- # Round amounts to eight places before rounding to the requested
- # number of places. This gets rid of errors due to internal floating
- # point representation.
- $amount = $self->round_amount($amount, 8) if $places < 8;
- $amount = $amount * (10**($places));
- $round_amount = int($amount + .5 * ($amount <=> 0)) / (10**($places));
-
- $main::lxdebug->leave_sub(2);
-
- return $round_amount;
+ # If you search for rounding in Perl, you'll likely get the first version of
+ # this algorithm:
+ #
+ # ($amount <=> 0) * int(abs($amount) * 10**$places) + .5) / 10**$places
+ #
+ # That doesn't work. It falls apart for certain values that are exactly 0.5
+ # over the cutoff, because the internal IEEE754 representation is slightly
+ # below the cutoff. Perl makes matters worse in that it really, really tries to
+ # recognize exact values for presentation to you, even if they are not.
+ #
+ # Example: take the value 64.475 and round to 2 places.
+ #
+ # printf("%.20f\n", 64.475) gives you 64.47499999999999431566
+ #
+ # Then 64.475 * 100 + 0.5 is 6447.99999999999909050530, and
+ # int(64.475 * 100 + 0.5) / 100 = 64.47
+ #
+ # Trying to round with more precision first only shifts the problem to rarer
+ # cases, which nevertheless exist.
+ #
+ # Now we exploit the presentation rounding of Perl. Since it really tries hard
+ # to recognize integers, we double $amount, and let Perl give us a representation.
+ # If Perl recognizes it as a slightly too small integer, and rounds up to the
+ # next odd integer, we follow suit and treat the fraction as .5 or greater.
+
+ my $sign = $amount <=> 0;
+ $amount = abs $amount;
+
+ my $shift = 10 ** ($places);
+ my $shifted_and_double = $amount * $shift * 2;
+ my $rounding_bias = sprintf('%f', $shifted_and_double) % 2;
+ $amount = int($amount * $shift) + $rounding_bias;
+ $amount = $amount / $shift * $sign;