Re: How do I print a large integer with thousands separators?

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

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Re: Answer: How do I print a large integer with thousands separators? by Roy Johnson (Monsignor) on Jan 09, 2004 at 22:20 UTC
To do it without capturing and resubstituting, you can add a lookbehind: `s/(?<=\d)(?=(?:\d{3})+\b)/$sep/g;` [download] This also works: `substr($n, pos($n), 0) = $sep while ($n =~ /\d(?=(?:\d{3})+\b)/g);` [download] As does: `substr($n, -($_3 + ($_-1)length($sep)), 0) = $sep for (1..int((lengt +h($n)-1)/3));` [download]	[reply] [d/l] [select]