Forums > MaxMSP

how to convert text from message to identical binary number

April 30, 2007 | 12:59 am

i’ve tried very hard (and yes i looked throug all the posts):

could not manage this: need to transport a symbol to a binary number
with the same value. how can i manage this. found no way (needs to be
done without any external)


———————————
rens veltman@mac.com
++43 650 7603596
———————————


April 30, 2007 | 4:47 am

Do you mean a symbol containing an integer value, or you want the binary equivalents of the ascii numbers? There’s always this sort of thing … example for positive values 0-255:

#P window setfont "Sans Serif" 9.;
#P window linecount 1;
#P newex 57 292 29 196617 >> 7;
#P newex 101 292 29 196617 >> 6;
#P newex 57 268 39 196617 & 128;
#P newex 101 268 33 196617 & 64;
#P newex 236 112 40 196617 unpack;
#P newex 236 91 52 196617 listfunnel;
#P newex 236 70 40 196617 atoi;
#P message 236 46 77 196617 "this is a test";
#P newex 55 118 64 196617 fromsymbol;
#P newex 55 71 51 196617 tosymbol;
#P user hslider 55 46 18 128 256 1 0 0;
#P newex 138 292 29 196617 >> 5;
#P newex 175 292 29 196617 >> 4;
#P newex 215 292 29 196617 >> 3;
#P newex 250 292 29 196617 >> 2;
#P newex 286 292 29 196617 >> 1;
#P newex 57 421 27 196617 i;
#P newex 57 399 64 196617 fromsymbol;
#P newex 57 447 32 196617 print;
#P newex 75 351 98 196617 loadmess separator;
#P number 138 207 35 9 0 0 0 3 0 0 0 221 221 221 222 222 222 0 0 0;
#P newex 138 268 33 196617 & 32;
#P newex 57 323 277 196617 sprintf %ld %ld %ld %ld %ld %ld %ld %ld;
#P newex 175 268 33 196617 & 16;
#P newex 215 268 27 196617 & 8;
#P newex 323 268 27 196617 & 1;
#P newex 250 268 27 196617 & 4;
#P newex 57 376 51 196617 tosymbol;
#P newex 286 268 27 196617 & 2;
#P connect 26 0 28 0;
#P connect 28 0 6 0;
#P connect 25 0 27 0;
#P connect 27 0 6 1;
#P connect 24 1 8 0;
#P connect 8 0 26 0;
#P connect 8 0 25 0;
#P connect 8 0 7 0;
#P connect 8 0 5 0;
#P connect 8 0 4 0;
#P connect 8 0 2 0;
#P connect 8 0 0 0;
#P connect 8 0 3 0;
#P connect 3 0 6 7;
#P connect 13 0 6 6;
#P connect 14 0 6 5;
#P connect 15 0 6 4;
#P connect 16 0 6 3;
#P connect 17 0 6 2;
#P connect 6 0 1 0;
#P connect 22 0 23 0;
#P connect 23 0 24 0;
#P connect 21 0 22 0;
#P connect 20 0 8 0;
#P connect 19 0 20 0;
#P connect 18 0 19 0;
#P connect 12 0 10 0;
#P connect 7 0 17 0;
#P connect 5 0 16 0;
#P connect 4 0 15 0;
#P connect 2 0 14 0;
#P connect 0 0 13 0;
#P connect 11 0 12 0;
#P connect 1 0 11 0;
#P connect 9 0 1 0;
#P window clipboard copycount 29;


April 30, 2007 | 2:47 pm

On Apr 30, 2007, at 6:47 AM, John Pitcairn wrote:

