On 17 May 2017 at 07:01, Edward Diener via Boost
On 5/16/2017 8:46 PM, Peter Dimov via Boost wrote:
Edward Diener wrote:
An alternative algorithm I have been considering is when the bit sizes are unequal is:
a) The bits comprising the common length are the low order bits.
The fact that bits are part of the common length does not mean that these bits have anything in common (semantically). It seems (what Edward D. seemed to think originally) non-sensical to me (not to say overly expensive).
d) if still equal, the larger bit size > the smaller bit size.
This makes 000111 > 111 instead of equal, so that equality (which I presume checks sizes) and equivalence match.
This means you can just as well start with the length comparison first. The Algorithm: 1. Compare length: shorter < longer. 2. If equal lengths, compare bitwise (in some way, f.e. memcmp()), but ALL the bits. Any use of ulong() as suggested, or f.e. hashing, will not result in a strict weak ordering, i.e. it's useless. degski -- "*Ihre sogenannte Religion wirkt bloß wie ein Opiat reizend, betäubend, Schmerzen aus Schwäche stillend.*" - Novalis 1798