Trying to write a regex matcher for roman numerals. In sed (which I think is considered 'standard' for regex?), if you have multiple options delimited by the alternation operator, it will match the longest. Namely, "I|II|III|IV" will match "IV" for "IV" and "III" for "III"
In Java, the same pattern matches "I" for "IV" and "I" for "III". Turns out Java chooses between alternation matches left-to-right; that is, because "I" appears before "III" in the regex, it matches. If I change the regex to "IV|III|II|I", the behavior is corrected, but this obviously isn't a solution in general.
Is there a way to make Java choose the longest match out of an alternation group, instead of choosing the 'first'?
A code sample for clarity:
public static void main(String[] args)
{
Pattern p = Pattern.compile("six|sixty");
Matcher m = p.matcher("The year was nineteen sixty five.");
if (m.find())
{
System.out.println(m.group());
}
else
{
System.out.println("wtf?");
}
}
This outputs "six"
No, it's behaving correctly. Java uses an NFA, or regex-directed flavor, like Perl, .NET, JavaScript, etc., and unlike sed, grep, or awk. An alternation is expected to quit as soon as one of the alternatives matches, not hold out for the longest match.
You can force it to continue by adding a condition after the alternation that can't be met until the whole token has been consumed. What that condition might be depends on the context; the simplest option would be an anchor ($) or a word boundary (\b).
"\\b(I|II|III|IV)\\b"
EDIT: I should mention that, while grep, sed, awk and others traditionally use text-directed (or DFA) engines, you can also find versions of some of them that use NFA engines, or even hybrids of the two.