How to fix an object in a certain position(x,y)?

Programming Questions

diastudent

How to fix an object in a certain position(x,y)?

in Programming Questions • 1 year ago

Hello,

I am a beginner in Processing, and I need help.

My question is: how to fix an object in a certain position(x,y)?

Here is the code I use. What I want is to take the first rectangle by pressing the 'a' key, move it with the mouse, and let it where it is. Then I want to take the second rectangle, move it, and let it where it is. And so on with the two next rectangles. I want to keep the possibility to reset the position of the rectangles when I want, and to control the position of each one independantly of each other.

It shouldn't be difficult, but I don't know how to do.

Thank you if somebody has the answer!

float x;

float y;

void setup() {

size (800,800);

smooth();

noStroke();

}

void draw(){

background(0);

x=mouseX -400;

y=mouseY -100;

fill(255);

//A rectangle

if (key == 'a' || key == 'A') {

rect(56+x, 42+y, 293, 61);

}

else {

rect (56,42,293,61);

}

//B rectangle

if (key == 'b' || key == 'B') {

rect(485+x, 42+y, 211, 61);

}

else {

rect(485, 42, 211, 61);

}

//C rectangle

if (key == 'c' || key == 'C') {

rect(358+x, 281+y, 119, 66);

}

else {

rect(358, 281, 119, 66);

}

//D rectangle

if (key == 'd' || key == 'D') {

rect(582+x, 355+y, 63, 118);

}

else {

rect(582, 355, 63, 118);

}

Replies(5)

tfguy44

Re: How to fix an object in a certain position(x,y)?

1 year ago

class Rectangle {
color c;
int x, y, w, h;
Rectangle( int _x, int _y, int _w, int _h ) {
x = _x;
y = _y;
w = _w;
h = _h;
c = color(255);
}
void draw() {
pushStyle();
fill(c);
noStroke();
rect(x, y, w, h);
popStyle();
}
void setColor(color _c) {
c = _c;
}
void adjustPosition(int _x, int _y ) {
x = x + _x;
y = y + _y;
}
}
int active;
Rectangle r1, r2, r3, r4;
// You could also have an array or arrayList of them!
// That way you could simplify a lot of the code below...
void setup() {
size(800, 800);
smooth();
r1 = new Rectangle( 56, 42, 293, 61 );
r2 = new Rectangle( 485, 42, 211, 61 );
r3 = new Rectangle( 358, 281, 119, 66 );
r4 = new Rectangle( 582, 355, 63, 118 );
active = 0;
}
void draw() {
background(0);
r1.draw();
r2.draw();
r3.draw();
r4.draw();
}
void mouseMoved() {
if ( active == 1 ) {
r1.adjustPosition( mouseX - pmouseX, mouseY - pmouseY );
}
else if ( active == 2 ) {
r2.adjustPosition( mouseX - pmouseX, mouseY - pmouseY );
}
else if ( active == 3 ) {
r3.adjustPosition( mouseX - pmouseX, mouseY - pmouseY );
}
else if ( active == 4 ) {
r4.adjustPosition( mouseX - pmouseX, mouseY - pmouseY );
}
}
void keyPressed() {
if (key == 'a' || key == 'A') {
active = active==1?0:1;
}
if (key == 'b' || key == 'B') {
active = active==2?0:2;
}
if (key == 'c' || key == 'C') {
active = active==3?0:3;
}
if (key == 'd' || key == 'D') {
active = active==4?0:4;
}
}

PhiLho

Re: How to fix an object in a certain position(x,y)?

1 year ago

As shown, you need a x, y pair of coordinates for each object. tfguy44 took your title literally and made a class (thus you create 'objects' in the OOP sense), but you can also store the coordinates in one array (using a PVector to store them) or in two arrays (one for x, one for y). The last solution is the most common among beginners, but the less effective (the arrays are not connected, must be kept synchronized if changed, etc.). The class approach is the best, and not so hard: here, you can see it as a way to define parameters for each object (color, position, dimensions) and functions specifics to this kind of object. It greatly reduces duplication of code and isolates (encapsulates) functionalities specific to this class of objects.
Later, you can define a class for "shape" (color, position for example) and inherit from it for rectangle, triangle, etc., each sub-class adding parameters (dimensions, etc.) to the base class. The base class can be abstract, meaning that you cannot instantiate it, as it has no defined shape, and with some functions that remain empty (like draw()). This way, you know you can call draw() on all Shape objects, but it is really defined only in the sub-classes.

diastudent

Re: How to fix an object in a certain position(x,y)?

1 year ago

Thank you very much, in the idea that's perfectly what I was searching for. But I think I made a mistake: to explain this more simply, I simplified the code. Now, I'm trying to mix it with the real code.

In this real one, let's say each rectangle has a different color (white, yellow, red and blue), and instead of control it with the mouse, I have to control it with an Accelerometer through Arduino.

