On 01/28/2013 06:29 PM, Tuomo Keskitalo wrote:
So, does this mean that whenever an algorithm relies on a=a+a-a or a=a*a/a, the algorithm is going to fail when high precision is required?
Sorry, meant a=a+b-b and a=a*b/b. -- [email protected] http://iki.fi/tuomo.keskitalo
