One more note on David's suggestion. I didn't really want to keep separately named subarrays around (here, a and b) and found the following variation to work. int meta[2][0]; [0, 1, 2, 3] @=> int b[]; [10, 11, 12, 13, 14, 15] @=> int c[]; b @=> meta[0]; c @=> meta[1]; for(int j; j< meta.size(); j++) { for (int i; i< meta[j].size(); i++) <<< " meta[", j, ",", i, "] ==", meta[j][i] >>>; } Regards, Dana On Fri, Jan 7, 2022 at 10:14 AM Dana Batali wrote:

Hey David, (mind blown).

Very interesting and useful trick! My c++ brain didn't allow for this possibility. Esthetically the misrepresentation of the array "shape" bothered me. I found that I could be less bothered by changing one line and it still works!

int meta[2][0];

Thanks for playing! Dana

On Fri, Jan 7, 2022 at 10:06 AM David Loberg Code wrote:

...

Dana,

I realize that this example is a 2D array, but does this workaround accomplish what you want?

int meta[2][2];

[0, 1, 2, 3] @=> int b[];

[10, 11, 12, 13, 14, 15] @=> int c[];

b @=> meta[0];

c @=> meta[1];

for (int i; i< b.size(); i++)

{ <<< " b["+i+"] ==", meta[0][i] >>>; }

for (int i; i< c.size(); i++)

{ <<< "c["+i+"] ==", meta[1][i] >>>; }

Santé, davd

-----------------

David Loberg Code

School of Music

Western Michigan

code@wmich.edu

*any pronouns*

------------------------------

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