Collections.sort（）引发比较方法违反了它的一般约定！例外 [英] Collections.sort() throws Comparison method violates its general contract! exception

问题描述

我正在尝试对List对象进行排序，并且抛出此异常（仅适用于大型列表）

I'm trying to sort a List<> object and I get this exception thrown (for large lists only though)

排序代码：

List<FinalSentence> sentenceList = finalRepresentation.getSentences();
Collections.sort(sentenceList); // <=== EXCEPTION THROWN HERE!!!

FinalSentence类标题：

FinalSentence class header:

public class FinalSentence implements Comparable<FinalSentence>{...}

compareTo （）实现：

compareTo() implementation:

@Override
public int compareTo(FinalSentence o) {
    if (this == o) {
        return 0;
    }
    if (this.score > o.score) {
        return 1;
    }
    if (this.score < o.score) {
        return -1;
    }
    return 0;
}

这是一个例外：

Exception in thread "main" java.lang.IllegalArgumentException: Comparison method violates its general contract!
at java.util.ComparableTimSort.mergeHi(Unknown Source)
at java.util.ComparableTimSort.mergeAt(Unknown Source)
at java.util.ComparableTimSort.mergeCollapse(Unknown Source)
at java.util.ComparableTimSort.sort(Unknown Source)
at java.util.ComparableTimSort.sort(Unknown Source)
at java.util.Arrays.sort(Unknown Source)
at java.util.Collections.sort(Unknown Source)
at feature.finalRepresentation.Summarizer.summarize(Summarizer.java:30)
at driver.Driver.main(Driver.java:114)

对于一个小的列表（少于50个元素）有效。对于较大的列表（它也应该与之配合使用），则会引发此异常。
List的实例类型是ArrayList，并不重要。

for a small list (less than 50 elements) it works. for a large list (it's supposed to work with those as well) it throws this exception. The instance type of the List is ArrayList, not that it should matter.

我不知道如何深入了解这一点。列表已满，元素是同一类型（那里没有多态性），但是对于大型列表，我却得到了这个怪异的例外。

I have no idea how to get to the bottom of this. The list is full, the elements are of the same type (no polymorphism there) and yet I get this weird exception for large lists.

有什么想法吗？

谢谢！

推荐答案

根据 OP的评论，我的建议使用

Double.compare(score, o.score)

解决了该问题。我的猜测是±0 s或 NaN s有问题。实际上，如果您查看 Double.compare（）的来源，您会发现它比您想象的要复杂一些，并专门处理以下情况：

fixed the issue. My guess is that there was either a problem with ±0s or NaNs. In fact, if you look at the source of Double.compare(), you will find that it's slightly more complicated than you might think, and treats these cases specifically:

958    public static int compare(double d1, double d2) {
959        if (d1 < d2)
960            return -1;           // Neither val is NaN, thisVal is smaller
961        if (d1 > d2)
962            return 1;            // Neither val is NaN, thisVal is larger
963
964        long thisBits = Double.doubleToLongBits(d1);
965        long anotherBits = Double.doubleToLongBits(d2);
966
967        return (thisBits == anotherBits ?  0 : // Values are equal
968                (thisBits < anotherBits ? -1 : // (-0.0, 0.0) or (!NaN, NaN)
969                 1));                          // (0.0, -0.0) or (NaN, !NaN)
970    }

（源）

道德是：比较双打时要小心！ ：）

Moral is: be careful when comparing doubles! :)

参考：

• Double.compare（）

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