>
> Do you mean a symbol containing an integer value, or you want the
> binary equivalents of the ascii numbers? There’s always this sort
> of thing … example for positive values 0-255:
>
> #P window setfont "Sans Serif" 9.;
> #P window linecount 1;
> #P newex 57 292 29 196617 >> 7;
> #P newex 101 292 29 196617 >> 6;
> #P newex 57 268 39 196617 & 128;
> #P newex 101 268 33 196617 & 64;
> #P newex 236 112 40 196617 unpack;
> #P newex 236 91 52 196617 listfunnel;
> #P newex 236 70 40 196617 atoi;
> #P message 236 46 77 196617 "this is a test";
> #P newex 55 118 64 196617 fromsymbol;
> #P newex 55 71 51 196617 tosymbol;
> #P user hslider 55 46 18 128 256 1 0 0;
> #P newex 138 292 29 196617 >> 5;
> #P newex 175 292 29 196617 >> 4;
> #P newex 215 292 29 196617 >> 3;
> #P newex 250 292 29 196617 >> 2;
> #P newex 286 292 29 196617 >> 1;
> #P newex 57 421 27 196617 i;
> #P newex 57 399 64 196617 fromsymbol;
> #P newex 57 447 32 196617 print;
> #P newex 75 351 98 196617 loadmess separator;
> #P number 138 207 35 9 0 0 0 3 0 0 0 221 221 221 222 222 222 0 0 0;
> #P newex 138 268 33 196617 & 32;
> #P newex 57 323 277 196617 sprintf %ld %ld %ld %ld %ld %ld %ld %ld;
> #P newex 175 268 33 196617 & 16;
> #P newex 215 268 27 196617 & 8;
> #P newex 323 268 27 196617 & 1;
> #P newex 250 268 27 196617 & 4;
> #P newex 57 376 51 196617 tosymbol;
> #P newex 286 268 27 196617 & 2;
> #P connect 26 0 28 0;
> #P connect 28 0 6 0;
> #P connect 25 0 27 0;
> #P connect 27 0 6 1;
> #P connect 24 1 8 0;
> #P connect 8 0 26 0;
> #P connect 8 0 25 0;
> #P connect 8 0 7 0;
> #P connect 8 0 5 0;
> #P connect 8 0 4 0;
> #P connect 8 0 2 0;
> #P connect 8 0 0 0;
> #P connect 8 0 3 0;
> #P connect 3 0 6 7;
> #P connect 13 0 6 6;
> #P connect 14 0 6 5;
> #P connect 15 0 6 4;
> #P connect 16 0 6 3;
> #P connect 17 0 6 2;
> #P connect 6 0 1 0;
> #P connect 22 0 23 0;
> #P connect 23 0 24 0;
> #P connect 21 0 22 0;
> #P connect 20 0 8 0;
> #P connect 19 0 20 0;
> #P connect 18 0 19 0;
> #P connect 12 0 10 0;
> #P connect 7 0 17 0;
> #P connect 5 0 16 0;
> #P connect 4 0 15 0;
> #P connect 2 0 14 0;
> #P connect 0 0 13 0;
> #P connect 11 0 12 0;
> #P connect 1 0 11 0;
> #P connect 9 0 1 0;
> #P window clipboard copycount 29;
>

On Apr 30, 2007, at 6:47 AM, John Pitcairn wrote:
Do you mean a symbol containing an integer value, or you want the
binary equivalents of the ascii numbers? There’s always this sort of
thing … example for positive values 0-255:

hi john,
thanks for replying.
parsing a row from a jit.cellblock (like: 0 0 0 255 255 255 255 255
255 0 0 0 < - ie. 5th row of a cellblock becomes -> 0 0 0 1 1 1 1 1 1
0 0 0 in a textedit box). i need to bring the same value (without
spaces) into a number box of type binary, so that i can change it
easily into its hex.

sort of green, i can not find out how this can be done with your
example batch. could you please explain.

———————————
rens.veltman@mac.com
++43 650 7603596
———————————

———————————
rens.veltman@mac.com
++43 650 7603596
———————————


April 30, 2007 | 4:21 pm

Hello, this may be helpful… though I don’t use Jitter. Connect your symbol to the top t-object (trigger). The number box at the bottom is set to binary, but you can easily set it to hexadecimal in its info window (select it and press command-i).
— Ludwig Sears

