On 03/07/2014 10:11 AM, Mikael wrote:
Why this result of (haxor-exact->inexact 0), just curious, does the RNRS dictate this?
No. But it allows it.
The explanation starts with: exact zero times any exact number is exact 0.
So if 1. (inexact 1) is the result of some computation X that returns a number with some uncertainty, the uncertainty doesn't matter---if X is changed to compute the result exactly, then exact 0 times that exact result will be exact 0, no matter what the result is.
Brad