path: root/doc/pddp/otherbinops-help.pd

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#X obj 494 28 |;
#X obj 574 28 &&;
#X obj 601 28 ||;
#X obj 7 25 >;
#X obj 36 25 >=;
#X obj 67 24 ==;
#X obj 125 24 <=;
#X obj 153 24 <;
#X obj 217 417 >;
#X obj 267 496 ==;
#X obj 96 24 !=;
#X obj 521 28 <<;
#X obj 548 28 >>;
#X obj 627 28 %;
#X text 464 5 THE LOGICAL OPERATORS -- A.K.A. "Bit Twiddling";
#X text 6 6 THE RELATIONAL OPERATORS;
#N canvas 0 22 454 304 understanding_%_modulus 0;
#X text 24 23 MODULUS - [%];
#X floatatom 28 187 0 0 0 0 - - -;
#X text 22 40 - this object has nothing to do with percentage!;
#X text 20 54 - a modulus is a number by which two given numbers can
be divided and produce the same remainder.;
#X text 21 81 - in the example below: 9 / 2 = 4.5 \, and 7 / 2 = 3.5.
Hence if 7 and 9 are divided by 2 \, then the remainder of both equations
is .5. Therefore \, the modulus of 7 and 9 is "2".;
#X msg 28 138 9;
#X obj 28 166 % 7;
#X floatatom 62 142 5 0 0 0 - - -;
#X text 20 222 Note that the modulus operator is not a "bitwise" operator
\, but a math function.;
#X connect 5 0 6 0;
#X connect 6 0 1 0;
#X connect 7 0 6 0;
#X restore 476 418 pd understanding_%_modulus;
#X text 478 252 Below is a brief explanation of each of these logical
operators.;
#X text 473 53 These objects are adopted from the mother of all object
oriented languages: C. They are "bitwise" operators which perform logical
and shift operations on 32-bit numbers.;
#X text 467 100 WHAT DOES "BITWISE" MEAN?;
#X text 478 208 Hence \, performing "bitwise" relational tests means
that Pd can compare "1101" to "1001" instead of operating with the
integers that are represented by those binary codes.;
#N canvas 81 197 456 306 understanding_&_AND 0;
#X obj 33 216 &;
#X floatatom 87 182 5 0 0 0 - - -;
#X floatatom 129 183 5 0 0 0 - - -;
#X msg 33 154 13;
#X msg 62 155 9;
#X text 18 18 [&] -- This is the bitwise AND operator which returns
a "1" for each bit position where the corresponding bits of both its
operands are "1". For example:;
#X text 22 67 13 = "1101";
#X text 28 79 9 = "1001";
#X text 15 92 Hence:"1001";
#X obj 33 114 bng 15 250 50 0 empty empty empty 0 -6 0 8 -262144 -1
-1;
#X obj 33 132 t b b;
#X text 101 66 When comparing the binary codes for 13 and 9 \, we can
see that the first and fourth digits of both codes are 1 Hence the
result will be "1001" -- in other words "9".;
#X floatatom 33 238 0 0 0 0 - - -;
#X connect 0 0 12 0;
#X connect 1 0 0 0;
#X connect 2 0 0 1;
#X connect 3 0 0 0;
#X connect 4 0 0 1;
#X connect 9 0 10 0;
#X connect 10 0 3 0;
#X connect 10 1 4 0;
#X restore 478 286 pd understanding_&_AND;
