Same floating point operation, different results

Question

I really can't wrap my head around the fact that this code gives 2 results for the same formula:

#include <iostream>
#include <cmath>

int main() {
  // std::cout.setf(std::ios::fixed, std::ios::floatfield);
  std::cout.precision(20);
  float a = (exp(M_PI) - M_PI);
  std::cout << (exp(M_PI) - M_PI) << "\n";
  std::cout << a << "\n";
  return (0);
}

I don't really think that the IEEE 754 floating point representation is playing a significant role here ...

Kerrek SB · Accepted Answer · 2013-08-06T08:43:32.793

3

The first expression (namely (exp(M_PI) - M_PI)) is a double, the second expression (namely a) is a float. Neither even have 20 decimal digits of precision, but the float has a lot less precision than the double.

edited Aug 06 '13 at 08:43

answered Aug 06 '13 at 08:22

Kerrek SB

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time spent with the floating point system and I forgot about the other basic type in C++, great! – user2485710 Aug 06 '13 at 08:25
@EricPostpischil what ? Can you explain this ? – user2485710 Aug 06 '13 at 08:40
@user2485710: By "first expression" I'm referring to `exp(M_PI) - M_PI`. I'll edit the post. – Kerrek SB Aug 06 '13 at 08:43

Yu Hao · Answer 2 · 2013-08-06T08:34:20.790

2

Because M_PI are of type double, so change a to double, you will have the same result:

#include <iostream>
#include <cmath>

int main() {
  // std::cout.setf(std::ios::fixed, std::ios::floatfield);
  std::cout.precision(20);
  double a = (exp(M_PI) - M_PI);
  std::cout << (exp(M_PI) - M_PI) << "\n";
  std::cout << a << "\n";
  return (0);
}

edited Aug 06 '13 at 08:34

answered Aug 06 '13 at 08:22

Yu Hao

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Same floating point operation, different results

2 Answers2