#P window setfont "Sans Serif" 9.;
#P window linecount 1;
#P comment 20 65 72 196617 Just to test it;
#P newex 33 45 51 196617 tosymbol;
#P message 33 28 26 196617 153;
#P window linecount 2;
#P comment 191 61 152 196617 Spell and Split objects find symbols representing numbers.;
#P window linecount 4;
#P comment 246 110 134 196617 Count how many digits are in number , convert ascii codes to numbers (-48) , and assemble them into a list.;
#P window linecount 5;
#P comment 211 195 118 196617 Because we start at left end of number , we first had to find largest 10′s place before converting to a single digit.;
#P window linecount 3;
#P comment 55 221 55 196617 Clear list and reset counters.;
#P window linecount 1;
#P comment 166 311 183 196617 Number object is set to binary form.;
#P newex 97 261 36 196617 b;
#P newex 105 44 63 196617 t 0 set b s;
#P message 92 284 33 196617 set 0;
#P newex 154 264 108 196617 expr pow (10\, $i1);
#P newex 135 283 29 196617 *;
#P newex 123 303 41 196617 accum;
#P newex 135 183 24 196617 iter;
#P newex 147 236 59 196617 !-;
#P message 143 141 56 196617 append $1;
#P newex 151 116 33 196617 – 48;
#P window linecount 0;
#P message 133 164 212 196617;
#P window linecount 1;
#P newex 151 88 62 196617 split 48 57;
#N counter;
#X flags 0 1;
#P newobj 143 212 47 196617 counter;
#N counter;
#X flags 0 1;
#P newobj 196 116 41 196617 counter;
#P number 121 327 283 9 0 0 2048 3 0 0 0 221 221 221 222 222 222 0 0 0;
#P newex 154 66 33 196617 spell;
#P connect 22 0 14 0;
#P connect 21 0 22 0;
#P connect 9 0 11 0;
#P connect 9 0 3 0;
#P fasten 14 0 15 0 110 258 102 258;
#P fasten 14 0 3 2 110 209 166 209;
#P fasten 14 0 2 2 110 113 215 113;
#P connect 4 0 6 0;
#P fasten 4 0 2 0 156 109 201 109;
#P connect 0 0 4 0;
#P connect 6 0 7 0;
#P connect 7 0 5 0;
#P connect 3 0 8 0;
#P connect 5 0 9 0;
#P connect 12 0 11 1;
#P connect 8 0 12 0;
#P connect 11 0 10 1;
#P connect 10 0 1 0;
#P connect 2 0 8 1;
#P connect 14 3 0 0;
#P fasten 14 2 5 0 138 71;
#P fasten 14 1 5 0 138 73;
#P connect 15 1 10 0;
#P connect 15 0 13 0;
#P fasten 13 0 10 0 97 304 128 304;
#P window clipboard copycount 24;


May 1, 2007 | 1:52 am

Quote: rens.veltman wrote on Tue, 01 May 2007 02:47
—————————————————-
> i can not find out how this can be done with your
> example batch.

It can’t – I misunderstood what you were asking. You didn’t mention that you were working with a list of RGB values…


May 1, 2007 | 6:05 pm

On May 1, 2007, at 3:52 AM, John Pitcairn wrote:

>
> Quote: rens.veltman wrote on Tue, 01 May 2007 02:47
> —————————————————-
>> i can not find out how this can be done with your
>> example batch.
>
> It can’t – I misunderstood what you were asking. You didn’t mention
> that you were working with a list of RGB values…

sorry for not being more specific. but you showed me the direction
with your patch already (if there is one..) my problem seems to be
that max is cutting the leading zeroes from my binary parsing. have
to find a way around.

———————————
rens.veltman@mac.com
++43 650 7603596
———————————


May 1, 2007 | 6:09 pm

Hello, this may be helpful… though I don’t use Jitter.
Connect your symbol to the top t-object (trigger). The number box at
the bottom is set to binary, but you can easily set it to hexadecimal
in its info window (select it and press command-i).
— Ludwig Sears

