Re: Simple adding numbers

[davido]

This is the result of storing floating point numbers in base two. Consider the fraction 1/3rd. Expressed in decimal, that's 0.33 (repeating forever). Add 0.33 three times, and you get 0.99, not one. You can extend to as many decimal places as you like, you'll never get .3333333333333333333333333 * 3 to add up to 1.0. Well, in base 2, the fraction 1/10th has a non-terminating expansion. So 0.1 has to be stored as 0.09999999999999999 or something similar, and when you add that up ten times you won't get to one.

It's only shocking because you grew up with base-ten and are so accustomed to seeing the problem that you don't notice it until it turns up in another base system.

Of course this is not unique to Perl. Consider the following C++ code:

#include <iostream>
#include <iomanip>


int main () {

  double n = 0.0;
  for( int i = 0; i != 10; i++ ) {
    n += 0.1;
    std::cout << std::setprecision(16) << std::fixed << n << std::endl
+;
  }

  if( n != 1.0 )
    std::cout << "Whoops, " << n << " isn't equal to 1.0." << std::end
+l;

  return 0;
}

...outputs...

0.1000000000000000
0.2000000000000000
0.3000000000000000
0.4000000000000000
0.5000000000000000
0.6000000000000000
0.7000000000000000
0.7999999999999999
0.8999999999999999
0.9999999999999999
Whoops, 0.9999999999999999 isn't equal to 1.0.