Java类型系统中的一个bug？

Question

想象一下有一个主类——模拟器——它使用另外两个类——生产者和评价者，分别实现接口IProducer和IEEvaluator。

IProducer实现产生结果，而IEEvaluator实现评估这些结果。模拟器通过查询IProducer实现，然后将结果传递给IEEvaluator实例，来控制执行流。

生产者和评估者的实际实现在运行时是已知的，在编译时我只知道他们的接口。检查下面的示例。

package com.test;

import java.util.ArrayList;
import java.util.HashMap;
import java.util.List;
import java.util.Map;
import java.util.Map.Entry;

/**
 * Producers produce results. I do not care what their actual type is, but the
 * values in the map have to be comparable amongst themselves.
 */
interface IProducer<T extends Comparable<T>> {
    public Map<Integer, T> getResults();
}

/**
 * This example implementation ranks items in the map by using Strings.
 */
class ProducerA implements IProducer<String> {
    @Override
    public Map<Integer, String> getResults() {
        Map<Integer, String> result = new HashMap<Integer, String>();
        result.put(1, "A");
        result.put(2, "B");
        result.put(3, "B");

        return result;
    }
}

/**
 * This example implementation ranks items in the map by using integers.
 */
class ProducerB implements IProducer<Integer> {
    @Override
    public Map<Integer, Integer> getResults() {
        Map<Integer, Integer> result = new HashMap<Integer, Integer>();
        result.put(1, 10);
        result.put(2, 30);
        result.put(3, 30);

        return result;
    }
}

/**
 * Evaluator evaluates the results against the given groundTruth. All it needs
 * to know about results, is that they are comparable amongst themselves.
 */
interface IEvaluator {
    public <T extends Comparable<T>> double evaluate(Map<Integer, T> results,
            Map<Integer, Double> groundTruth);
}

/**
 * This is example of an evaluator, metric Kendall Tau-B. Don't bother with
 * semantics, all that matters is that I want to be able to call
 * r1.compareTo(r2) for every (r1, r2) that appear in Map<Integer, T> results.
 */
class KendallTauB implements IEvaluator {
    @Override
    public <T extends Comparable<T>> double evaluate(Map<Integer, T> results,
            Map<Integer, Double> groundTruth) {
        int concordant = 0, discordant = 0, tiedRanks = 0, tiedCapabilities = 0;

        for (Entry<Integer, T> rank1 : results.entrySet()) {
            for (Entry<Integer, T> rank2 : results.entrySet()) {
                if (rank1.getKey() < rank2.getKey()) {
                    final T r1 = rank1.getValue();
                    final T r2 = rank2.getValue();
                    final Double c1 = groundTruth.get(rank1.getKey());
                    final Double c2 = groundTruth.get(rank2.getKey());

                    final int ranksDiff = r1.compareTo(r2);
                    final int actualDiff = c1.compareTo(c2);

                    if (ranksDiff * actualDiff > 0) {
                        concordant++;
                    } else if (ranksDiff * actualDiff < 0) {
                        discordant++;
                    } else {
                        if (ranksDiff == 0)
                            tiedRanks++;

                        if (actualDiff == 0)
                            tiedCapabilities++;
                    }
                }
            }
        }

        final double n = results.size() * (results.size() - 1d) / 2d;

        return (concordant - discordant)
                / Math.sqrt((n - tiedRanks) * (n - tiedCapabilities));
    }
}

/**
 * The simulator class that queries the producer and them conveys results to the
 * evaluator.
 */
public class Simulator {
    public static void main(String[] args) {
        // example of a ground truth
        Map<Integer, Double> groundTruth = new HashMap<Integer, Double>();
        groundTruth.put(1, 1d);
        groundTruth.put(2, 2d);
        groundTruth.put(3, 3d);

        // dynamically load producers
        List<IProducer<?>> producerImplementations = lookUpProducers();

        // dynamically load evaluators
        List<IEvaluator> evaluatorImplementations = lookUpEvaluators();

        // pick a producer
        IProducer<?> producer = producerImplementations.get(0);

        // pick an evaluator
        IEvaluator evaluator = evaluatorImplementations.get(0);

        // evaluate the result against the ground truth
        double score = evaluator.evaluate(producer.getResults(), groundTruth);

        System.out.printf("Score is %.2f\n", score);
    }

    // Methods below are for demonstration purposes only. I'm actually using
    // ServiceLoader.load(Clazz) to dynamically discover and load classes that
    // implement interfaces IProducer and IEvaluator
    public static List<IProducer<?>> lookUpProducers() {
        List<IProducer<?>> producers = new ArrayList<IProducer<?>>();
        producers.add(new ProducerA());
        producers.add(new ProducerB());

        return producers;
    }

    public static List<IEvaluator> lookUpEvaluators() {
        List<IEvaluator> evaluators = new ArrayList<IEvaluator>();
        evaluators.add(new KendallTauB());

        return evaluators;
    }
}

这段代码编译时没有警告，并且运行方式也应该如此。这是我之前提出的问题的答案，所以这是一个后续问题。

使用上面的代码，假设您想将producer.get结果（）调用的结果存储在一个变量中（稍后将在调用evaluator.evaluate（结果，地真相）调用中使用）。这个变量的类型是什么？

地图

public static void main(String[] args) {
    // example of a ground truth
    Map<Integer, Double> groundTruth = new HashMap<Integer, Double>();
    groundTruth.put(1, 1d);
    groundTruth.put(2, 2d);
    groundTruth.put(3, 3d);

    // dynamically load producers
    List<IProducer<?>> producerImplementations = lookUpProducers();

    // dynamically load evaluators
    List<IEvaluator> evaluatorImplementations = lookUpEvaluators();

    // pick a producer
    IProducer<?> producer = producerImplementations.get(0);

    // pick an evaluator
    IEvaluator evaluator = evaluatorImplementations.get(0);

    // evaluate the result against the ground truth
    Map<Integer, ?> data = producer.getResults(); // this type works
    double score = evaluator.evaluate(data, groundTruth); // but now this call does not


    System.out.printf("Score is %.2f\n", score);
}

好像是制片人。getResults（）返回不能用Java静态表示的内容。这是虫子，还是我遗漏了什么？

Answer 1

这不是一个bug，而是一个限制。在类型系统社区中，众所周知，带有通配符的Java具有不能用Java语法表示的类型。您的示例展示了这样一种情况，并证明了通配符本质上与F-有界多态性不兼容（即T形式的类型参数扩展了某些内容）

坦率地说，通配符是一种可怕的系统黑客。它们不应该被放到Java中。人们真正想要的，以及使您的示例易于表达的，是适当的存在类型（其中通配符是一种有限的特殊变体）。不幸的是，Java没有它们（尽管Scala有）。

Answer 2

在我回答之前，请注意：您所有的T扩展都是可比较的

这并不是Java类型系统的一个真正的“bug”，它只是一个不便之处。Java并不特别喜欢将交集类型作为类型声明的一部分。

我发现解决这个问题的一种方法是创建一种“不安全”的方法，这种方法在正常情况下不应使用：

@SuppressWarnings("unchecked")
private static <T extends Comparable<? super T>> Map<Integer, T> cast(Map<Integer, ?> map) {
    return (Map<Integer, T>) map;
}

只要确保使用实际上是地图的Map调用此方法

使用此方法，您可以执行以下操作：

IProducer<?> producer = ...
IEvaluator evaluator = ...
Map<Integer, ?> product = producer.getResults();
evaluator.evaluate(cast(product), truth);

然后Java将自动为您推断正确的类型参数。

而且，I前缀在Java社区中通常是不受欢迎的。