Sorry dude, it's already been done. 
printf("%.$ARGV[1]f\n",$ARGV[0]);

[download]
printf takes a sting that describes what you want to print. In your case a floating point number (f) with a precision that you specify as the second arg ($ARGV[1]) using the value in $ARGV[0]

/\/\averick
OmG! They killed tilly! You *bleep*!!

In reply to Re: Perl rounder... by maverick
in thread Perl rounder... by abaxaba

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