$number =~ s/(\d{1,3}?)(?=(\d{3})+$)/$1,/g;

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The lookahead (?=(\d{3})+$) ensures that the number of digits before each underscore we insert is a multiple of 3.

An extra set of parentheses fools Perl because the regex parser barks when there are two quantifiers in a row, which is perfectly legitimate here.

This issue was also discussed in the thread Splitting every 3 digits?.

In reply to Re: How do I print a large integer with thousands separators? by stefp
in thread How do I print a large integer with thousands separators? by miblo

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