Arduino first timer!! Why don`t this program work

Projects Programming
1

String Innkommende;
int Skallsikring1 = 1;
int Skallsikring2 = 2;
int Skallsikring3 = 3;
int UtgangAlarm = 13;
void setup() {
Serial.begin(9600);
pinMode(Skallsikring1, INPUT);
pinMode(Skallsikring2, INPUT);
pinMode(Skallsikring3, INPUT);
pinMode(UtgangAlarm, OUTPUT);
}

void loop() {
if (Serial.available()) {
if ((Skallsikring1 != LOW) || (Skallsikring2 != LOW)|| (Skallsikring3 != LOW) == HIGH){
digitalWrite(UtgangAlarm, HIGH);
Serial.print("Situasjon:");
Serial.println(7); //Whatever input status - program will always go this way???

} else {
digitalWrite(UtgangAlarm, LOW);
Serial.print("Situasjon:");
Serial.println(6);
}

delay(1000);
}

}

2

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I'm guessing this is your problem:

(Skallsikring3 != LOW) == HIGH
4

Thank you for your help.

This is my first attempt at programming of Arduino. Also my first posting on this forum. Sorry for missing information in my frustrating question.

The software are ment to have three inputs and one output. If one of the inputs are active it shall activate the output.

When the output is HIGH then send a serial print
When the output is LOW then send a serial print

6

Thanks.

You gave me progress with that by the way!. Now the output is not always HIGH any more, but not possible to get HIGH.

I have used many hours (do not dear to tell how many) trying to get a grip of this programming metod. This is totally new for me.

8

Hope that this are posted correctly?

I am very grateful for your help.

  String Innkommende;
  int Skallsikring1 = 2;
  int Skallsikring2 = 3;
  int Skallsikring3 = 4;
  int UtgangAlarm = 13;
  void setup() {
      Serial.begin(9600);
      pinMode(Skallsikring1, INPUT);
      pinMode(Skallsikring2, INPUT);
      pinMode(Skallsikring3, INPUT);
      pinMode(UtgangAlarm, OUTPUT);
  }

  void loop() {
      if (Serial.available()) {
            if ((Skallsikring1 == HIGH) || (Skallsikring2 == HIGH)|| (Skallsikring3 == HIGH) == HIGH){
              digitalWrite(UtgangAlarm, HIGH);
                Serial.print("Situasjon:");
                Serial.println(7); 
          } else {
              digitalWrite(UtgangAlarm, LOW);
                Serial.print("Situasjon:");
                Serial.println(6);
          }

          
           
          delay(1000);
      }


  }
10 
  int Skallsikring1 = 2;
  
            if ((Skallsikring1 == HIGH)

If I tell you that HIGH has the value 1, how likely do you think it is that that "if" expression will ever be true?

11

I would like to see how it’s wired up and if your using something that requires a pull up or a pull down resistor.

You have inputs but they may not be wired correctly or you need to use input pull-up in addition to code corrections.

Also get rid of Serial available. What do you need it for?

The reason your having an issue is because you have to monitor the state change of your inputs.

You may keep your first set of ints as they were originally but you must use a state variable and tell that to be the read of your PIN number whether the pin is 1 or 2 or whatever.

Then when you write the of statement use the state variable as your high and low. NOT THE PIN INT.

12 
  int Skallsikring1 = 2;
  int Skallsikring2 = 3;
  int Skallsikring3 = 4;
  int UtgangAlarm = 13;
  void setup() {
      Serial.begin(9600);
      pinMode(Skallsikring1, INPUT);
      pinMode(Skallsikring2, INPUT);

So SkallsikringX are Arduino pin numbers?

      if (Serial.available()) {
            if ((Skallsikring1 == HIGH) || (Skallsikring2 == HIGH)|| (Skallsikring3 == HIGH) == HIGH){
              digitalWrite(UtgangAlarm, HIGH);

[/code]

But here, you are accessing only the pin numbers; to test the values of the pins themselves, you need to access them via the digitalRead() function:

           if ((digitalRead(Skallsikring1) == HIGH) || (digitalRead(Skallsikring2) == HIGH) ...

Also, you check Serial.available(), but never actually read the data, so once you've sent data, Serial.available() will be true forever...

13

Thanks for all help.

Finally the program works as it should.