接上篇http://www.javashuo.com/article/p-kxvfgusz-ez.htmljava
加权轮询,我第一次没理解,我的以为很差理解。因而先仿照源码抽象出逻辑模型,代码以下:算法
public static void main(String[] args) { //存储调用的方法和总的调用次数的map final ConcurrentMap<String, AtomicInteger> sequences = new ConcurrentHashMap<String, AtomicInteger>(); //统计invoker 被调用次数map final ConcurrentMap<String, AtomicInteger> result = new ConcurrentHashMap<String, AtomicInteger>(); //模拟方法三个invoker 分别为 a,b,c List<String> invokes=new ArrayList<String>(3); invokes.add("a"); invokes.add("b"); invokes.add("c"); //存储invoker和权重的对应map final LinkedHashMap<String, AtomicInteger> invokerToWeightMap = new LinkedHashMap<String,AtomicInteger>(); for(int i=0;i<21;i++){ //每次调用都把模拟的权重从新放入 invokerToWeightMap.put("a",new AtomicInteger(3)); invokerToWeightMap.put("b",new AtomicInteger(6)); invokerToWeightMap.put("c",new AtomicInteger(9)); select(invokes,invokerToWeightMap,sequences,result); } //打印调用结果统计 for(Map.Entry<String, AtomicInteger> r : result.entrySet()){ System.out.println(r.getKey()+"被调用:"+r.getValue()+"次"); } } private static void select(List<String> invokes, LinkedHashMap<String, AtomicInteger> invokerToWeightMap, ConcurrentMap<String, AtomicInteger> sequences, ConcurrentMap<String, AtomicInteger> result){ //假设调用servcie.hello方法 AtomicInteger sequence = sequences.get("service.hello"); if (sequence == null) { //默认调用次数为0 sequences.putIfAbsent("servcie.hello", new AtomicInteger(0)); sequence = sequences.get("servcie.hello"); } //调用次数+1 int currentSequence = sequence.getAndIncrement(); System.out.print("currentSequence:" + currentSequence); int maxWeight=9;//最大权重 int minWeight=3;//最小权重 int weightSum=18;//总权重 if (maxWeight > 0 && minWeight < maxWeight) { // 走权重不同逻辑。 int mod = currentSequence % weightSum; System.out.print(" mod:" + mod); for (int i = 0; i < maxWeight; i++) { for (Map.Entry<String, AtomicInteger> each : invokerToWeightMap.entrySet()) { final String k = each.getKey(); final AtomicInteger v = each.getValue(); if (mod == 0 && v.intValue() > 0) { System.out.println(" selected:"+k); AtomicInteger count = result.get(k); if (count == null) { result.putIfAbsent(k, new AtomicInteger(1)); }else{ count.incrementAndGet(); } return; } if (v.intValue() > 0) { v.decrementAndGet(); mod--; } } } } }
输出结果,分析:app
currentSequence:0 mod:0 selected:a
currentSequence:1 mod:1 selected:b
currentSequence:2 mod:2 selected:c
currentSequence:3 mod:3 selected:a
currentSequence:4 mod:4 selected:b
currentSequence:5 mod:5 selected:c
currentSequence:6 mod:6 selected:a
currentSequence:7 mod:7 selected:b
currentSequence:8 mod:8 selected:c
//前9次调用一直是简单轮询
currentSequence:9 mod:9 selected:b
currentSequence:10 mod:10 selected:cthis
currentSequence:11 mod:11 selected:b
currentSequence:12 mod:12 selected:curl
currentSequence:13 mod:13 selected:b
currentSequence:14 mod:14 selected:c
//10到16次调用只在b c间轮询
currentSequence:15 mod:15 selected:c
currentSequence:16 mod:16 selected:c
currentSequence:17 mod:17 selected:c
//17到19次调用只调c
currentSequence:18 mod:0 selected:a
currentSequence:19 mod:1 selected:b
currentSequence:20 mod:2 selected:c
//因为mod取值归零,20到21次新一轮的轮询
最后调用总结:
a被调用:4次
b被调用:7次
c被调用:10次.net
体现出加权轮询,这就是duboo的加权轮询算法。code
理解上面的代码,再看源代码,就容易理解不少。
源码以下:blog
public static final String NAME = "roundrobin"; private final ConcurrentMap<String, AtomicPositiveInteger> sequences = new ConcurrentHashMap<String, AtomicPositiveInteger>(); protected <T> Invoker<T> doSelect(List<Invoker<T>> invokers, URL url, Invocation invocation) { String key = invokers.get(0).getUrl().getServiceKey() + "." + invocation.getMethodName(); int length = invokers.size(); // 总个数 int maxWeight = 0; // 最大权重 int minWeight = Integer.MAX_VALUE; // 最小权重 final LinkedHashMap<Invoker<T>, IntegerWrapper> invokerToWeightMap = new LinkedHashMap<Invoker<T>, IntegerWrapper>(); int weightSum = 0; for (int i = 0; i < length; i++) { int weight = getWeight(invokers.get(i), invocation); maxWeight = Math.max(maxWeight, weight); // 累计最大权重 minWeight = Math.min(minWeight, weight); // 累计最小权重 if (weight > 0) { invokerToWeightMap.put(invokers.get(i), new IntegerWrapper(weight)); weightSum += weight; } } AtomicPositiveInteger sequence = sequences.get(key); if (sequence == null) { sequences.putIfAbsent(key, new AtomicPositiveInteger()); sequence = sequences.get(key); } //总的调用次数 int currentSequence = sequence.getAndIncrement(); if (maxWeight > 0 && minWeight < maxWeight) { // 权重不同 //取模操做,保证mod值域在[0,weightSum) int mod = currentSequence % weightSum; for (int i = 0; i < maxWeight; i++) {//maxWeight选最大权重,保证加上子循环 // 即 maxWeight*(invoker个数)>mod 值。这个mod就能够减到0 for (Map.Entry<Invoker<T>, IntegerWrapper> each : invokerToWeightMap.entrySet()) { final Invoker<T> k = each.getKey(); final IntegerWrapper v = each.getValue(); //在一次选择过程后(mod--,权重--,mod==0结束),选下一个权重大于0的 if (mod == 0 && v.getValue() > 0) { return k; } if (v.getValue() > 0) { //基于mod,来讲以前选过的,权重-1.mod-- v.decrement(); mod--; } } } } // 权重同样 取模轮循 简单轮询 return invokers.get(currentSequence % length); } private static final class IntegerWrapper { private int value; public IntegerWrapper(int value) { this.value = value; } public int getValue() { return value; } public void setValue(int value) { this.value = value; } public void decrement() { this.value--; } }
我理解,这个算法的思想,就是让权重小的候选者,经过-1 老是很快较早的不参与某个轮次的选择。ci