Mutli-dimensional array initialisation completion about compiler HOT 7 CLOSED

Just to clarity, what is arr[2] in your second example supposed to be? Should be it be 0, 4, 8, ... or 0, 2, 4, ... (i.e. same as arr[1])? In the current implementation it would be the latter.

Edit:

The second option probably doesn't make much sense but just to be sure....

from compiler.

I would think the first one. If you want the second do:

new arr[10][20] = {{0, 1, 2, ...}, {0, 2, 4, ...}, {0, 2, 4, ...}, ...};

As with one dimension {0, 1, 1} to fill with almost all 1s using just the last two specified values.

from compiler.

Forgive me for the dumb question, but I'm still kind of trying to figure out how to calculate arr[2] via arr[0] and arr[1], so far I came up with this:

d = diff_2 - diff_1
arr[2] = { arr[1][0] + d, arr[1][1] + d, ...}

where diff_1 and diff_2 are the differences between the elements of arr[0] and arr[1] respectively. And the question is:

Does this look like the correct formula? I'm asking because this would yield arr[2] as 0, 3, 6 instead of 0, 4, 8 like I wrote before.

Edit:

If instead we do

d = diff_2 / diff_1
arr[2] = { arr[1][0] * d, arr[1][1] * d, ...}

I believe this would give 0, 4, 8 as intended. Still not sure what to do here: for the 1d case we subtracted/added things, but here we are dividing/multiplying (kind of confusing).

from compiler.

I would expect {0, 3, 6}, since the increase is linear.

from compiler.

Actually, I would do it:

arr[0][2] = arr[0][1] + diff(arr[0][0], arr[0][1]); // As in 1-d.
arr[2][0] = arr[1][0] + diff(arr[0][0], arr[1][0]); // Same in the other dimension
arr[2][1] = arr[1][1] + diff(arr[0][1], arr[1][1]);

Thus:

arr[2][2] = arr[2][1] + diff(arr[2][0], arr[2][1]);

Which gives:

{{0, 1, 2, 3, 4,...}, {0, 2, 4, 6, 8, ...},{0, 3, 6, 9,12, ...}}

from compiler.

Seems working now, if you need an executable for testing you can download it here:
https://ci.appveyor.com/project/Zeex/pawn/branch/master/artifacts

from compiler.

Awesome, thanks.

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