Regex - substitute

Anonymous Monk has asked for the wisdom of the Perl Monks concerning the following question:

Wise monks, I'm here to seek your wisdom. Try to learn the Perl Regex feature, I tried a few examples, and came across this weird thing(?), and cannot find any good answer: The example is like this:

$s = "goodwood";

$s =~ s/o*/e/;

print "$s \n";

# got - 'egoodwood'

It really stuck me - since there is no match, why the 'e' got prefixed into the string? I understand that o* - can match nothing or more, but should the * quantifier be greedy and match the 'oo' in the good? Thanks.

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Re: Regex - substitute by graff (Chancellor) on Feb 17, 2003 at 03:38 UTC
To get the match that you're looking for, use the "+" instead of the "*": `s/o+/e/;` [download] will turn "goodwood" into "gedwood" (only the first occurrence is replaced); "+" refers to "one or more". To replace all occurrences of "o+" in the string, use the "g" modifier as well: `s/o+/e/g;` [download] will turn "goodwood" into "gedwed".	[reply] [d/l] [select]
Re: Regex - substitute by Paladin (Vicar) on Feb 17, 2003 at 03:30 UTC
Perl RE's match "leftmost, longest". "Leftmost" takes precedence over "longest". `/o*/` can match zero characters, as you said, and the leftmost place it finds 0 characters is right at the start of the string.	[reply] [d/l]
Re: Regex - substitute by Enlil (Parson) on Feb 17, 2003 at 03:33 UTC
What happens is that the regular expression will start searching the first position of the string and attempt to match o and not finding an o it will try the other alternative at this position. The * after the o gives it the option of matching nothing. (which it does). And, since it has found a match it replaces it with an e. Regular expressions try all the possible alternatives at a position in a string before moving on to the next position (left most first). so it never reaches the oo part as it can match nothing much earlier. -enlil	[reply]