每次使用.put()时都会重写HashMap

我的申请与股票市场有关.我有一个不断更新名为Price的对象的Feed.Price有一个存储安全代码(String)和价格(Double)的HashMap.每当新价格出现时,此对象都会更新.

该应用程序应该扫描大幅移动的价格.我有一个名为Poller的独立类,它每秒轮询Price对象并拍摄价格的快照.快照是如上所述的HashMap.然后我想将这个HashMap的价格与pollNumber一起存储在另一个HashMap中我以后可以传递pollNumber并在与该pollNumber对应的时间输出价格.

但相反,我得到所有以前的价格被覆盖,输出类似于下面的价格.

0:{MSFT = 17.67,AAPL = 93.85,GOOG = 333.86} {0 = {MSFT = 17.67,AAPL = 93.85,GOOG = 333.86}}

1:{MSFT = 17.64,AAPL = 93.85,GOOG = 334.02} {0 = {MSFT = 17.64,AAPL = 93.85,GOOG = 334.02},1 = {MSFT = 17.64,AAPL = 93.85,GOOG = 334.02}}

2:{MSFT = 17.64,AAPL = 93.85,GOOG = 334.08} {0 = {MSFT = 17.64,AAPL = 93.85,GOOG = 334.08},1 = {MSFT = 17.64,AAPL = 93.85,GOOG = 334.08},2 = {MSFT = 17.64,AAPL = 93.85,GOOG = 334.08}}

3:{MSFT = 17.65,AAPL = 93.83,GOOG = 334.08} {0 = {MSFT = 17.65,AAPL = 93.83,GOOG = 334.08},1 = {MSFT = 17.65,AAPL = 93.83,GOOG = 334.08},2 = {MSFT = 17.65,AAPL = 93.83,GOOG = 334.08},3 = {MSFT = 17.65,AAPL = 93.83,GOOG = 334.08}}

4:{MSFT = 17.64,AAPL = 93.83,GOOG = 334.07} {0 = {MSFT = 17.64,AAPL = 93.83,GOOG = 334.07},1 = {MSFT = 17.64,AAPL = 93.83,GOOG = 334.07},2 = {MSFT = 17.64,AAPL = 93.83,GOOG = 334.07},3 = {MSFT = 17.64,AAPL = 93.83,GOOG = 334.07},4 = {MSFT = 17.64,AAPL = 93.83,GOOG = 334.07}}

正如您所看到的那样,当我打印出应该具有不同价格序列的整个HashMap时,它们都是相同的.

基本上.put()函数以某种方式覆盖旧条目.

如果您知道如何修复该行为,以便HashMap(大的)每次都有一个新的价格序列条目.

public class Poller {

    private final int period=1000;
    private final int delay=1000;

    private static int pollNumber=0;
    private static HashMap> 
        polledPrice = new HashMap>();

    public void pollPrice(){

    Timer timer = new Timer();
    timer.scheduleAtFixedRate(new TimerTask() {
                public void run() {
                //    System.out.println(Price.getPricesMap());
System.out.println(pollNumber+" : "+Price.getPricesMap());
                    polledPrice.put(pollNumber, Price.getPricesMap());
                    System.out.println(polledPrice);
                    pollNumber = pollNumber+1;

                    Time atime = new Time();
                        atime.addToTimeMap(pollNumber);

                }
            }, delay, period);
    }
}

Nick Fortesc.. 6

您需要复制HashMap,否则看起来您只是一遍又一遍地存储相同的Map,这当然会被覆盖.使用此行:

polledPrice.put(pollNumber, new HashMap(Price.getPricesMap()));

作为最简单的修复.

您需要复制HashMap,否则看起来您只是一遍又一遍地存储相同的Map,这当然会被覆盖.使用此行:

polledPrice.put(pollNumber, new HashMap(Price.getPricesMap()));

作为最简单的修复.