#P window setfont "Sans Serif" 9.;
#P window linecount 1;
#P comment 20 65 72 196617 Just to test it;
#P newex 33 45 51 196617 tosymbol;
#P message 33 28 26 196617 153;
#P window linecount 2;
#P comment 191 61 152 196617 Spell and Split objects find symbols
representing numbers.;
#P window linecount 4;
#P comment 246 110 134 196617 Count how many digits are in number ,
convert ascii codes to numbers (-48) , and assemble them into a list.;
#P window linecount 5;
#P comment 211 195 118 196617 Because we start at left end of number
, we first had to find largest 10′s place before converting to a
single digit.;
#P window linecount 3;
#P comment 55 221 55 196617 Clear list and reset counters.;
#P window linecount 1;
#P comment 166 311 183 196617 Number object is set to binary form.;
#P newex 97 261 36 196617 b;
#P newex 105 44 63 196617 t 0 set b s;
#P message 92 284 33 196617 set 0;
#P newex 154 264 108 196617 expr pow (10\, $i1);
#P newex 135 283 29 196617 *;
#P newex 123 303 41 196617 accum;
#P newex 135 183 24 196617 iter;
#P newex 147 236 59 196617 !-;
#P message 143 141 56 196617 append $1;
#P newex 151 116 33 196617 – 48;
#P window linecount 0;
#P message 133 164 212 196617;
#P window linecount 1;
#P newex 151 88 62 196617 split 48 57;
#N counter;
#X flags 0 1;
#P newobj 143 212 47 196617 counter;
#N counter;
#X flags 0 1;
#P newobj 196 116 41 196617 counter;
#P number 121 327 283 9 0 0 2048 3 0 0 0 221 221 221 222 222 222 0 0 0;
#P newex 154 66 33 196617 spell;
#P connect 22 0 14 0;
#P connect 21 0 22 0;
#P connect 9 0 11 0;
#P connect 9 0 3 0;
#P fasten 14 0 15 0 110 258 102 258;
#P fasten 14 0 3 2 110 209 166 209;
#P fasten 14 0 2 2 110 113 215 113;
#P connect 4 0 6 0;
#P fasten 4 0 2 0 156 109 201 109;
#P connect 0 0 4 0;
#P connect 6 0 7 0;
#P connect 7 0 5 0;
#P connect 3 0 8 0;
#P connect 5 0 9 0;
#P connect 12 0 11 1;
#P connect 8 0 12 0;
#P connect 11 0 10 1;
#P connect 10 0 1 0;
#P connect 2 0 8 1;
#P connect 14 3 0 0;
#P fasten 14 2 5 0 138 71;
#P fasten 14 1 5 0 138 73;
#P connect 15 1 10 0;
#P connect 15 0 13 0;
#P fasten 13 0 10 0 97 304 128 304;
#P window clipboard copycount 24;

hello and thank you for the example.
maybe i should name my goal in more detail. i already have parsed my
(12bit) binary number into a textedit box looks like 0011000000110.
now i need to transfer it into a hex number.

———————————
rens.veltman@mac.com
++43 650 7603596
———————————


May 1, 2007 | 9:10 pm

But that’s the simplest part of the problem, unless I misunderstand you. Number boxes will automatically convert any input number into any one of six Display Styles you choose in the info window. Just select ‘Hex’ from the Display Style menu and you’re good to go.


May 1, 2007 | 9:14 pm

Oops, my bad.

I just tried connecting that Hex number box to a print object… and it was printed in decimal format. Sorry, I never had reason to notice that before! I guess that’s the gist of your question there.
Hmm… gimmee a while to work on this, though someone smarter will probably beat me to it.

El Searso


May 1, 2007 | 9:47 pm

On May 1, 2007, at 11:14 PM, David Wright wrote:

> I just tried connecting that Hex number box to a print object…
> and it was printed in decimal format. Sorry, I never had reason to
> notice that before!
> Hmm… gimmee a while to work on this, though someone smarter will
> probably beat me to it.

thats my problem.

———————————
rens.veltman@mac.com
++43 650 7603596
———————————


May 2, 2007 | 12:03 am

At 8:09 PM +0200 5/1/07, rens veltman wrote:
>maybe i should name my goal in more detail. i already have parsed my (12bit) binary number into a textedit box looks like 0011000000110. now i need to transfer it into a hex number.

Could you post the relevant portion of your patch?

-C


Chris Muir | "There are many futures and only one status quo.
cbm@well.com | This is why conservatives mostly agree,
http://www.xfade.com | and radicals always argue." – Brian Eno


May 2, 2007 | 12:08 am

>> I just tried connecting that Hex number box to a print object…
>> and it was printed in decimal format.

sprintf %x converts ints to hex symbolics

hth,
a


May 2, 2007 | 1:19 am

I’m still not sure I completely understand your problem, but here’s a
brute force method of doing something:

