Hi,


I have written a script in which I enter the invoice amount, the tax amount, and the invoice total. Then I make the script check that the invoice amount + the tax amount equal the invoice total. It works well most of the time, but yesterday one invoice had these amounts (invoice amount = 17554.61, tax = 2194.33, and total = 19748.94) and the script informed me that the two figures were not equal. As you have explained, it does so because the internal base 2 representations of the two numbers are not the same. Is there any way of getting around this hurdle? How can perl be made to see that the two figures are the same in the decimal format?

One general approach is to maintain and operate on amounts internally as fixed-point integer quantities; e.g., as the number of cents, or possibly mills (tenths of a cent). While the numeric representation used (at least by Perl) is still double-precision floating point, the goal is to never produce a non-zero fractional result. Addition, subtraction and multiplication on integers always yield integer results. Operations that can produce fractional results (e.g., division) must sooner or later (and sooner is usually better) be reconciled into the fixed point representation that has been selected.

Or maybe check out something like integer or Math::FixedPrecision.

printf ("\$%2.2f\n", $val);

Hi,


Thanks. What I understand now is that the internal base 2 representation is not likely to cause any problems if I am dealing with only integers. In which case, I can convert the amounts to cents in the section of the code where I check whether the two amounts are equal. Thanks again.

What I understand now is that the internal base 2 representation is not likely to cause any problems if I am dealing with only integers.

Yes, as long as they're not too large. If you keep the numbers small than ±9,007,199,254,740,992, you'll be ok.

>perl -e"printf qq{%.0f\n}, 9007199254740992 + 3"
9007199254740996    even? how odd!
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Have fun calculating tax without using decimals, though. What you want to do then is perform rounding:

# If working with cents
$tax_amount = sprintf('%.0f', $amount * $tax_rate);
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Hi,


I'm afraid that didn't work out as I expected. I multiplied the amounts by 100 to get the amounts in cents. But when I compare the two, my script is still telling me that they are not equal./p>

use warnings;
use strict;

my($amt1,$amt2,$mytot,$perltot,$diff);

print "Enter amt1 : ";
chomp($amt1 = <STDIN>);
print "\nEnter amt2 : ";
chomp($amt2 = <STDIN>);
print "\nEnter the total : ";
chomp($mytot = <STDIN>);

$mytot = $mytot * 100;        # convert to cents

$perltot = ($amt1 * 100) + ($amt2 * 100);    #convert to cents

$diff = $perltot - $mytot;

print "$mytot - $perltot = $diff\n
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            _____
61/100 = 1.3851EB * 2**(-1)
           _____
33/100 = 1.51EB8  * 2**(-2)
           _____
94/100 = 1.E147A  * 2**(-1)
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Hi,

But

perl -e "print 100.61 + 100.33 - 200.94"
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works correctly.

Sometimes the errors cancel out. Floats also do a bit of rounding for you to try to avoid problems. You need to do some of your own.

see

Humans have too many fingers

Re: eq vs ==

HTH! 8)

Thanks. Removing the decimal using the substitute function helped. That tip was in the footnotes. I was earlier removing it by multiplying by 100.

As previous posts have mentioned, floating point numbers are only stored in a limited precision. If your number can be written in a binary non-repeating decimal , your ok, but for repeating floating point numbers in binary way, some accuracy is lost. For example, 1/4th can be written as a binary non-repeating decimal, (1/4 = 0.01 binary), but 1/5th is repeating in binary form (1/5 = 0.00110011...) and will get truncated. This means that for a computer, 1/5 is not 0.2!

The important lesson is that you should never (I mean it!) compare floating point numbers for equality. By the way, this goes for most programming languages.

To solve your problem, write the numbers as a formatted string wide enough, and compare those. For example:

my $val1 = 17554.61 - 2194.33;
my $val2 = 19748.94;

my $val1_str = sprintf("12.2f", $val1);
my $val2_str = sprintf("12.2f", $val2);

print ("equal\n") if ($val1_str eq $val2_str);
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You've just run into a problem that computer programmers have dealt with for decades: floating point numbers do not form a closed set under the operations of normal arithmetic. The reason is, of course, that floating point numbers only represent a finite subset of rational numbers.

Ignore!