实现一致性哈希(Consistent Hashing)java版本

实现一致性哈希(Consistent Hashing)java版本

--转自 【http://afghl.github.io/2016/11/19/implement-consistent-hashing.html】,仅供学习

一致性哈希算法是分布式系统中重要的路由算法。

这篇文章主要说说它的实现。首先,是几个关键的抽象:

  • Entry,要放入cache服务器中的对象。
  • Server,真正存放缓存对象的cache服务器。
  • Cluster,服务器集群,维护一组Servers,相当于这一组servers的代理,接受putget请求,通过一定算法(普通取余或一致性哈希)把请求转发到特定的server。

首先来看看不使用一致性哈希算法的情况,会出现什么问题:

原始版本

Entry:

public class Entry {
    private String key;

    Entry(String key) {
        this.key = key;
    }

    @Override
    public String toString() {
        return key;
    }
}

Server:

public class Server {
    private String name;
    private Map<Entry, Entry> entries;

    Server(String name) {
        this.name = name;
        entries = new HashMap<Entry, Entry>();
    }

    public void put(Entry e) {
        entries.put(e, e);
    }

    public Entry get(Entry e) {
        return entries.get(e);
    }
}

Cluster:

public class Cluster {
    private static final int SERVER_SIZE_MAX = 1024;

    private Server[] servers = new Server[SERVER_SIZE_MAX];
    private int size = 0;

    public void put(Entry e) {
        int index = e.hashCode() % size;
        servers[index].put(e);
    }

    public Entry get(Entry e) {
        int index = e.hashCode() % size;
        return servers[index].get(e);
    }

    public boolean addServer(Server s) {
        if (size >= SERVER_SIZE_MAX)
            return false;

        servers[size++] = s;
        return true;
    }
}

EntryServerCluster是对这三个抽象的实现,看代码应该是非常清晰的。

其中,Cluster类是实现路由算法的类,也就是根据entry的key决定entry放入哪个server中,在最简单的实现里,直接用取余的方法:e.hashCode() % size

然后看看测试:

public class Main {

    public static void main(String[] args) {
        Cluster c = createCluster();

        Entry[] entries = {
                    new Entry("i"),
                    new Entry("have"),
                    new Entry("a"),
                    new Entry("pen"),
                    new Entry("an"),
                    new Entry("apple"),
                    new Entry("applepen"),
                    new Entry("pineapple"),
                    new Entry("pineapplepen"),
                    new Entry("PPAP")
                };

        for (Entry e : entries)
            c.put(e);

        c.addServer(new Server("192.168.0.6"));

        findEntries(c, entries);

    }

    private static Cluster createCluster() {
        Cluster c = new Cluster();
        c.addServer(new Server("192.168.0.0"));
        c.addServer(new Server("192.168.0.1"));
        c.addServer(new Server("192.168.0.2"));
        c.addServer(new Server("192.168.0.3"));
        c.addServer(new Server("192.168.0.4"));
        c.addServer(new Server("192.168.0.5"));
        return c;
    }

    private static void findEntries(Cluster c, Entry[] entries) {
        for (Entry e : entries) {
            if (e == c.get(e)) {
                System.out.println("重新找到了entry:" + e);
            } else {
                System.out.println("entry已失效:" + e);
            }
        }
    }
}

测试里,先构建一个6个服务器的集群,然后把一组entries逐个放入集群,然后向集群里添加一个新的server,看有多少个entry失效了,结果:

重新找到了entry: i
entry已失效: have
entry已失效: a
entry已失效: pen
entry已失效: an
entry已失效: apple
entry已失效: applepen
entry已失效: pineapple
entry已失效: pineapplepen
重新找到了entry: PPAP

可见,在普通取余路由算法的实现,几乎所有的entry都会被映射到新的server中,大部分缓存都失效了。

实现consistent-hashing

首先,为了servers和entries在hash环上足够分散,重写它们的hashCode方法,简单起见,复用String的hashCode算法:

public int hashCode() {
    return name.hashCode();
}

然后,就可以选择几个命名的服务器名字,确保它们不会集中在环上的某一段上。

然后,在Cluster中,用SortMap存储servers:

public class Cluster {
    private static final int SERVER_SIZE_MAX = 1024;

    private SortedMap<Integer, Server> servers = new TreeMap<Integer, Server>();
    private int size = 0;

    public boolean addServer(Server s) {
        if (size >= SERVER_SIZE_MAX)
            return false;

        servers.put(s.hashCode(), s);

        size++;
        return true;
    }
}

重写Cluster的routeServer方法:

public Server routeServer(int hash) {
    if (servers.isEmpty())
        return null;

    if (!servers.containsKey(hash)) {
        SortedMap<Integer, Server> tailMap = servers.tailMap(hash);
        hash = tailMap.isEmpty() ? servers.firstKey() : tailMap.firstKey();
    }
    return servers.get(hash);
}

这里传入的参数hash是entry的hashcode,根据entry的hashCode,向上找一个和它最接近的servers并返回。

再测试一下这个一致性hash的表现:

public class Main {
    public static void main(String[] args) {
        Cluster c = createCluster();