#N canvas 190 317 454 304 understanding_|_OR 0;
#X floatatom 32 247 0 0 0 0 - - -;
#X floatatom 86 191 5 0 0 0 - - -;
#X floatatom 128 192 5 0 0 0 - - -;
#X msg 32 163 13;
#X msg 61 164 9;
#X text 21 76 13 = "1101";
#X text 27 88 9 = "1001";
#X obj 32 123 bng 15 250 50 0 empty empty empty 0 -6 0 8 -262144 -1
-1;
#X obj 32 141 t b b;
#X text 18 18 [|] -- This is the bitwise OR operator which returns
a "1" for each bit position where one OR both of the corresponding
bits of both its operands is a "1". For example:;
#X text 14 101 Hence:"1101";
#X text 98 76 When comparing the binary codes for 13 and 9 \, we can
see that the first and fourth digits of both codes are both 1 and the
second position of 13 is a one. Hence the result will be "1101" --
in other words "13".;
#X obj 32 225 |;
#X connect 1 0 12 0;
#X connect 2 0 12 1;
#X connect 3 0 12 0;
#X connect 4 0 12 1;
#X connect 7 0 8 0;
#X connect 8 0 3 0;
#X connect 8 1 4 0;
#X connect 12 0 0 0;
#X restore 478 307 pd understanding_|_OR;
#N canvas 0 22 454 304 understanding_<<_LEFT-SHIFT 0;
#X obj 46 142 bng 15 250 50 0 empty empty empty 0 -6 0 8 -262144 -1
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#X obj 46 160 t b b;
#X msg 46 181 13;
#X obj 46 222 <<;
#X floatatom 46 244 5 0 0 0 - - -;
#X msg 74 181 2;
#X floatatom 112 193 5 0 0 0 - - -;
#X floatatom 160 193 5 0 0 0 - - -;
#X text 29 25 [<<] -- This is the left shift operator and it works
by shifting the digits of the binary representation of the first operand
(left inlet) to the left by the number of places specified by the second
operand (right inlet). The spaces created to the right are filled by
zeros \, and any digits falling off the left are discarded. The following
code returns 52 as the binary of 13 ("1101") is shifted two places
to the left giving "110100":;
#X connect 0 0 1 0;
#X connect 1 0 2 0;
#X connect 1 1 5 0;
#X connect 2 0 3 0;
#X connect 3 0 4 0;
#X connect 5 0 3 1;
#X connect 6 0 3 0;
#X connect 7 0 3 1;
#X restore 477 328 pd understanding_<<_LEFT-SHIFT;
#N canvas 0 22 456 380 understanding_>>_RIGHT-SHIFT 0;
#X obj 41 155 bng 15 250 50 0 empty empty empty 0 -6 0 8 -262144 -1
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#X obj 41 173 t b b;
#X floatatom 41 257 5 0 0 0 - - -;
#X floatatom 107 206 5 0 0 0 - - -;
#X floatatom 155 206 5 0 0 0 - - -;
#X msg 41 194 13;
#X obj 41 235 >>;
#X msg 69 194 2;
#X text 33 21 [>>] -- This is the sign-propagating right shift operator
which shifts the digits of the binary representation of the first operand
(left inlet) to the right by the number of places specified by the
second operand (right inlet) \, discarding any shifted off to the right.
