为什么containsKey没找到钥匙？

Question

我有以下代码:

payoffs2exchanges.put(point, exchange);
if (!payoffs2exchanges.containsKey(point) ) {
   game.log.fine("yes");
} else {
   game.log.fine("no");
}

它输出"否".换句话说,我将键值对添加到地图中,然后,在此之后我立即检查密钥是否存在并发现它不存在.为什么？

我仍然有钥匙的问题.以下代码说每次添加密钥时我都会添加一个新密钥.我知道事实并非如此.

        Integer[] point = new Integer[2];
        point[0] = proposerBestScore;
        point[1] = responderBestScore;
        game.log.fine("In the getCloudOfPayoffs: found payoffs:" + point[0] + "," + point[1] + ". Exchange: " + exchange[0]+","+exchange[1]+","+exchange[2]+","+exchange[3]+","+exchange[4]);
        // With the following block we ensure that every options (pair of payoffs) is represented by exchange with minimal number of moves.
        if (!payoffs2exchanges.containsKey(point)) {
            payoffs2exchanges.put(point, exchange); 
            game.log.fine("In the getCloudOfPayoffs: this option is new. We add it to the map.");
        } else {
            game.log.fine("In the getCloudOfPayoffs: this option is old.");
            Integer[] exchangeFromMap = payoffs2exchanges.get(point);
            Integer newSum = 0;
            Integer oldSum = 0;
            for (int i=0;i<Design.nColors;i++) {
                newSum = newSum + Math.abs(exchange[i]);
                oldSum = oldSum + Math.abs(exchangeFromMap[i]);
            }
            if (newSum<oldSum) {
                game.log.fine("In the getCloudOfPayoffs: the new exchange is better than the old one.");
                payoffs2exchanges.put(point, exchange);
            }
        }

Answer 1

它正在做正确的事情.containsKey返回true,!运算符否定它false,因此它输出no(else子句).

Answer 2

你Integer[]在地图中使用了一个键.这是一件坏事,因为Java数组没有实现equals,hashCode正如您所料.看这个例子:

public class Test {
    public static void main(String[] args) {
        Integer[] arr1 = { 1, 2 };
        Integer[] arr2 = { 1, 2 };

        System.out.println(arr1.equals(arr2));
        System.out.println(arr1.hashCode() + " / " + arr2.hashCode());
    }
}

在我的电脑上打印:

false
1476323068 / 535746438

我的建议是创建一个自定义Point,妥善覆盖类equals和hashCode(或可能重用java.awt.Point,如果你认为这是有道理的).