Re: [boost] Infinite precision integer draft

From: Bronek Kozicki (brok_at_[hidden])
Date: 2006-05-28 07:00:47

Ben Artin wrote:
> If we are talking about infinite precision integers, then some example of 
> reasonable invariants on integers that do not apply to unsigned integers might 
> be:
> 
> a + (-a) = 0

Hold on for a second. We pretty well know what (-a) means for built-in 
unsigned integers (whose precision is limited), but that meaning is based on 
number of bits used to represent the number. Once you take "infinite 
precision" integer, that meaning is lost because we have no predefined number 
of bits used to represent the number. I mean that infinite precision integer 
cannot reproduce behaviour of built-in types. In other words, if you can 
"infinitely" extend given positive number towards +inifnity, does it make any 
sense to disallow its extension past zero? It makes me think that concept of 
"unsigned integer with infinite precision" is self-contradictory.

B.