Bug with floor()?

#include <math.h>
#include <stdio.h>

int main() {
  double a = 16.99;
  double b = 100.0;

  printf("a is %f\n", a);
  printf("b is %f\n", b);

  printf("a*b is %f\n", a*b);
  printf("floor(a*b) is %f\n", floor(a*b));
  return 0;
}


a is 16.990000
b is 100.000000
a*b is 1699.000000
floor(a*b) is 1698.000000

double vs. float... floorf() is your friend :)

floorf(a*b) is 1699.000000

printf('%.16f',17.99 * (10**2))

1798.9999999999997726

That would be your problem :)

So, the real reason I am mucking about with floor() is because that's how ruby's Fixnum.to_i function works...
http://www.ruby-doc.org/core/classes/Float.src/M000240.html

floorf sort of fixes it... for a certain threshold. Here's the output for with more precision (and some more debugging)

a is 16.989999999999998436805981327779591083526611328125000000000000000000000000000000000000000000000000000
b is 100.000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000
a_b is 1698.999999999999772626324556767940521240234375000000000000000000000000000000000000000000000000000000000
a_b < c
floorf(a*b) is 1699.000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000