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我正在使用以下引发 IllegalArgumentException 的代码：

// Copyright (c) 2003-2014, Jodd Team (jodd.org). All Rights Reserved.

package jodd.util;

import java.util.Comparator;

/**
 * Compares two strings in natural, alphabetical, way.
 */
public class NaturalOrderComparator<T> implements Comparator<T> {

    protected final boolean ignoreCase;

    public NaturalOrderComparator() {
        ignoreCase = false;
    }

    public NaturalOrderComparator(boolean ignoreCase) {
        this.ignoreCase = ignoreCase;
    }

    /**
     * Compare digits at certain position in two strings.
     * The longest run of digits wins. That aside, the greatest
     * value wins.
     */
    protected int compareDigits(String str1, int ndx1, String str2, int ndx2) {
        int bias = 0;

        while (true) {
            char char1 = charAt(str1, ndx1);
            char char2 = charAt(str2, ndx2);

            boolean isDigitChar1 = CharUtil.isDigit(char1);
            boolean isDigitChar2 = CharUtil.isDigit(char2);

            if (!isDigitChar1 && !isDigitChar2) {
                return bias;
            }
            if (!isDigitChar1) {
                return -1;
            }
            if (!isDigitChar2) {
                return 1;
            }

            if (char1 < char2) {
                if (bias == 0) {
                    bias = -1;
                }
            } else if (char1 > char2) {
                if (bias == 0) {
                    bias = 1;
                }
            } else if (char1 == 0 && char2 == 0) {
                return bias;
            }

            ndx1++;
            ndx2++;
        }
    }

    public int compare(T o1, T o2) {
        String str1 = o1.toString();
        String str2 = o2.toString();

        int ndx1 = 0, ndx2 = 0;
        int zeroCount1, zeroCount2;
        char char1, char2;

        int result;

        while (true) {
            // only count the number of zeroes leading the last number compared
            zeroCount1 = zeroCount2 = 0;

            char1 = charAt(str1, ndx1);
            char2 = charAt(str2, ndx2);

            // skip over leading spaces or zeros in both strings

            while (Character.isSpaceChar(char1) || char1 == '0') {
                if (char1 == '0') {
                    zeroCount1++;
                } else {
                    zeroCount1 = 0; // counts only last 0 prefixes, space char interrupts the array of 0s
                }
                ndx1++;
                char1 = charAt(str1, ndx1);
            }

            while (Character.isSpaceChar(char2) || char2 == '0') {
                if (char2 == '0') {
                    zeroCount2++;
                } else {
                    zeroCount2 = 0;
                }
                ndx2++;
                char2 = charAt(str2, ndx2);
            }

            // process digits

            boolean isDigitChar1 = CharUtil.isDigit(char1);
            boolean isDigitChar2 = CharUtil.isDigit(char2);

            if (isDigitChar1 && isDigitChar2) {
                result = compareDigits(str1, ndx1, str2, ndx2);
                if (result != 0) {
                    // not equals, return
                    return result;
                }
                // equal numbers
                if (zeroCount1 != zeroCount2) {
                    return zeroCount1 - zeroCount2;
                }
            }

            if (char1 == 0 && char2 == 0) {
                // the end; the strings are the same, maybe compare ascii?
                return zeroCount1 - zeroCount2;
            }

            // check when one of the numbers is just zeros
            if (isDigitChar1 || isDigitChar2) {
                if (zeroCount1 != zeroCount2) {
                    return zeroCount2 - zeroCount1;
                }
            }

            // checks when both numbers are zero
            if (zeroCount1 != zeroCount2) {
                return zeroCount1 - zeroCount2;
            }

            // compare chars
            if (ignoreCase) {
                char1 = Character.toLowerCase(char1);
                char2 = Character.toLowerCase(char2);
            }
            if (char1 < char2) {
                return -1;
            }
            if (char1 > char2) {
                return 1;
            }

            ndx1++;
            ndx2++;
        }
    }

    /**
     * Safe charAt.
     */
    private static char charAt(String s, int i) {
        if (i >= s.length()) {
            return 0;
        }
        return s.charAt(i);
    }
}

抛出异常：

java.lang.IllegalArgumentException: Comparison method violates its general contract!
    at java.util.TimSort.mergeLo(TimSort.java:747)
    at java.util.TimSort.mergeAt(TimSort.java:483)
    at java.util.TimSort.mergeCollapse(TimSort.java:410)
    at java.util.TimSort.sort(TimSort.java:214)
    at java.util.TimSort.sort(TimSort.java:173)
    at java.util.Arrays.sort(Arrays.java:659)
    at java.util.Collections.sort(Collections.java:217)

这由以下函数调用：

@Override
    public int compare(final T o1, final T o2) {
        int result;
        final MyObject obj1 = (MyObject) o1;
        final MyObject obj2 = (MyObject) o2;

                   return     result = compareStringId(obj1.getStringId(),obj2.getStringId());           

    }

private int compareStringId(final String Id1, final String Id2) {
        return super.compare((T) Id1, (T) Id2);
    }

它在我们的本地机器上运行良好，但是在生产中失败了，我们无法连接到那台机器来找出原因。你能帮忙吗

4

1 回答 1

2

问题在于Comparator. 根据 Java 文档，aComparator必须是自反的和传递的。在这种情况下，不能保证传递性。之前的 Java 8 不是一个大问题，即排序实现 ( MergeSort) 不会抛出异常。Java8 将默认排序实现更改TimSort为对具有无效合同的比较器更加敏感，因此它可能会引发异常。

但是，这对您解决问题没有多大帮助。如何在这里查看同一类的最新版本。它已升级以支持比较器合约，而且它在某些边缘情况下效果更好，更不用说对口音的初始支持了。

于 2017-01-20T08:15:05.003 回答