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