Pattern matches always find the leftmost possible match. Greediness does not change this, it only affects how soon a quantifier will consider itself satisfied. In your case, / finds the leftmost slash in the string, .*? then tries to match nothing, which fails because the following $ needs the end of string to succeed, and then successively keeps trying to match only one more character, which keeps failing because the $ does not succeed, until the .*? has matched all of the string after the first slash. Since all parts of the pattern then succeed, you have a match.

You can use this knowledge to combine the leftmost match behaviour with greediness to make them work for you. Try this:

m{ .* / (.*) $ }msx

[download]

What happens here is that the first greedy .* will match the entire string. But then the / wants a slash, and that fails, since there’s no slash after the end of the string; so the .* is forced to concede one character. The / will keep failing and the .* will keep conceding characters, until it is matching only up to the character before the last slash in the string, at which point the / can match that slash and thus succeed. The parenthesised .* will then swallow the rest of the string, which means the $ immediately succeeds.

Now the parenthesised quantifier has matched exactly what you wanted.

Makeshifts last the longest.

In reply to Re: Regex: Matching last of repeated character to end of string by Aristotle
in thread Regex: Matching last of repeated character to end of string by loris

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