Why perl regex '*?' stay greedy?

Question

I run a simple program:

my $_ =  '/login/.htaccess/.htdf';
s!(/\.ht.*?)$!/!;
print "$_ $1";

OUT
/login/ /.htaccess/.htdf

I want this regex to match only /.htdf.

Example 2:

my $_ =  'abcbc';
m/(b.*?)$/;
print "$_ $1\n";

OUT
abcbc bcbc

I expect bc.

Why is *? still greedy? (I want the minimal match.)

Answer 1

Atoms are matched in sequence, and each atom after the first must match at the position where the previous atom left off matching. (The first atom is implicitly preceded by \A(?s:.)*?.) That means that .*/.*? doesn't get to decided where it starts matching; it only gets to decided where it stops matching.

Example 1

It's not being greedy. \.ht brings the match to position 10, and at position 10, the minimum .*? can match and still have the rest of the pattern match is access/.htdf. In fact, it's the only thing .*? can match at position 10 and still have the rest of the pattern match.

I think you want to remove that last part of the path if it starts with .ht, leaving the preceding / in place. For that, you can use either of the following:

s{/\.ht[^/]*$}{/}

or

s{/\K\.ht[^/]*$}{}

Example 2

It's not being greedy. b brings the match to position 2, and at position 2, the minimum .*? can match and still have the rest of the pattern match is cbc. In fact, it's the only thing .*? can match at position 2 and still have the rest of the pattern match.

You are probably looking for

/b[^b]*$/

or

/b(?:(?!b).)*$/    # You'd use this if "b" was really more than one char.