#P window setfont "Sans Serif" 9.;
#P window linecount 1;
#P newex 56 353 50 196617 +;
#P newex 136 313 50 196617 +;
#P newex 216 275 50 196617 +;
#P newex 296 237 50 196617 +;
#P number 16 438 66 9 0 0 0 3 0 0 0 221 221 221 222 222 222 0 0 0;
#P number 16 420 66 9 0 0 16 3 0 0 0 221 221 221 222 222 222 0 0 0;
#P number 16 402 66 9 0 0 2048 3 0 0 0 221 221 221 222 222 222 0 0 0;
#P newex 456 161 50 196617 * 2048;
#P newex 416 161 44 196617 * 1024;
#P newex 416 180 50 196617 +;
#P newex 376 180 38 196617 * 512;
#P newex 336 199 38 196617 * 256;
#P newex 336 218 50 196617 +;
#P newex 296 218 38 196617 * 128;
#P newex 256 237 32 196617 * 64;
#P newex 256 256 50 196617 +;
#P newex 216 256 32 196617 * 32;
#P newex 176 275 32 196617 * 16;
#P newex 176 294 50 196617 +;
#P newex 136 294 27 196617 * 8;
#P newex 96 313 27 196617 * 4;
#P newex 96 333 50 196617 +;
#P newex 56 333 27 196617 * 2;
#P newex 16 373 50 196617 +;
#P number 16 138 35 9 0 0 0 3 0 0 0 221 221 221 222 222 222 0 0 0;
#P newex 16 117 27 196617 int;
#P newex 41 66 40 196617 t b l b;
#P number 456 138 35 9 0 0 0 3 0 0 0 221 221 221 222 222 222 0 0 0;
#P number 416 138 35 9 0 0 0 3 0 0 0 221 221 221 222 222 222 0 0 0;
#P number 376 138 35 9 0 0 0 3 0 0 0 221 221 221 222 222 222 0 0 0;
#P number 336 138 35 9 0 0 0 3 0 0 0 221 221 221 222 222 222 0 0 0;
#P number 296 138 35 9 0 0 0 3 0 0 0 221 221 221 222 222 222 0 0 0;
#P number 256 138 35 9 0 0 0 3 0 0 0 221 221 221 222 222 222 0 0 0;
#P number 216 138 35 9 0 0 0 3 0 0 0 221 221 221 222 222 222 0 0 0;
#P number 176 138 35 9 0 0 0 3 0 0 0 221 221 221 222 222 222 0 0 0;
#P number 136 138 35 9 0 0 0 3 0 0 0 221 221 221 222 222 222 0 0 0;
#P number 96 138 35 9 0 0 0 3 0 0 0 221 221 221 222 222 222 0 0 0;
#P number 56 138 35 9 0 0 0 3 0 0 0 221 221 221 222 222 222 0 0 0;
#P newex 56 115 415 196617 bucket 11;
#P newex 56 87 25 196617 iter;
#P message 41 32 115 196617 0 0 0 0 1 1 1 1 1 1 0 0;
#P newex 376 199 50 196617 +;
#P newex 307 392 62 196617 prepend set;
#P number 307 375 66 9 0 0 0 3 0 0 0 221 221 221 222 222 222 0 0 0;
#P number 307 357 66 9 0 0 16 3 0 0 0 221 221 221 222 222 222 0 0 0;
#P number 307 339 112 9 0 0 2049 3 0 0 0 221 221 221 222 222 222 0 0 0;
#P comment 93 67 236 196617 This relies on there being exactly 12
binary digits;
#P connect 2 0 3 0;
#P connect 1 0 2 0;
#P connect 3 0 4 0;
#P fasten 4 0 1 0 312 415 301 415 301 332 312 332;
#P connect 29 0 28 0;
#P connect 12 0 29 0;
#P connect 13 0 30 0;
#P connect 30 0 44 0;
#P connect 11 0 27 0;
#P connect 27 0 45 0;
#P connect 10 0 26 0;
#P connect 26 0 25 0;
#P connect 9 0 24 0;
#P connect 24 0 46 0;
#P connect 32 0 31 0;
#P connect 14 0 32 0;
#P connect 33 0 43 0;
#P connect 15 0 33 0;
#P connect 43 0 31 1;
#P connect 46 0 23 1;
#P connect 25 0 46 1;
#P connect 45 0 25 1;
#P connect 28 0 45 1;
#P connect 44 0 28 1;
#P connect 31 0 44 1;
#P connect 34 0 43 1;
#P connect 5 0 34 1;
#P connect 36 0 5 0;
#P connect 37 0 5 1;
#P connect 41 0 42 0;
#P connect 40 0 41 0;
#P connect 23 0 40 0;
#P connect 19 0 39 0;
#P connect 18 0 38 0;
#P connect 17 0 36 0;
#P connect 16 0 35 0;
#P connect 39 0 37 1;
#P connect 38 0 37 0;
#P connect 35 0 34 0;
#P connect 22 0 23 0;
#P connect 7 0 21 1;
#P connect 7 0 8 0;
#P connect 21 0 22 0;
#P connect 8 0 9 0;
#P connect 8 1 10 0;
#P connect 8 2 11 0;
#P connect 8 3 12 0;
#P connect 8 4 13 0;
#P connect 8 5 14 0;
#P connect 8 6 15 0;
#P connect 8 7 16 0;
#P connect 8 8 17 0;
#P connect 8 9 18 0;
#P connect 8 10 19 0;
#P connect 20 0 21 0;
#P connect 20 1 7 0;
#P connect 6 0 20 0;
#P window clipboard copycount 47;


