Re: a=0; a([1,1])++ -> a == 2

[Etienne Grossmann]

  Hello,

#  From address@hidden Wed Aug  2 19:49:40 2000
#  On  2-Aug-2000, Etienne Grossmann <address@hidden> wrote:
#
#  |   sorry if the answer to my question is a clear-cut "no" or if it has
#  | already been discussed.
#  | 
#  |   Wouldn't it make sense to have the code 
#  | 
#  |        a=0; b = a([1,1])++ ;
#  | 
#  |   be equivalent to either
#  | 
#  |        a=0; b=a([1,1]); b++; for i in [1,1], a(i)++ ; end
#  | 
#  |        yielding       a == 2, b == [1;1]
#
#  Hmm.  I can agree with what is inside the loop, but not the b++ that
#  is outside.  The way a statement like
#
#    b = a([1,1])++
#
#  is evaluated requires that the operation on the right hand side
#  produce a value, and then that result is copied to the left hand side
#  of the assignment.  So, what should the value of a([1,1])++ be?  Since
#  a([1,1]) references the same location twice, I suppose it is
#  reasonable to think that it should increment twice, but it also seems
#  reasonable that it should only increment once, since you have one `++'
#  operator operating on one location.  It also makes sense when you
#  think about it this way:
#
#    a = 0; a([1,1])++
#
#  is equivalent to
#
#    a = 0; t = a([1,1]); a([1,1]) += 1; t
#
#  or
#
#    a = 0; t = a([1,1]); a([1,1]) = a([1,1]) + 1; t

  Yes, that is true (and unfortunate, imho).

#  With the value of the expression being `t'.
#
#  |   or 
#  |        a=0; b=zeros(size(a)); for i in [1,1], b(i) = a(i)++ ; end
#  |     
#  |        yielding       a == 2, b == [1;2]

  This b==[1;2] is clearly a mistake; ++a(i) could lead to b==[1;2].

#  I don't think this is the correct interpretation of what this code
#  would currently do in Octave.  With my current sources (approximately
#  the same as the latest 2.1.x release), I see
#
#    octave:11> a=0; b=zeros(size(a)); for i = [1,1], b(i) = a(i)++ ; end, a, b
#    a = 2
#    b = 1
#
#  |   The current behavior yields :
#  | 
#  |       a == 1, b == [1;1]
                         ^^^
             My mistake again, it's [0;0]. Prefer_column_vector==1.
                           

#  Hmm.  Not what I see:
#
#    octave:9> a=0; b = a([1,1])++; a, b 
#    a = 1
#    b =
#
#      0  0
#
#  Are you using some different version of Octave that behaves
#  differently?

  No.

  Maybe a function a2 = loop_add (a1, indices, b) that does 
 
  a2=a1; for i=indices, a1(i)+=b(i); end

  would make me happy. I will have a look when I have time. Maybe
together with loop_mul and loop_exp. 

  Cheers,

  Etienne



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