Re^3: "Commifying" a number

that almost works

Nice catch; should have been obvious when I wrote it.

Btw, your lookahead assertion needn’t be quantified: (?=\d) will work exactly as (?=\d+) does, but won’t waste time matching extra digits.

Makeshifts last the longest.

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Re^4: "Commifying" a number by Anonymous Monk on Aug 13, 2009 at 14:47 UTC
Here is a version without using reverse function. sub commify { (my $number = shift) =~ s/\G(-?)(\d{1,3})(?=(?:\d\d\d)+(?:\.\|$))/$1$2,/g; return $number; } Manikanthan Velayutham programmed to think BIG	[reply]
Re^5: "Commifying" a number by Anonymous Monk on Aug 13, 2009 at 14:56 UTC
My earlier post did not appear properly. `sub commify { (my $number = shift) =~ s/\G([+\|-]?)(\d{1,3})(?=(?:\d\d\d)+(?:\.\|$ +))/$1$2,/g; return $number; }` [download] Manikanthan Velayutham programmed to think BIG	[reply] [d/l]