The copies of the leftmost bit are added on from the left \, thereby
preserving the sign of the number. This next examples returns 3 ("11")
as the two right-most bits of 13 ("1101") are shifted off to the right
and discarded.;
#X text 33 284 Note that this object preserves negative values for
negative operands. ("sign-propagating").;
#X connect 0 0 1 0;
#X connect 1 0 5 0;
#X connect 1 1 7 0;
#X connect 3 0 6 0;
#X connect 4 0 6 1;
#X connect 5 0 6 0;
#X connect 6 0 2 0;
#X connect 7 0 6 1;
#X restore 477 350 pd understanding_>>_RIGHT-SHIFT;
#N canvas 56 51 528 425 understanding_&&_LOGICAL-AND 0;
#X msg 56 269 5;
#X obj 25 319 &&;
#X floatatom 25 339 5 0 0 0 - - -;
#X floatatom 194 277 5 0 0 0 - - -;
#X text 12 26 [&&] - This is the logical AND operator \, which returns
a Boolean true (a one) if both operands are true. Logically it follows
that if the first operand is false \, then the whole expression is
false \, and this is how the objects works: It first evaluates the
left hand operand (left inlet) and if this returns false (zero) then
\, without going any further \, it returns a false (a zero). Otherwise
it returns the value of the second operand (right inlet).;
#X floatatom 237 277 5 0 0 0 - - -;
#X text 25 364 Note that this is not a bitwise operator. It compares
floats.;
#X obj 25 227 bng 15 250 50 0 empty empty empty 0 -6 0 8 -262144 -1
-1;
#X obj 25 245 t b b;
#X msg 25 269 17;
#X text 12 145 In other words \, IF the left inlet is zero \, THEN
output zero. ELSEIF the left inlet is non-zero AND the right inlet
is zero \, then output zero. ELSEIF the left inlet is non-zero AND
the right inlet is non-zero \, THEN output non-zero!;
#X obj 91 227 bng 15 250 50 0 empty empty empty 0 -6 0 8 -262144 -1
-1;
#X obj 91 245 t b b;
#X msg 91 269 17;
#X msg 122 269 0;
#X connect 0 0 1 1;
#X connect 1 0 2 0;
#X connect 3 0 1 0;
#X connect 5 0 1 1;
#X connect 7 0 8 0;
#X connect 8 0 9 0;
#X connect 8 1 0 0;
#X connect 9 0 1 0;
#X connect 11 0 12 0;
#X connect 12 0 13 0;
#X connect 12 1 14 0;
#X connect 13 0 1 0;
#X connect 14 0 1 1;
#X restore 477 373 pd understanding_&&_LOGICAL-AND;
#N canvas 244 51 530 427 understanding_||_LOGICAL-OR 0;
#X msg 56 269 5;
#X floatatom 25 339 5 0 0 0 - - -;
#X floatatom 196 280 5 0 0 0 - - -;
#X floatatom 239 280 5 0 0 0 - - -;
#X text 25 364 Note that this is not a bitwise operator. It compares
floats.;
#X obj 25 227 bng 15 250 50 0 empty empty empty 0 -6 0 8 -262144 -1
-1;
#X obj 25 245 t b b;
#X msg 25 269 17;
#X text 17 21 [||] -- This is the logical OR operator and it returns
a value of true (non-zero) if one or both of the operands is true.
It works by first evaluating the left-hand operand (left inlet) and
\, if this is true \, diregarding the right-hand operand (right inlet)
and returning a non-zero. If \, however \, the left-hand operand (left
inlet) is false \, then it returns the value of the right-hand operand
(right inlet).;
#X text 12 145 In other words \, IF the left inlet is non-zero \, THEN
output non-zero. ELSEIF the left inlet is zero AND the right inlet
is zero \, then output zero. ELSEIF the left inlet is zero AND the
right inlet is non-zero \, THEN output non-zero!;
#X obj 25 319 ||;
#X obj 96 226 bng 15 250 50 0 empty empty empty 0 -6 0 8 -262144 -1
-1;
#X obj 96 244 t b b;
#X msg 96 268 0;
#X msg 127 268 0;
#X connect 0 0 10 1;
#X connect 2 0 10 0;
#X connect 3 0 10 1;
#X connect 5 0 6 0;
#X connect 6 0 7 0;
#X connect 6 1 0 0;
#X connect 7 0 10 0;
#X connect 10 0 1 0;
#X connect 11 0 12 0;
#X connect 12 0 13 0;
#X connect 12 1 14 0;
#X connect 13 0 10 0;
#X connect 14 0 10 1;
#X restore 477 395 pd understanding_||_LOGICAL-OR;
#X obj 432 12 bng 15 250 50 0 empty empty empty 0 -6 0 8 -262144 -1
-1;
#X obj 432 607 bng 15 250 50 0 empty empty empty 0 -6 0 8 -262144 -1
-1;
#X obj 54 186 == 42;
#X floatatom 54 165 5 0 0 0 - - -;
#X floatatom 28 212 2 0 0 0 - - -;
#X obj 53 211 bng 15 250 50 0 empty empty empty 0 -6 0 8 -262144 -1
-1;
#X msg 24 161 42;
#X text 9 143 For example: IF 42 is equal to x \, then "1" (True);
#X text 73 203 Note that the object outputs 1 or 0 with every incoming
message.;
#X text 10 233 All of these objects operate the same way. The right
inlet or creation argument sets the "condition" to which the incoming
messages are compared. The left inlet accepts numbers or a "bang" --
a number will reset the value and output a true or false (1 or 0) depending
on whether or not the incoming value meets the necessary condition.
