(Ref), lager:info("igonre route_up:~p ~p ~p", [Name, Route, Ref]), State end.
2. 使用自己的gen_call 替代 rpc:call 调用,节省monitor 资源消耗
route(Name, Key) -> case hash_ring:find_node({route, Name}, neo_util:to_binary(erlang:phash2(Key))) of {ok, Route} -> {ok, binary_to_term(Route)}; Error -> Error end. call(Name, Key, Req, Timeout) -> {ok, Dst} = route(Name, Key), gen_call(Dst, Req, Timeout). %参考 whatsapp 做法 %使用场景: %1. Process 需要被长期monitor %2. Process 为node内常驻服务进程,不是临时进程 %3. 调用前需要先确认Process alive 状态 %优势: %不需要monitor, 节省两次网络交互 %gen:call 需要monitor -> call -> demonitor dst node monitor 操作消耗资源 %影响: %process down 瞬间, call 应答都会超时,但此调用返回的是timeout % gen_call(Process, Request) -> gen_call(Process, Request, 5000). gen_call(Process, Request, Timeout) -> Ref = erlang:make_ref(), catch erlang:send(Process, {'$gen_call', {self(), Ref}, Request}), receive {Ref, Reply} -> Reply after Timeout -> exit({timeout, {?MODULE, call, [Process, Request, Timeout]}}) end.
问题原因是:https://github.com/chrismoos/hash-ring/blob/master/sort.c#52
快排算法中STACK_SIZE=1024 移除节点时,实现会对剩余节点做一次排序。节点过多时,数组就越界了。
解决方案:
sort.h 快排算法是有问题的,但先不急着优化排序算法。
其实一致性hash的添加加点和删除节点也能能够做到O(1) 的,为此我提交了一个patch修复该问题。
方案:
1. 移除节点,每个虚拟节点保存物理节点的引用,删除时只需要将空位campact
2. 添加节点,只需要计算新加节点并做快排,而不要所有节点再次排序,做一次二路归并即可
quicksort((void**)adds, ring->numReplicas, item_sort); + //ring->numNodes * ring->numReplicas + if(ring->items == NULL || ring-> numNodes == 1) { + ring->items = adds; + } else { + size_t size_new = sizeof(hash_ring_item_t*) * ring->numNodes * ring->numReplicas; + hash_ring_item_t **news = (hash_ring_item_t **)malloc(size_new); + hash_ring_item_t **olds = ring->items; + if(news == NULL) { + return HASH_RING_ERR; + } + int oldlen = (ring->numNodes - 1) * ring->numReplicas; + int addlen = ring->numReplicas; + int i=0, j=0, k=0; + while(1) { + if(i == oldlen && j == addlen) { + break; + } + if(j == addlen) { + news[k++] = olds[i++]; + continue; + } + if(i == oldlen) { + news[k++] = adds[j++]; + continue; + } + int ret = item_sort(olds[i], adds[j]); + if(ret == 0) { + news[k++] = olds[i++]; + news[k++] = adds[j++]; + } else if (ret < 0) { + news[k++] = olds[i++]; + } else { + news[k++] = adds[j++]; + } + }; + free(adds); + free(olds); + ring->items = news;
另:
这个hash 算法其实还是不太可靠的,虚拟节点数字相同怎么办?
一般来说所有节点添加,删除次序相同即可保证一致性,当然最好能够通过nodename 对相同虚节点做排序。