Chris Muir | "There are many futures and only one status quo.
cbm@well.com | This is why conservatives mostly agree,
http://www.xfade.com | and radicals always argue." – Brian Eno


May 2, 2007 | 9:10 am

rens veltman schrieb:
> sorry for not being more specific. but you showed me the direction with
> your patch already (if there is one..) my problem seems to be that max
> is cutting the leading zeroes from my binary parsing. have to find a way
> around.

cutting the leading zeros should be no problem, they mean nothing. My
patch below will accept whatever length you give it (up to 256 bit)…

but beware, in a 32- bit world it would not make sense to go beyond 32
(though if the first bits are all set to zero, still no problem… ;-)

#P window setfont "Sans Serif" 9.;
#P window linecount 1;
#N vpatcher 20 74 620 474;
#P window setfont "Sans Serif" 9.;
#P window linecount 1;
#P newex 86 70 36 196617 zl rev;
#P window linecount 2;
#P newex 158 133 43 196617 prepend set;
#P window linecount 1;
#P newex 50 178 83 196617 accum;
#P newex 86 154 24 196617 < <;
#P newex 50 50 120 196617 t b l 0 0;
#N counter;
#X flags 0 0;
#P newobj 100 133 54 196617 counter;
#P newex 86 112 24 196617 t i b;
#P newex 86 91 24 196617 iter;
#P inlet 50 30 15 0;
#P outlet 50 200 15 0;
#P connect 1 0 5 0;
#P fasten 8 0 7 0 163 174 55 174;
#P connect 5 0 7 0;
#P connect 7 0 0 0;
#P connect 5 1 9 0;
#P connect 9 0 2 0;
#P connect 2 0 3 0;
#P connect 3 0 6 0;
#P connect 6 0 7 1;
#P connect 3 1 4 0;
#P connect 4 0 6 1;
#P connect 5 2 4 2;
#P connect 5 3 8 0;
#P pop;
#P newobj 32 85 57 196617 p bin2num;
#P message 113 63 50 196617 0 1 0 1;
#P number 32 109 35 9 0 0 0 3 0 0 0 221 221 221 222 222 222 0 0 0;
#P message 32 63 78 196617 0 0 0 1 0 1 1 1;
#P connect 0 0 3 0;
#P fasten 2 0 3 0 118 81 37 81;
#P connect 3 0 1 0;
#P window clipboard copycount 4;


Stefan Tiedje————x——-
–_____———–|————–
–(_|_ —-|—–|—–()——-
– _|_)—-|—–()————–
———-()——–www.ccmix.com


May 2, 2007 | 10:02 am

At 11:10 AM +0200 5/2/07, Stefan Tiedje wrote:
>My patch below will accept whatever length you give it (up to 256 bit)…

Lovely solution.

-C


Chris Muir | "There are many futures and only one status quo.
cbm@well.com | This is why conservatives mostly agree,
http://www.xfade.com | and radicals always argue." – Brian Eno


May 2, 2007 | 1:45 pm

On May 2, 2007, at 2:03 AM, Chris Muir wrote:

> Could you post the relevant portion of your patch?

just want to web in all your helpfull suggestion. coming out then.

