以下程序抛出以下异常:
java.lang.IllegalArgumentException: Comparison method violates its general contract!
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我理解了这个问题Comparator.请参阅无法复制:"比较方法违反了其总合同!"
我不明白为什么只有List32或更大的s 才会失败.谁能解释一下?
class Experiment {
private static final class MyInteger {
private final Integer num;
MyInteger(Integer num) {
this.num = num;
}
}
private static final Comparator<MyInteger> COMPARATOR = (r1, r2) -> {
if (r1.num == null || r2.num == null)
return 0;
return Integer.compare(r1.num, r2.num);
};
public static void main(String[] args) {
MyInteger[] array = {new MyInteger(0), new MyInteger(1), new MyInteger(null)};
Random random = new Random();
for (int length = 0;; length++) {
for (int attempt = 0; attempt < 100000; attempt++) {
List<MyInteger> list = new ArrayList<>();
int[] arr = new int[length];
for (int k = 0; k < length; k++) {
int rand = random.nextInt(3);
arr[k] = rand;
list.add(array[rand]);
}
try {
Collections.sort(list, COMPARATOR);
} catch (Exception e) {
System.out.println(arr.length + " " + Arrays.toString(arr));
return;
}
}
}
}
}
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fgb*_*fgb 13
它取决于实现,但在openjdk 8中,数组的大小是根据MIN_MERGE检查的,它等于32.这避免了对mergeLo/ mergeHi抛出异常的调用.
来自JDK/jdk/openjdk/7u40-b43 8-b132 7-b147 - 8-b132/java.util.TimSort:
Run Code Online (Sandbox Code Playgroud)static <T> void sort(T[] a, int lo, int hi, Comparator<? super T> c, T[] work, int workBase, int workLen) { assert c != null && a != null && lo >= 0 && lo <= hi && hi <= a.length; int nRemaining = hi - lo; if (nRemaining < 2) return; // Arrays of size 0 and 1 are always sorted // If array is small, do a "mini-TimSort" with no merges if (nRemaining < MIN_MERGE) { int initRunLen = countRunAndMakeAscending(a, lo, hi, c); binarySort(a, lo, hi, lo + initRunLen, c); return; } /** * March over the array once, left to right, finding natural runs, * extending short natural runs to minRun elements, and merging runs * to maintain stack invariant. */ TimSort<T> ts = new TimSort<>(a, c, work, workBase, workLen); int minRun = minRunLength(nRemaining); do { // Identify next run int runLen = countRunAndMakeAscending(a, lo, hi, c); // If run is short, extend to min(minRun, nRemaining) if (runLen < minRun) { int force = nRemaining <= minRun ? nRemaining : minRun; binarySort(a, lo, lo + force, lo + runLen, c); runLen = force; } // Push run onto pending-run stack, and maybe merge ts.pushRun(lo, runLen); ts.mergeCollapse(); // Advance to find next run lo += runLen; nRemaining -= runLen; } while (nRemaining != 0); // Merge all remaining runs to complete sort assert lo == hi; ts.mergeForceCollapse(); assert ts.stackSize == 1; }
Jun*_*sor 10
Java 8使用TimSort算法对输入进行排序.在TimSort长度至少为32时发生合并阶段.当长度低于32时,则使用可能不检测合同违规的简单排序算法.让源代码评论TimSort.java为自己说话:
class TimSort<T> {
/**
* This is the minimum sized sequence that will be merged. Shorter
* sequences will be lengthened by calling binarySort. If the entire
* array is less than this length, no merges will be performed.
*
* This constant should be a power of two. It was 64 in Tim Peter's C
* implementation, but 32 was empirically determined to work better in
* this implementation. In the unlikely event that you set this constant
* to be a number that's not a power of two, you'll need to change the
* {@link #minRunLength} computation.
*
* If you decrease this constant, you must change the stackLen
* computation in the TimSort constructor, or you risk an
* ArrayOutOfBounds exception. See listsort.txt for a discussion
* of the minimum stack length required as a function of the length
* of the array being sorted and the minimum merge sequence length.
*/
private static final int MIN_MERGE = 32;
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