Github user ahgittin commented on a diff in the pull request:
https://github.com/apache/brooklyn-server/pull/740#discussion_r124509326
--- Diff:
utils/common/src/main/java/org/apache/brooklyn/util/text/NaturalOrderComparator.java
---
@@ -140,6 +147,32 @@ public int compare(String a, String b) {
if ((result = compareRight(a.substring(ia),
b.substring(ib))) != 0) {
return result;
}
+ // numeric portion is the same; previously we incremented
and checked again
+ // which is inefficient (we keep checking remaining
numbers here);
+ // also if we want to compare on 0's after comparing
remainder of string
+ // we need to recurse here after skipping the digits we
just checked above
+ do {
+ ia++; ib++;
+ boolean aDigitsDone = ia >= a.length() ||
!Character.isDigit(charAt(a, ia));
+ boolean bDigitsDone = ib >= b.length() ||
!Character.isDigit(charAt(b, ib));
+ if (aDigitsDone != bDigitsDone) {
+ // sanity check
+ throw new IllegalStateException("Digit sequence
lengths should have matched, '"+a+"' v '"+b+"'");
+ }
+ if (aDigitsDone) break;
+ } while (true);
+ if (ia < a.length() || ib < b.length()) {
+ if (ia >= a.length()) return -1; // only b has more
chars
+ if (ib >= b.length()) return 1; // only a has more
chars
+ // both have remaining chars; neither is numeric due
to compareRight; recurse into remaining
+ if ((result = compare(a.substring(ia),
b.substring(ib))) != 0) {
+ return result;
+ }
+ }
+ // numbers are equal value, remaining string compares
identical;
+ // comes down to whether there are any leading zeroes here
+ return nzb-nza;
+
} else if ((Character.isDigit(ca) || nza>0) &&
(Character.isDigit(cb) || nzb>0)) {
// both sides are numbers, but at least one is a sequence
of zeros
if (nza==0) {
--- End diff --
his `charAt` returns 0 if we've gone beyond the string length, i think it
might rely on that
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