Do you know how to do it? It could be super great! If you need me to explain it a bit more precise or to paste the real code here, just ask me =)

Anyway, thank you again!

diastudent

Re: How to fix an object in a certain position(x,y)?

1 year ago

I think it's better to understand if I post the real code:

(Sorry if it looks a bit barbarian!)

I'm not sure to use easing, it's maybe complexifying things for nothing. But, if possible, I would still like to control each rectangle independantly from the others, without taking care about the colors, but keeping 4 different ones.

import processing.serial.*;

import cc.arduino.*;

Arduino arduino;

//coords

int cordx;

int cordy;

int cordz;

//floats_easing

float x = cordx;

float y = cordy;

float z = cordz;

float targetX, targetY, targetZ;

float easing = 0.1;

void setup()

{

println(Arduino.list());

arduino = new Arduino(this, Arduino.list()[1], 57600);

size(1430,820);

noStroke();

smooth();

}

void draw(){

background(0);

translate (300,120);

cordx = arduino.analogRead(1);

cordy = arduino.analogRead(2);

cordz = arduino.analogRead(3);

targetX = cordx-354;

float dx = targetX - x;

if(abs(dx)>1) {

x+= dx * easing;

}

targetY = cordy-343;

float dy= targetY - y;

if(abs(dy) >1) {

y += dy * easing;

}

targetZ = cordz-320;

float dz = targetZ - z;

if(abs(dz)>1) {

z+= dz * easing;

}

println("cordx=" +cordx);

println("cordy=" +cordy);

println("cordz=" +cordz);

//white

fill(240);

if (key == 'w' || key == 'W') {

rect(10-x,10-y, 38, 93);

rect(56-x, 10-y, 293, 24);

rect(358-x, 10-y, 338, 24);

rect(358-x, 42-y, 119, 61);

rect(704-x, 10-y, 42, 262);

rect(56-x, 111-y, 88, 236);

rect(10-x, 356-y, 38, 152);

rect(10-x, 516-y, 177, 44);

rect(195-x, 111-y, 112, 161);

rect(314-x, 111-y, 35, 161);

rect(485-x, 281-y, 160, 66);

rect(653-x, 281-y, 43, 66);

rect(240-x, 355-y, 193, 118);

rect(441-x, 355-y, 133, 118);

rect(704-x, 355-y, 86, 118);

rect(582-x, 481-y, 208, 79);

}

else{

rect(10,10, 38, 93);

rect(56, 10, 293, 24);

rect(358, 10, 338, 24);

rect(358, 42, 119, 61);

rect(704, 10, 42, 262);

rect(56, 111, 88, 236);

rect(10, 356, 38, 152);

rect(10, 516, 177, 44);

rect(195, 111, 112, 161);

rect(314, 111, 35, 161);

rect(485, 281, 160, 66);

rect(653, 281, 43, 66);

rect(240, 355, 193, 118);

rect(441, 355, 133, 118);

rect(704, 355, 86, 118);

rect(582, 481, 208, 79);

}

//blue

fill(0, 0, 120);

if (key == 'b' || key == 'B') {

rect(56-x*2, 42-y*2, 293, 61);

}

else {

rect (56,42,293,61);

}

if (key == 'n' || key == 'N') {

rect(485-x*2, 42-y*2, 211, 61);

}

else {

rect(485, 42, 211, 61);

}

if (key == 'v' || key == 'V') {

rect(358-x*2, 281-y*2, 119, 66);

}

else {

rect(358, 281, 119, 66);

}

if (key == 'c' || key == 'C') {

rect(582-x*2, 355-y*2, 63, 118);

}

else {

rect(582, 355, 63, 118);

}

//yellow

fill(250, 210, 0);

if (key == 'y' || key == 'Y') {

rect(195-x*4, 281-0.5*y, 37, 66);

rect(195-x*4, 355-0.5*y, 37, 205);

rect(358-x*4, 111-0.5*y, 119, 161);

rect(653-x*4, 355-0.5*y, 43, 118);

rect(754-x*4, 10-0.5*y, 36, 262);

}

else {

rect(754, 10, 36, 262);

rect(195, 281, 37, 66);

rect(195, 355, 37, 205);

rect(358, 111, 119, 161);

rect(653, 355, 43, 118);

}

//red

fill(200, 0, 0);

if (key == 'r' || key == 'R') {

rect(240+x, 481+y, 107, 79);

rect(240+x, 281+y, 109, 66);

rect(56+x, 355+y, 88, 152);

rect(485+x, 111+y, 160, 161);

rect(653+x, 111+y, 43, 161);

}

else {

rect(240, 481, 107, 79);

rect(240, 281, 109, 66);

rect(56, 355, 88, 152);

rect(485, 111, 160, 161);

rect(653, 111, 43, 161);

}

diastudent

Re: How to fix an object in a certain position(x,y)?

1 year ago

I forgot to say: control each rectangle by keeping the fact to press a key touch assigned to each one, as in the first code. There are 30 rectangles, so I don't know if there is enough key touches for each one.

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