wog is right. My mistake was to put the right anchor outside the zero width look-ahead assertion.
The correct
substitution code is:
s/(\d{1,3}?)(?=(\d{3})+$)/$1_/g;
The lookahead makes sure that the number of digits before each underscore we insert is a multiple of 3
The lookahead: (?=(\d{3})+$)
I needed an extra set of parenthesis
to fool Perl
because the regexp
parser barks if there is two quantifiers in a row, which is perfectly legitimate here.
There is a general lesson to be learned here: unchecked idiotism are for idiots. So much for me :)
When dealing with
new material (here regexp assertion that I have not used much), one must learn to
reassess idiotisms that may not work in a new larger context. Here, I used the idotism: force the match to the end of the string => add a $ at the the very end of the regexp. It did not work here because
I wanted the lookahead to match to the end of the string.
Compare the previous code with the easy way
-- stefp | 