Yes, I've seen and skimmed this document, but, not to be rude, but I'm not interested in becoming a computer scientist in order to write a script to do basic math. Adding together 0.001 40 times is pretty basic and if my calculator can do it, I not sure I understand why Perl won't.

It's not a Perl problem, it's the problem of representing a non-terminating series for base 2 using a finite number of binary digits. You are accustomed to seeing it in base ten when you try to represent 1/3rd, yet I hear no complaints that your ten counting fingers are malfunctioning. Your calculator gets it right by rounding to the eight or ten significant digits that you see on its little LCD display. In other words, it really doesn't get it right; it just covers up the ugliness. And, in fact, I see a nearly identical question every day in reference to C, C++, PHP, and myriad other programming languages over on StackOverflow. It's not a problem unique to Perl.

I understand that the "What every computer scientist should know..." article is a little beyond what someone who just wants to get the job done might want to digest. That's fine, the article goes into painful details. Try this response (shameless plug, I wrote it), which tries to spell it out in less technical terms: Re: shocking imprecision.




Dave

If you don't like the WhatEvery... article -- and I agree that it is written more to impress than inform; and is way over promoted -- then try this. Short, sweet & clear.

---

With the rise and rise of 'Social' network sites: 'Computers are making people easier to use everyday'

Examine what is said, not who speaks -- Silence betokens consent -- Love the truth but pardon error.

"Science is about questioning the status quo. Questioning authority".

In the absence of evidence, opinion is indistinguishable from prejudice.

The start of some sanity?

Yes, I've seen and skimmed this document, but, not to be rude, but I'm not interested in becoming a computer scientist in order to write a script to do basic math. Adding together 0.001 40 times is pretty basic and if my calculator can do it, I not sure I understand why Perl won't.

:) Try site:perlmonks.org What Every Computer Scientist Should Know About Floating-Point Arithmetic and you can learn from others who weren't satisfied with that document

• Simple math gone wrong
• Floating Point Looping In Perl
• Math 101 anyone?
• adding numbers and floating point errors
• sprintf rounding convention
• perl subtraction
• RFC: Large Floating Point Numbers - Rounding Errors
• perl subtraction
• Using (s)printf()
• http://en.wikipedia.org/wiki/Floating_point#Addition_and_subtraction
• limit of for loop!!
• Why does perl math suck?
• Perl 5.8.x floating point representation error
• Strange rounding Error in Perl
• Bug? 1+1 != 2
• [Win32, C, and way OT] C floats, doubles, and their equivalence
• My floating point comparison does not work. Why ?
• cos (100000000.0)
• Floating point problems
• Ever heard of pi? 22/7? Guess what, 22/7 exceeds p
• General Decimal Arithmetic
• Decimal Arithmetic FAQ(Frequently Asked Questions)

 

Now you say you're studying chemical diffusion so I assume you would have heard of significant figures? Surely your professor, when discussing significant figures, would have explained the basic limitations of adding machines (calculators/computers)?

I was hoping, after reading that document, you would ask explicitly how to round numbers for display purposes in perl.

• perlfaq4#Does Perl have a round() function? What about ceil() and floor()? Trig functions?
• Math::BigRat
• Math::BigInt
• Math::BigFloat
• bigint/bigrat/bignum

While you can create a calculator using perl (like Tk::Calculator::RPN::HP ), and expect it to do rounding like your pocket calculator, perl itself, not being a calculator, won't hide the details of floating point arithmetic from you, so it is good knowledge to have.

Any scientist using computers for calculations needs to know the limits of his tools.

If I were doing it with a calculator, I'd surely end up with too many or too few entries of 0.001. I might mess it up building an array. Are you sure you have the right number of elements?

You can check easily enough:

say scalar(@arr);  # 40

say $#arr;            # 39
[download]

For reasons described in excruciating detail in the document already cited, there will be errors, but according to a quick test on my system, they are nowhere near as large as you claim:

$ perl -e '$x=0.001; $sum += $x for 1..40; printf "%.20f", $sum/40'
0.00100000000000000067
[download]

Looks more like a "one off" error to me (i.e. summing over one more than you divide by):

$ perl -e '$x=0.001; $sum += $x for 0..40; printf "%.20f", $sum/40'
0.00102500000000000074
[download]

:D Looks like two separate off-by-one error (OBOE) errors to me :)

First you start with non-zero and add 40 times (one too many), then you start with non-zero and add 41 times (one too many twice).

If you start with non-zero you need to add only 39 times, or start with zero and add 40 times :)

In short

perl -MData::Dump -e " @f = map { 0.001 } 1 .. 40; dd\@f; $o = 0; for(
+@f){ dd $o+=$_; } dd int @f; dd $o/int(@f); "

perl -MData::Dump -e "  $o = 0; for(1 .. 40){ dd $o+= 0.001; } dd $o/4
+0; "
[download]

It didn't dawn on me to check wickedjesters (or your) math until ww raised the quesiton

... you start with non-zero ...

Not sure what you're talking about.

n times adding x to zero is mathematically (but not necessarily numerically) the same as n * x.

$ perl -le '$sum += 1 for 1..40; print $sum'
40
[download]

So where is the problem?  I think you overlooked that $sum is initially undef/zero.

wickedjester:

You don't show your code, but I'm pretty sure you're not doing what you think you're doing. Specifically, I believe you're adding 41 copies of 0.001, as that's the only way I can reproduce your results:

$ cat t.pl
#!/usr/bin/perl
my @a = (0.001) x 41;
my $sum=0;
$sum += $_ for @a;
print "Avg: ", $sum/40, "\n";

$ perl t.pl
Avg: 0.001025
[download]

...roboticus

When your only tool is a hammer, all problems look like your thumb.

• sprintf
• perlfaq4 Why am I getting long decimals (eg, 19.9499999999999) instead of the numbers I should be getting (eg, 19.95)?

#!/usr/bin/perl -l

use strict;
use warnings;
use Array::Average;

print average(
0.001, 0.001, 0.001, 0.001, 0.001, 0.001, 0.001, 0.001, 0.001, 0.001,
0.001, 0.001, 0.001, 0.001, 0.001, 0.001, 0.001, 0.001, 0.001, 0.001,
0.001, 0.001, 0.001, 0.001, 0.001, 0.001, 0.001, 0.001, 0.001, 0.001,
0.001, 0.001, 0.001, 0.001, 0.001, 0.001, 0.001, 0.001, 0.001, 0.001,
);
[download]

You can dispense with the addition, division,...

And how do you think the module arrives at its result?  What it does is exactly addition and division — which of course suffers from the same floating point issues.  Here's the relevant code snippet:

  if (@data) {
    my $sum=0;
    $sum+=$_ foreach @data;
    return $sum/scalar(@data);
  } else {
    return undef;
  }
[download]

Anyhow, as has already been pointed out, the OP's problem has likely nothing whatsoever to do with those general floating point issues, but is presumably simply the result of having computed the sum incorrectly.