———————————
rens.veltman@mac.com
++43 650 7603596
———————————


May 2, 2007 | 2:31 pm

On May 2, 2007, at 3:19 AM, Chris Muir wrote:

> I’m still not sure I completely understand your problem, but here’s
> a brute force method of doing something:

thanks chris (and all). helpfull indeed (so many ways to rome…).
but my problem persists, because native handling of the number box
(here as binary box) in max truncate the leading zeros, chanching btw
a 12bit num into a 10bit num. that means i have to substitute the
lost 0′s, because i have to deal with the originally
value. never declared why, i think.

i’ will be posting a patch (hopefully soon) to show what i want to
reach. still at the beginning (struggling with my skills).

what if done so far.

step 1:
reading an black and white images (24 by 32 pixels, representing
letters from a-z from a whole font, at least the meaningfull signs)
into jit.op. -> eliminating all halftonesof a plane -> unpacking the
values (at that time only 0′s and 255′s are left) after having them
parsed into a bitlist, row by row into a jit.cellblock, with
corresponding (to the imagesize) cells and rows. now i can see a
picture of my fontletter in the cellblock. now with the help of all
of you i am able to convert my binary rows (one row for example looks
like "0 1 1 0 0 0 0 0 0 1 1 0", the top row of my "H") to a table of
hexadecimal values. has to be portable to a microcontroller (sort of
led-display, so called "pov – persistance of vision". < - lots of
projects on the web).

so far so good. but my (faraway, i know) goal is to simulate this pov
- phenomenon via jitter. for better understanding how a pov can be
done in reality and testing purposes and to learn some math and
maxmspjitter.

yours rv

———————————
rens.veltman@mac.com
++43 650 7603596
———————————

———————————
rens.veltman@mac.com
++43 650 7603596
———————————


May 3, 2007 | 6:57 am

>but my problem persists, because native handling of the
>number box (here as binary box) in max truncate the leading
>zeros, chanching btw a 12bit num into a 10bit num.
>that means i have to substitute the lost 0′s, because i
>have to deal with the originally value.

It’s really hard to help without actually seeing your problem. Please post the part of your patch that you’re having trouble with.

-C



dnz
January 18, 2008 | 1:01 am

Hi

I’m trying to read coll list into textedit, problem is that it
doesn’t recognise commas & semi colons. Can anyone shed some light on
this?

My use is pretty simple as i want to edit the coll object while
running in the main patch, without having to open the coll window.

best

denniz
i’m on OS 10.4.10, mac 4.6.3


January 18, 2008 | 10:02 am

Sonic Kooking schrieb:
> My use is pretty simple as i want to edit the coll object while running
> in the main patch, without having to open the coll window.

I would do this with jit.cellblock, you can link it to a coll…

Stefan


Stefan Tiedje————x——-
–_____———–|————–
–(_|_ —-|—–|—–()——-
– _|_)—-|—–()————–
———-()——–www.ccmix.com



dnz
January 18, 2008 | 4:48 pm

ah. i could do it this way as well.

thanks

dennis
Am 18.01.2008 um 11:02 schrieb Stefan Tiedje:

> Sonic Kooking schrieb:
>> My use is pretty simple as i want to edit the coll object while
>> running in the main patch, without having to open the coll window.
>
> I would do this with jit.cellblock, you can link it to a coll…
>
> Stefan
>
> —
> Stefan Tiedje————x——-
> –_____———–|————–
> –(_|_ —-|—–|—–()——-
> — _|_)—-|—–()————–
> ———-()——–www.ccmix.com
>
>


January 21, 2008 | 8:14 am

On 20 janv. 08, at 17:43, Werner Funk wrote:
> Am 18.01.2008 um 02:01 schrieb Sonic Kooking:
>>
>> I’m trying to read coll list into textedit, problem is that it
>> doesn’t recognise commas & semi colons. Can anyone shed some light
>> on this?
>>
>> My use is pretty simple as i want to edit the coll object while
>> running in the main patch, without having to open the coll window.
>
> If you have any spreadsheet such as iWork, AppleWorks, OpenOffice,
> MS Works or Exsel etc, you easily edit colls by using columns and
> ‘copy downward’ etc., even with functions to calculate entries.

It’s just too bad that [coll]‘s readagain message doesn’t work since
Max 4.3 when the file is changed in an external editor. Now you have
to use "read mycoll.txt" instead, which is not so nice.

_____________________________
Patrick Delges

Centre de Recherches et de Formation Musicales de Wallonie asbl

http://www.crfmw.be/max


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