Hi list,

To elaborate on Claude's answer, here is how I would do it in vanilla :

#N canvas 355 282 450 300 12; #X obj 214 124 % 10; #X obj 96 126 / 10; #X obj 182 91 t f f; #X floatatom 182 32 5 0 0 0 - - -; #X obj 96 148 i; #X obj 96 172 sel 0; #X obj 214 147 list append; #X obj 214 170 t a a; #X obj 182 54 t f b; #X obj 96 214 list store; #X obj 96 239 print; #X connect 0 0 6 0; #X connect 1 0 4 0; #X connect 2 0 1 0; #X connect 2 1 0 0; #X connect 3 0 8 0; #X connect 4 0 5 0; #X connect 5 0 9 0; #X connect 5 1 2 0; #X connect 6 0 7 0; #X connect 7 0 9 1; #X connect 7 1 6 1; #X connect 8 0 2 0; #X connect 8 1 6 1; #X connect 9 0 10 0;

The inconvenient is that your number cannot start with extra zeros, but I guess this is not an issue here. Also, I have no idea how to perform this in other bases ... but it works just fine in base 10 for positive numbers.

Cheers, Joseph

Le 17/01/2021 à 20:24, pd-list-request@lists.iem.at a écrit :

Hi Antoine,

On 17/01/2021 18:15, Antoine Rousseau wrote:

...

• transform "float 512" into "symbol 512" using [makefilename %d]

I would not recommend this, because symbols stick around forever and Pd gets slower the more symbols there are.  AKA "symbol table pollution".

Instead, better to use [mod 10] to get the least significant digit, and [div 10] to get the most significant digits (then repeat until you have all the digits).  This approach also works for bases other than 10.