Regex pattern matching behavior for basic implementation

Question

I'm attempting to count the number of syllables in a given word using regex. I've referred to some other posts on stack overflow but am getting strange behavior when I run my program.

Here is what I'm attempting

int n=0;
Pattern p = Pattern.compile("[aeiouy]+[^$e]");
Matcher m = p.matcher("Harry");

while(m.find()) {
     n++;
}

System.out.println(n);

Now, it's printing out 1. But there is both an "a" and "y" within the string. Where have I gone wrong?

Community · Accepted Answer · 2017-05-23 12:33:12Z

2

At the end of the String there are no more letters to match so remove the expression [^$e]

Pattern p = Pattern.compile("[aeiouy]+");

Although from your comment you seem to want to treat e as a special case. You could do

Pattern p = Pattern.compile("[aiouy]|(?!^)e(?<!$)");

edited May 23, 2017 at 12:33

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answered Oct 1, 2016 at 15:33

Reimeus

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4 Comments

ClownInTheMoon Over a year ago

The purpose of the [^$e] was to ignore any occurrences of "e" at the end of the String.

Titus Over a year ago

@ClownInTheMoon if that is what you want to do, you should be using [^e]$ instead of [^$e]. Also, [aeiouy]+ creates groups of one or more vowels because of the + at the end. I'm not sure if that is what you're looking for.

ClownInTheMoon Over a year ago

Intersting. When I use '[^e]$' then the value of 'n' comes out as zero. Yes I'm looking for groups of one or more vowels and then attempting to not count silent e's.

Reimeus Over a year ago

Possibly what you're really looking for is negative assertions. See the update

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