A "bang" will force the object to output a true or false (1 or 0) based
on the value that is already stored in the left inlet.;
#X floatatom 25 378 5 0 0 0 - - -;
#X floatatom 98 345 5 0 0 0 - - -;
#X text 10 376 a;
#X text 138 344 b;
#X text 63 439 Is a greater than b?;
#X floatatom 242 478 0 0 0 0 - - -;
#X text 17 478 Is a greater than or equal to b?;
#X obj 242 457 >=;
#X text 136 517 Is a equal to b?;
#X obj 295 534 !=;
#X floatatom 295 554 0 0 0 0 - - -;
#X obj 325 367 r a_b;
#X obj 325 386 unpack f f;
#X obj 25 395 pack f f;
#X obj 25 415 s a_b;
#X obj 98 361 bang;
#X text 142 555 Is a NOT equal to b?;
#X floatatom 321 592 0 0 0 0 - - -;
#X text 185 592 Is a less than b?;
#X obj 321 572 <;
#X floatatom 346 631 0 0 0 0 - - -;
#X obj 346 611 <;
#X text 138 631 Is a less than or equal to b?;
#X text 464 583 This document was updated for Pd version 0.35 test
29 by Dave Sabine as part of a project called pddp proposed by Krzysztof
Czaja to build comprehensive documentation for Pd.;
#X text 461 460 RELATED OBJECTS;
#X obj 853 477 +;
#X text 460 477 Visit the Help document for MATH for more math objects:
;
#N canvas 0 22 452 302 related_objects_from_other_libraries 0;
#X obj 47 34 strcomp;
#X text 102 33 Relational tests for strings.;
#X text 29 104 These objects are offered in Pd only if you have downloaded
and properly installed the appropriate library. These objects may or
may not exist in a single library.;
#X text 28 153 The best places to find information about Pd's libraries
is:;
#X text 25 175 www.puredata.org and click on "Downloads" then "Software"
;
#X text 27 190 or;
#X text 27 205 iem.kug.ac.at/pdb/;
#X restore 482 501 pd related_objects_from_other_libraries;
#X text 478 120 Well \, these objects perform "relational" tests on
the binary forms of 32-bit numbers. For example \, the number 13 is
represented in your computer's operating system in binary code by "1101"
and the number 9 is "1001". Each of those binary digits is an 8-bit
word: 8 bits * 4 digits = 32-bits!;
#X obj 179 24 mod;
#X obj 206 24 div;
#X text 8 84 Most relational operators output a boolean value: true
or false (1 or 0) depending on the relation between the input (left
inlet) and the condition (right inlet or creation argument).;
#N canvas 7 22 514 656 understanding_MOD_and_DIV 0;
#X text 24 5 [mod] and [div] are helpful objects to determine whether
or not a fraction produces a remainder \, or to determine the value
of the remainder.;
#X text 24 80 while \, 4 / 3 = 1 with a remainder of 1;
#X text 25 51 For example \, 3 / 3 = 1 with a remainder of zero (i.e.
no remainder).;
#X floatatom 26 190 0 0 0 0 - - -;
#X floatatom 26 232 0 0 0 0 - - -;
#X floatatom 138 192 0 0 0 0 - - -;
#X text 58 191 divided by;
#X text 173 193 has a remainder of;
#X floatatom 300 193 0 0 0 0 - - -;
#X obj 26 211 mod;
#X text 22 103 [mod] takes a number in its left inlet and will divide
that number by either the creation argument or the number given at
its left inlet and will produce the value of the remainder at its outlet.
