0xCAFEBABE
0xCAFEBABE

Reputation: 5666

Why doesn't this regex work as expected in Java?

trivial regex question (the answer is most probably Java-specific):

"#This is a comment in a file".matches("^#")

This returns false. As far as I can see, ^ means what it always means and # has no special meaning, so I'd translate ^# as "A '#' at the beginning of the string". Which should match. And so it does, in Perl:

perl -e "print '#This is a comment'=~/^#/;"

prints "1". So I'm pretty sure the answer is something Java specific. Would somebody please enlighten me?

Thank you.

Upvotes: 9

Views: 4273

Answers (3)

Joachim Sauer
Joachim Sauer

Reputation: 308001

Matcher.matches() checks to see if the entire input string is matched by the regex.

Since your regex only matches the very first character, it returns false.

You'll want to use Matcher.find() instead.

Granted, it can be a bit tricky to find the concrete specification, but it's there:

Upvotes: 17

codaddict
codaddict

Reputation: 454960

The matches method matches your regex against the entire string.

So try adding a .* to match rest of the string.

"#This is a comment in a file".matches("^#.*")

which returns true. One can even drop all anchors(both start and end) from the regex and the match method will add it for us. So in the above case we could have also used "#.*" as the regex.

Upvotes: 2

Andreas Dolk
Andreas Dolk

Reputation: 114757

This should meet your expectations:

"#This is a comment in a file".matches("^#.*$")

Now the input String matches the pattern "First char shall be #, the rest shall be any char"


Following Joachims comment, the following is equivalent:

"#This is a comment in a file".matches("#.*")

Upvotes: 0

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