I've spent the last several days pouring through all of that information and it proved very insightful. Holger you are correct in terms of the 3D transform. This is what I had reasoned but again I started confusing myself for some reason. If I perform the same transpose using the wdith*height of the 2D array does exactly what it should.
Thanks again,
Dave
David W. Gohara, Ph.D.
Harvard Medical School http://www.scianafilms.com
617-432-1216 (p)
617-432-4360 (f)
-----Original Message-----
From: Holger Bettag [mailto:email@hidden]
Sent: Sat 10/23/2004 2:30 PM
To: Gohara, David
Cc: email@hidden
Subject: Re: Moving Data in Memory Part II
On Fri, 22 Oct 2004, Gohara, David wrote:
> Regarding a 3D array this is similar to my last question regarding
> transposing a 2D array. Except this time I was wondering if it's
> possible to do this across the third dimension (that is all values in
> the third direction would become a stride of 1). I thought about this a
> while now, but I've now gotten myself completely confused and am asking
> for help. Previously I took the easy way and used mtrans() as was
> kindly suggested.
>
Well, if you have a solution for 2D, you can generalize it to 3D by
regarding the volume as a stack of slices parallel to the XZ plane,
stacked along the Y axis.
The primary stride of those slices would be 1, and the secondary stride
would be along Z (i.e. index += width*height) rather than along Y (i.e.
index =+ width).
You could transpose each slice exactly the way you did in 2D. Or is there
another complication?
Holger
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