Re: Allocatable arrays in derived types

glen herrmannsfeldt <gah@xxxxxxxxxxxxxxxx> wrote:

Richard Maine wrote:

Another way to get an array of stride other than 1 is to start with a
rank 2 array and take a slice that is a row. For example, if xx is an n
by n array and you take the slice x(i,:) for some particular value of i,
that slice will have stride n-1. Note that this is a case that
Yugoslav's description doesn't catch.

Won't the stride be n?

Oops. Yes, of course. I guess I was thinking about the gaps instead of
the stride... or something.

Also, if you have N by N array x, with N greater than 3, and
take, for example, x(2:3,:) the element spacing isn't uniform.

That's just an application of the same principle to higher ranks. The
spacing of the columns is uniform, and the spacing of the elements in
each columnn is uniform. I figured I had blathered along for long enough
without going into the elaboration for higher rank. Besides, the average
user doesn't need that elaboration detailed. He just needs to understand
the general concept. That's even more so with someone struggling with
the basic concept in rank 1. There seems little benefit in throwing in
the extra stuff for higher ranks until the rank 1 case is thoroughly
understood. We are *NOT* talking to "people designing computer memory
hardware" here.

--
Richard Maine | Good judgement comes from experience;
email: last name at domain . net | experience comes from bad judgement.
domain: summertriangle | -- Mark Twain
.

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