        Entry[] entries = {
                    new Entry("i"),
                    new Entry("have"),
                    new Entry("a"),
                    new Entry("pen"),
                    new Entry("an"),
                    new Entry("apple"),
                    new Entry("applepen"),
                    new Entry("pineapple"),
                    new Entry("pineapplepen"),
                    new Entry("PPAP")
                };

        for (Entry e : entries)
            c.put(e);

        c.addServer(new Server("1"));
        findEntries(c, entries);

    }

    private static Cluster createCluster() {
        Cluster c = new Cluster();
        c.addServer(new Server("international"));
        c.addServer(new Server("china"));
        c.addServer(new Server("japan"));
        c.addServer(new Server("Amarica"));
        c.addServer(new Server("samsung"));
        return c;
    }

    private static void findEntries(Cluster c, Entry[] entries) {
        // omitted...
    }
}

结果:

重新找到了entry: i
重新找到了entry: have
重新找到了entry: a
重新找到了entry: pen
重新找到了entry: an
重新找到了entry: apple
entry已失效: applepen
重新找到了entry: pineapple
重新找到了entry: pineapplepen
重新找到了entry: PPAP

大部分的缓存都没有失效!至此我们验证了当节点数量改变时,一致性hash能够使失效的缓存数量尽可能少。

 

更多代码(参考上面自己实现):--能直接执行

/**
 * @author zhangbaozhe
 * @date 2018/4/13
 */
public class Entry {
    private String key;

    public Entry(String key){
        this.key = key;
    }

    public String getKey() {
        return key;
    }

    public void setKey(String key) {
        this.key = key;
    }

    public int hashCode(){
        return this.key.hashCode();
    }
}

 

节点类

import java.util.HashMap;
import java.util.Map;

/**
 * @author zhangbaozhe
 * @date 2018/4/13
 */
public class Server {
    private String name;
    private Map<String , Entry > entries;

    public Server(String name){
        this.name = name;
        entries = new HashMap<String, Entry>();
    }

    public void put(Entry entry){
        entries.put(entry.getKey(),entry);
    }

    public Entry get(String key){
        return entries.get(key);
    }

    public String getName() {
        return name;
    }

    public int hashCode(){
        return this.name.hashCode();
    }
}

  

集群管理类:

import java.util.SortedMap;
import java.util.TreeMap;

/**
 * @author zhangbaozhe
 * @date 2018/4/13
 */
public class Cluster {
    private static final int SERVER_SIZE_MAX = 1024;

    private SortedMap<Integer,Server> servers = new TreeMap<>();
    private int size = 0;

    public boolean addServer(Server s){
        if(size +1 > SERVER_SIZE_MAX){
            return false;
        }
        servers.put(s.hashCode(),s);
        size++;
        return true;
    }

    public boolean removeServer(Server s) {
        if(servers!=null && servers.containsKey(s.hashCode())){
            servers.remove(s.hashCode());
            return true;
        }
        return false;
    }

    public Server route(Integer hash){
        if(servers==null){
            return null;
        }
        if(!servers.containsKey(hash)){
            SortedMap<Integer, Server> tailMap = servers.tailMap(hash);
            hash = tailMap.isEmpty()?servers.firstKey():tailMap.firstKey();
        }
        return servers.get(hash);
    }

    public Server put(Entry entry){
        Server s = route(entry.hashCode());
        s.put(entry);
        return s;
    }

    public Entry get(Entry entry){
        Server s = route(entry.hashCode());
        if(s!=null){
            return s.get(entry.getKey());
        }
        return null;
    }

    public Server getServer(Entry entry){
        Server s = route(entry.hashCode());
        if(s!=null){
            return s;
        }
        return null;
    }

}

  

测试类:

import java.util.ArrayList;
import java.util.Arrays;
import java.util.List;

/**
 * @author zhangbaozhe
 * @date 2018/4/13
 */
public class ConsistentHashMain {

    public static void main(String[] args){
        Cluster c = new Cluster();

        c.addServer(new Server("170.16.12.1"));
        c.addServer(new Server("192.168.0.1"));
        c.addServer(new Server("10.1.58.42"));
        c.addServer(new Server("172.168.12.6"));
        c.addServer(new Server("192.168.10.1"));
        c.addServer(new Server("192.168.30.1"));

        List<Entry> entryList = Arrays.asList(new Entry("a"),
                new Entry("dhergsaf"),
                new Entry("yty243u^&"),
                new Entry("34yhERQrtety$%hb"),
                new Entry("fvGFGJYHTSrrd"),
                new Entry("57g%$^U"),
                new Entry("87654rgjk876"),
                new Entry("nsdfeft"),
                new Entry("vv"));

        entryList.forEach(entry ->{
            c.put(entry);
        });


        //c.addServer(new Server("127.0.0.4"));
        //c.addServer(new Server("127.0.0.5"));
        //c.removeServer(new Server("192.168.0.1"));
        findEntry(c,entryList);
    }

    public static void findEntry(Cluster c, List<Entry> entryList){
        entryList.forEach(entry ->{
            if(c.get(entry) == entry){
                System.out.println("Found entry::"+entry.getKey()+" on Server::"+c.getServer(entry).getName());
            }else{
                System.out.println("Not Found entry::"+entry.getKey());
            }

        });
    }
}

  

  

posted @ 2018-04-13 16:10  pautcher  阅读(383)  评论(0编辑  收藏  举报