Re: [sc-users] 2D array question

[jamshark70@xxxxxxxxx]

On Wed, Mar 6, 2019 at 10:38 AM James Harkins <jamshark70@xxxxxxxxx> wrote:
> On Wed, Mar 6, 2019 at 1:02 AM <brianlheim@xxxxxxxxx> wrote:
> > It's the other way around -- the implementation mirrors the semantics exposed to the programmer.
>
> I do understand what you say, and it's both formally correct and
> interesting. From a teaching perspective, though, I think it
> introduces unnecessary confusion.

Not a super-urgent topic, but because (effectively) one of the
functions of the mailing list is teaching, I think it's worth
following up.

x = 5;        // A
y = x;        // B
x = x + 1;    // C
[x, y]
-> [ 6, 5 ]

We could explain this to nonprogrammers in pass-by-value terms as:

A. x is a sheet of paper with 5 on it
B. y is a sheet of paper with 5 on it
C. you take the 5 from x's sheet of paper, add 1, and now you have
another sheet of paper with 6 on it. Then you throw away x's old 5,
and replace it with the new 6. You've done nothing to y.

Then:

x = [0, 1];    // A
y = x;        // B
x[0] = 2;    // C
[x, y]
-> [ [ 2, 1 ], [ 2, 1 ] ]

At this point, nonprogrammers will want to extend the analogy:

A. x is a sheet of paper with [0, 1] on it
B. y is a sheet of paper with [0, 1] on it
C. you've done something to x's sheet of paper, and this has magically
affected y's sheet of paper.

So the analogy introduces "pass by value" (`x + 1` operates directly
on data) but it confuses.

Now, if you explain the first case as:

A. x is an arrow pointing to "integer 5" somewhere
B. y is an arrow pointing to the same "integer 5" that x is pointing to
C. resolve x --> 5; add 1; you get 6, and now change x's arrow to
point to this 6 (wherein it becomes clear that changing the target of
x's arrow doesn't change the target of y's arrow)

... and you can explain the second case using the same analogy.

A. x is an arrow pointing to [0, 1]
B. y is an arrow pointing to the same [0, 1] that x is pointing to
C. you've done something to the thing that x is pointing to. But y is
pointing to the same thing. So if you look at x, you see the change,
and if you look at y, you also see the change, because they're
pointing in the same direction.

One analogy that explains both scenarios is better than introducing a
distinction between two cases that doesn't actually arise in usage. So
I think I *am* keeping it simple ;)

hjh

_______________________________________________
sc-users mailing list

info (subscription, etc.): http://www.birmingham.ac.uk/facilities/ea-studios/research/supercollider/mailinglist.aspx
archive: https://listarc.bham.ac.uk/marchives/sc-users/
search: https://listarc.bham.ac.uk/lists/sc-users/search/