How to return to background in processing if a button has not been pressed for a while?

edited April 17 in Arduino

I am working on a simple project which is a bit challenging for a beginner like me. It is a project with processing and Arduino. In general, I made two buttons, A and B, in Arduino. Once the button A is pressed, Arduino will send a serial number to processing, and an image A will be displayed until the button B is pressed, which will make Arduino send another serial number to processing to display image B.

Here comes my question. I want to make an effect that** if none of buttons has been pressed for a while, the window of processing will automatically return to its background**. I have been stuck in this problem for long.. Now I have no clue about it.

This is my code for Arduino:

int switchPin2 =2;
int switchPin3 =3;

void setup() {
  pinMode(switchPin2, INPUT);
  pinMode(switchPin3, INPUT);
  Serial.begin(9600);
 }


void loop() {
 if(digitalRead(switchPin2)==HIGH && digitalRead(switchPin3)==LOW)
 {
   Serial.write(2);
   delay(1000);
 }

 else if(digitalRead(switchPin3)==HIGH && digitalRead(switchPin2)==LOW)
 {
  Serial.write(3);
  delay(1000);
 }

  delay(100);
 }

This is my code for processing:

    import processing.serial.*;
    Serial port;
   PImage img2;
   PImage img3;
   int val=0;

 void setup(){
   size(400,400);
   img2=loadImage("a.jpg");
   img3=loadImage("b.jpg");

   port= new Serial(this,"/dev/cu.usbmodem1451", 9600);
 }

 void draw(){
   background(255);
   if(0<port.available()){
      val=port.read();
   }
  if(val==2){
      image(img2,0,0);
   }

 if(val==3){
      image(img3,0,0);
   }

  }

I long for some hints ! Thank you in advance!

Answers

  • Answer ✓

    This is untested code but it should do the trick.

    Kf

    import processing.serial.*;
    Serial port;
    PImage img2;
    PImage img3;
    int val=0;
    
    int targetTime=0;
    final int kTimeToReset=2000;  //2 seconds
    
    void setup() {
      size(400, 400);
      img2=loadImage("a.jpg");
      img3=loadImage("b.jpg");
    
      port= new Serial(this, "/dev/cu.usbmodem1451", 9600);
    }
    
    void draw() {
      background(255);
      if (0<port.available()) {
        val=port.read();
        targetTime=millis()+kTimeToReset; //Resets next targetTime when data arrives
      }
    
      if (val==2) {
        image(img2, 0, 0);
      }
    
      if (val==3) {
        image(img3, 0, 0);
      }
    
      if(millis()>targetTime){
        val=-1;
      }
    
    }
    
  • @ckuoping --

    Setting aside the serial communication, consider this very simple case of an event timer (of the kind that @kfrajer shared). Any time it is activated, it sets an expire-time in the future. The event is then true until the expire time passes -- the event then becomes false.

    int timer;
    int delay = 1000;
    void draw() {
       if(millis() > timer){
         background(0);   // off
       } else {
         background(255); // on
       }
    }
    void mousePressed(){
       timer = millis() + delay;
    }
    
  • @kfrajer Thank u so much. It is really well activated.

  • edited April 21

    @kfrajer Thank you so much! it works pretty well! awesome!! and also thank @jeremydouglass it is good for me to realise Kfrajer's code with your clear explanation! There is one thing left I want to ask again: what is the function/meaning behind the script: val=-1. Is it just a random number or must it be set to -1?

  • Answer ✓

    If you check the code, the value of val drives the action in draw. If val is 2 or 3, it will display the respective image. Now, when the time runs out, we want to "not" show an image at all. For that, we need to change val to a different value. I choose this negative value assuming val will not take any negative value but only in this case, when no images are displayed. To make my code clear I could have use this next definition in a global scope in the code::

    final int SHOW_BACKGROUND = -1;

    and then:

      if(millis()>targetTime){
        val=SHOW_BACKGROUND;
      }
    

    Kf

  • @kfrajer! thanks again! Now everything is clear!

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