If no creation argument is given \, then the default value is 1;
#X obj 78 173 loadbang;
#X msg 138 173 1;
#X text 23 255 [div] takes a number in its left inlet and will divide
that number by either the creation argument or the number given at
its left inlet and will produce the result without a remainder. If
no creation argument is given \, then the default value is 1;
#X floatatom 28 341 0 0 0 0 - - -;
#X floatatom 28 383 0 0 0 0 - - -;
#X floatatom 140 343 0 0 0 0 - - -;
#X text 60 342 divided by;
#X floatatom 256 344 0 0 0 0 - - -;
#X obj 80 324 loadbang;
#X msg 140 324 1;
#X obj 28 362 div;
#X text 176 343 is equal to;
#X text 294 343 with no remainder.;
#X obj 257 371 /;
#X floatatom 257 391 0 0 0 0 - - -;
#X text 227 389 or;
#X text 297 392 with a remainder.;
#X text 23 408 In the following example \, I've built a metronome which
counts bar numbers and beat numbers: default time signature is 4/4
(Common Time).;
#X obj 23 489 metro 500;
#X obj 23 470 tgl 15 0 empty empty Start-Stop 0 -6 0 8 -262144 -1 -1
0 1;
#X obj 48 510 + 1;
#X floatatom 23 530 0 0 0 0 - - -;
#X text 52 532 Total Beat Count;
#X obj 23 559 div 4;
#X obj 134 560 mod 4;
#X floatatom 219 601 0 0 0 0 - - -;
#X floatatom 108 600 0 0 0 0 - - -;
#X obj 23 510 f 1;
#X msg 107 468 1;
#X obj 23 579 + 1;
#X obj 134 579 + 1;
#X text 131 468 Reset;
#X text 34 599 Bar number;
#X text 147 601 Beat Count;
#X floatatom 339 511 0 0 0 0 - - -;
#X text 176 511 How many beats per bar?;
#X connect 3 0 9 0;
#X connect 4 0 8 0;
#X connect 5 0 9 1;
#X connect 9 0 4 0;
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#X connect 16 0 21 1;
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#X connect 20 0 16 0;
#X connect 21 0 15 0;
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#X connect 31 0 38 1;
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#X connect 34 0 40 0;
#X connect 35 0 41 0;
#X connect 38 0 31 0;
#X connect 38 0 32 0;
#X connect 39 0 38 1;
#X connect 40 0 37 0;
#X connect 41 0 36 0;
#X connect 45 0 35 1;
#X connect 45 0 34 1;
#X restore 9 58 pd understanding_MOD_and_DIV;
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#X connect 12 0 1 0;
#X connect 30 0 31 0;
#X connect 32 0 34 0;
#X connect 32 0 35 0;
#X connect 33 0 32 0;
#X connect 36 0 32 0;
#X connect 40 0 53 0;
#X connect 41 0 53 1;
#X connect 41 0 55 0;
#X connect 47 0 45 0;
#X connect 49 0 50 0;
#X connect 51 0 52 0;
#X connect 52 0 11 0;
#X connect 52 0 47 0;
#X connect 52 0 12 0;
#X connect 52 0 49 0;
#X connect 52 0 59 0;
#X connect 52 0 61 0;
#X connect 52 1 11 1;
#X connect 52 1 47 1;
#X connect 52 1 12 1;
#X connect 52 1 49 1;
#X connect 52 1 59 1;
#X connect 52 1 61 1;
#X connect 53 0 54 0;
#X connect 55 0 40 0;
#X connect 59 0 57 0;
#X connect 61 0 60 0;