From: "Peña, Botp" <botp@...>
Date: 2007-08-30T16:46:36+09:00
Subject: Re: Bug in % (Float)?

From: Morton Goldberg [mailto:m_goldberg@ameritech.net] 
# On Aug 29, 2007, at 10:14 PM, Pe�a, Botp wrote:
# >Returns an array containing the quotient and modulus 
# >obtained by dividing num by aNumeric. If q, r = x.divmod(y), 
# >then
# >          q = floor(float(x)/float(y)
# >
# > irb(main):345:0> 1.divmod 0.1
# > => [9, 0.1]
# > irb(main):347:0> 9*0.1+0.1
# > => 1.0
# 
# If Numeric#divmod adhered to the above, the result would be [10.0,  
# 0.0], so the result is just plain wrong, not a matter of float  
# inaccuracy. Proof:
# 
# x, y = 1.0, 0.1
# (x/y).floor.to_f # => 10.0

indeed, that is the effect you are testing. what i'm saying is that,

A. result of q from q,r=x.divmod(y)
   may _not (always) guarantee that q == (x/y).floor

B. there are too many results of that kind above, so inferred this 
   is not plain ruby algo error.
   see,

irb(main):069:0> (0.5/0.1).floor
=> 5
irb(main):070:0> 0.5.divmod(0.1)
=> [4, 0.1]
irb(main):071:0> (0.6/0.1).floor
=> 5
irb(main):072:0> 0.6.divmod(0.1)
=> [5, 0.1]
irb(main):073:0> (0.7/0.1).floor
=> 6
irb(main):074:0> 0.7.divmod(0.1)
=> [6, 0.0999999999999999]
irb(main):075:0> (1.2/0.1).floor
=> 11
irb(main):076:0> 1.2.divmod(0.1)
=> [11, 0.0999999999999999]
irb(main):077:0> (1.3/0.1).floor
=> 13
irb(main):078:0> 1.3.divmod(0.1)
=> [12, 0.1]

i checked divmod second value is same as modulo result (fr which i based my previous statement to not trust float division; of course, i may be wrong). so i reckon, ruby gets modulo first, then generate q by q=(num-r)/div.  Thus ruby can only guarantee num = q*div + r. (maybe modulo is wrong, and the documentation can be wrong too). i think.

i'm sorry. i tend to generalize. i don't read ruby source. correct me if i'm wrong, pls. 

kind regards -botp