If a Smalltalk system did have a "cyclic" indexing

method that did "self at: (index - 1) \\ self size + 1",

where would it be?

Since it makes sense for all and only sequences,

we'd expect it to be in SequenceableCollection.

Since it is like #at:, and #at: is an 'accessing'

method, we'd expect to find it in the 'accessing'

methods of SequenceableCollection.

And indeed, #atWrap:[put:] are exactly where

they ought to be.

SequenceableCollection selectorsInCategory: 'accessing'

=> #(#after: #after:ifAbsent: #allButFirst #allButFirst: #allButLast #allButLast: #anyOne #at:ifAbsent: #at:incrementBy: #atAll: #atAll:put: #atAll:putAll: #atAllPut: #atLast: #atLast:ifAbsent: #atLast:put: #atPin: #atWrap: #atWrap:put: #before: #before:ifAbsent: #eighth #fifth #first #first: #fourth #from:to:put: #identityIndexOf: #identityIndexOf:ifAbsent: #indexOf: #indexOf:ifAbsent: #indexOf:startingAt: #indexOf:startingAt:ifAbsent: #indexOfAnyOf: #indexOfAnyOf:ifAbsent: #indexOfAnyOf:startingAt: #indexOfAnyOf:startingAt:ifAbsent: #indexOfSubCollection:startingAt: #indexOfSubCollection:startingAt:ifAbsent: #integerAt: #integerAt:put: #last #last: #lastIndexOf: #lastIndexOf:ifAbsent: #lastIndexOf:startingAt:ifAbsent: #lastIndexOfAnyOf:startingAt:ifAbsent: #middle #ninth #replaceAll:with: #replaceFrom:to:with: #replaceFrom:to:with:startingAt: #second #seventh #sixth #size #swap:with: #third)

On Fri, 29 Mar 2019 at 22:31, Tim Mackinnon <tim@testit.works> wrote:

Hey thanks guys - while it certainly makes sense when you think about it, I was kind of hoping we had something that was much more readable and obvious. It seems strange that when we have lots of esoteric things in collection, that something which is quite common to do isn��t there. I was kind of hoping we might have some circular list or something.

As it stands, if you really want to communicate this clearly you are almost better off just doing, the more obvious:

(Index := index - 1) = 0 ifTrue: [index := items size].

Anyway, an interesting one.

Tim

> On 28 Mar 2019, at 16:55, Peter Kenny <peter@pbkresearch.co.uk> wrote:
>
> Tim
>
> I found myself puzzling as to *why* James's solution works. This longer
> explanation helped me to understand.
>
> First, generalize moving forward to moving an arbitrary number of steps,
> still with wrapping.
>
> ^list at: (index + move - 1 \\ size + 1
>
> Second, realize that moving backward one step with wrapping is exactly
> equivalent to moving forward (size - 1) steps - because of wrapping, a move
> of size steps is null.
>
> Finally, substitute move = size - 1.
>
> ^list at: (index + size - 2 \\ size + 1.
>
> This can of course easily generalize to backward moves of any size.
>
> HTH
>
> Peter Kenny
>
>
>
> --
> Sent from: http://forum.world.st/Pharo-Smalltalk-Users-f1310670.html
>