[Boost-users] Re: MultiArray questions

From: garcia (garcia_at_[hidden])
Date: 2004-01-24 13:17:45

Dr Mark H Phillips <mark_at_[hidden]> wrote:

>Hi,
>
>I've recently found out about the existence of MultiArray and I
>am quite impressed!  I was thinking of implementing something
>similar myself but now hopefully I don't have to!  However I have
>a few questions that I'm hoping someone can answer.
>
>  
>
Hi Mark.  Thank you for the compliments on MultiArray.  I hope that it 
meets your needs, and if not, I hope I can be of some service.

>Question 1
>----------
>
>In the user documentation there is an example which has
>some example code:
>
>  for(index i = 0; i != 3; ++i)
>    for(index j = 0; j != 4; ++j)
>      for(index k = 0; k != 2; ++k)
>        A[i][j][k] = values++;
>
>Why can't you change the != into <?  Ie, do:
>
>  for(index i = 0; i < 3; ++i)
>    for(index j = 0; j < 4; ++j)
>      for(index k = 0; k < 2; ++k)
>        A[i][j][k] = values++;
>
>or can you?  This would be more efficient I should think, and
>requiring that index types be ordered shouldn't be too onerous
>should it?
>
>  
>
You can change the != into a < if you like.  I don't see any good reason 
why one should be more efficient than the other.  I personally use != 
because it more closely matches how I write for loops using iterators.  
So in short, that's a matter of personal preference.

>Question 2
>----------
>
>If I use the built-in "no-frills" C array type to represent a
>2-dimensional array in row-major order, then I would access
>an element (i, j) by doing:
>
>  a[ &a[0] + i*n + j ]
>
>where a is my m by n array.  And if I wish to traverse a, in
>row-major order, I can do:
>
>  for (int* ip=&(a[0]); ip<&(a[n*m]); ip+=n) {
>    for (int* ijp=&(ip[0]); ijp<&(ip[n]); ++ijp) {
>      visit(ijp);
>    } 
>  }
>
>These are a very efficient ways of doing things, but it lacks
>somewhat in ease-of-generalization and in readbility.  This is where
>the MultiArray library is useful.  But do I lose in efficiency by
>going to MultiArray?  Ie, in the simple case where base is 0, stride
>is 1, and when we traverse in row-major order, is the MultiArray
>library smart enough to do compile-time optimizations to generate
>code which is as efficient as the above?
>
>  
>
The MultiArray implementation doesn't currently do anything special for 
the case that the base is 0.  I'm in the midst of doing some performance 
analysis on the library, and so far, to be honest, it looks like there 
is a performance hit.  I'm investigating what is happening, and I hope 
to translate my findings into performance improvements. I would liket 
hose to be available by May of this year. 

>Is it possible to construct a new
>multi_array/multi_array_ref/const_multi_array_ref from an array_view
>by doing a deep copy of the view?
>
>A view is a nice way of looking at different parts of a MultiArray
>object in a readable way, but sometimes you might wish to turn this
>view into a more-efficiently represented object.  Is this easy to do
>in the MultiArray library?
>  
>
It is my intention that a new multi_array be constructible from an 
existing view, but on looking back through my code and test cases, the 
library does not currently support it.  This is an oversight on my 
part.  I hope to have this functionality available in CVS by the end of 
this coming week (which coincides with the end of the month).  I'm 
afraid that the functionality won't make the upcoming Boost release however.

>Question 4
>----------
>
>Sometimes you might have an object starts off being mostly
>accessed in row-major order, so you want a representation which works
>efficiently for this use.  But down the track you might wish to change
>to mostly column-major access.  Does MultiArray provide easy ways to
>transform from one underlying representation to another?
>
>  
>
There is no direct means to change the underlying representation of an 
existing array.  The layout of an array is determined at construction. 
You can however assign a row-major multi_array to a column-major 
multi_array. 

>Thanks,
>
>  
>
My pleasure.  Thank you for your questions and comments. Incidentally, I 
am now subscribed to Boost-Users.  I apologize for any discussion I have 
missed or inaccessibility resulting from my absence.

Cheers,

ron