Pascal Schumacher created LANG-1241:
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Summary: StringUtils.ordinalIndexOf broken
Key: LANG-1241
URL: https://issues.apache.org/jira/browse/LANG-1241
Project: Commons Lang
Issue Type: Bug
Components: lang.*
Affects Versions: 3.4
Reporter: Pascal Schumacher
Quoting rousej from the discussion of
[https://github.com/apache/commons-lang/pull/93]
I agree with britter. StringUtils.ordinalIndexOf("aaaaaa", "aa", 2) == 1 would
be true because the sequence 'aa' is at every index. There seems to be
confusion around this method, but it seems to me the original code had it
correct.
{code:java}
int index = lastIndex ? str.length() : INDEX_NOT_FOUND;
do {
if (lastIndex) {
index = CharSequenceUtils.lastIndexOf(str, searchStr, index - 1);
} else {
index = CharSequenceUtils.indexOf(str, searchStr, index + 1);
}
{code}
I'm not sure why StringUtils.ordinalIndexOf("aaaaaa", "aa", 2) == 3 would ever
be true.
StringUtils.ordinalIndexOf("aaaaaa", "aa", 2) == 2 is easier to see where it's
coming from, but StringUtils.ordinalIndexOf("aaaaaa", "aa", 2) == 1 is still
the correct answer, and it also works for other tests. I found my way here
because I wanted to use this method in a project, but found that the current
release is broken by commit
[e5a3039|https://github.com/apache/commons-lang/commit/e5a3039f7a1e727fca40db7357a9191b6a7cf41d]
. With the current release if the first index is between 0 and
searchStr.length() -1 the method will return the index for ordinal + 1.....in
other words the wrong index.
This fact is missed in a test like
assertEquals(3, StringUtils.ordinalIndexOf("aaaaaa", "aa", 2));
/* Note the above test from commit
[e5a3039|https://github.com/apache/commons-lang/commit/e5a3039f7a1e727fca40db7357a9191b6a7cf41d]
is incorrect since the array of chars begins at 0.*/
The following test should pass, but will fail in the current release due to the
broken method:
{code:java}
assertEquals(0, StringUtils.ordinalIndexOf("abaabaab", "ab", 1);
assertEquals(3, StringUtils.ordinalIndexOf("abaabaab", "ab", 2);
assertEquals(6, StringUtils.ordinalIndexOf("abaabaab", "ab", 3);
{code}
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