From help-octave-request at che dot utexas dot edu  Mon Jan 30 19:20:34 1995
Subject: Re: Zero-One indexing a length-one vector? 
From: John Eaton <jwe at che dot utexas dot edu>
To: Keh-Cheng Chu <kehcheng at cea dot Berkeley dot EDU>
cc: help-octave at che dot utexas dot edu
Date: Mon, 30 Jan 95 19:20:29 CST

Keh-Cheng Chu <kehcheng at cea dot Berkeley dot EDU> wrote:

: In Octave, can I use zero-one indexing on a length-one vector?
: I have some functions that work correctly in MATLAB but fail
: in Octave when the arguments are length-one vectors.
: 
: Here is an example:
: 
:     octave:15> b1 = [1]; b2 = [1 2]; b3 = [1 2 3];
:     octave:16> b1 (b1<0), b2 (b2<0), b3 (b3<0)
:     ans = [](0x0)
:     ans = [](1x0)
:     ans = [](1x0)
:     octave:17> b2 (b2<0) = b2 (b2<0) + 1 
:     b2 =
: 
:       1  2
:
:     octave:18> b1 (b1<0) = b1 (b1<0) + 1
:     error: invalid  index = 0
:     error: evaluating assignment expression near line 18, column 11

This bug is fixed in my sources and should not appear in version
1.1.1, which I plan to release soon.

: One related (?) question that I have is this:
: 
:    Why is b1(b1<0) a [](0x0) and not a [](1x0) or a [](0x1)?  After
:    all, b1 can be either a row or a column vector, so why isn't
:    one of the dimensions of the empty matrix 1?

I don't know.  I suppose `prefer_column_vectors' could be used to
determine the result.  Hmmm.

jwe