double type variables

[rupeshbhurke]

Hi,

Please check following code:

#include <iostream>
#include <iomanip>

int main(int argc, char *argv[])
{
  double someVal = 6.6f;
  std::cout << std::fixed << std::setprecision (9) << someVal << "\n";
  return 0;
}

shows output as:

6.599999905

Can somebody kindly explain why does this happen?

Thank you.

Best Regards
Rupesh

[Peter87]

Many floating point values are approximations and can't be stored exactly. float has lower precision than double so if you change
double someVal = 6.6f;
to
double someVal = 6.6;
you will probably get the correct output. Note that someVal is probably still an approximation of 6.6.

[Catfish2]

This has to do with how numbers are encoded, with an emphasis for performance, as long as the accuracy is "good enough".

http://en.wikipedia.org/wiki/Floating_point#Accuracy_problems
http://randomascii.wordpress.com/2012/02/25/comparing-floating-point-numbers-2012-edition/

[Zephilinox]

http://www.cplusplus.com/forum/articles/3827/

basically, you can't represent floating point numbers very well in binary, there has to be some rounding, that is the result of the rounding.