Question

In source code I have to read (not from me) I found a regex that is not working as expected (or at least with what it does the code is not doing what it should).

The Idea of the regex is to make sure the string it processes has exactly one slash at the end afterwards.

So here we go:

$string =~ s/\/*$/\//g; # Add a trailing slash

I read this as “find any number of slashes at the end of the string (0 included) and replace them with a single slash”. But this is not what happens.

The regex doubles an/multiple existing trailing slash(es). So I end up with two slashes if there already is one or more. If there is none I get one (as it would be the idea).

I can reproduce it with Regex tester for example. Just enter http://example.com/foo/ (and vary it with different slash count at the end) in the top box, the regex in the second, “/” as the replace pattern and hit “send”.

It seems to revolve around the regex also replacing “nothing” (or the line ending?) with a slash (so it replaces “nothing” with one slash and the existing slash with one slash which makes up for the two slashes I see…).

I used

$string =~ s/\/{1,0}$/\//g; # Add a trailing slash

also but this only works if there is at least one slash from the beginning.

EDIT

Just in case anyone is wondering:

$string =~ s/\/*\z/\//g; # Add a trailing slash

produces the same thing.

And here is a direct link to the test page

Answer 1

The problem is the /g modifier, which says to replace all occurrences. When you have a slash at the end of the string, Perl makes two replacements. First, it replaces the slash with another slash. Then (because $ is a zero-width assertion, and does not consume any characters) it replaces the empty string at the end with a slash. Remove the /g and it will work, because only the first match will be replaced:

$string =~ s!/*$!/!; # Add a trailing slash

(I also changed the delimiter to make it more readable, although that had nothing to do with your problem.)