in reply to Simple adding numbers
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.
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.
Dave
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