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Troubleshooting why Arduino logical operators doesn't compute

colporteur

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How would you write the expression to test sensors Departure (A or B) HIGH AND Arrival (A or B) HIGH?

The first else if statement shown below doesn't work. Departure A OR Departure B HIgh AND Arrival A OR Arrival B High. The OR doesn't work. Only when both Departure A AND B are HIGH the serial print. The same applies to the third else if statement both Departure A and B have to be low for the statement to function. What am I coding wrong for the OR?

      if ((digitalRead(s1DeparturePinA) == LOW || digitalRead(s1DeparturePinB) == LOW) &&
          (digitalRead(s1ArrivalPinA) == HIGH || digitalRead(s1ArrivalPinB) == HIGH))
        {
         Serial.println(F("Train on siding 1 status DEPARTURE")); 
         delay(1);
        }
      else if ((digitalRead(s1DeparturePinA) == HIGH || digitalRead(s1DeparturePinB) == HIGH)&&
               (digitalRead(s1ArrivalPinA) == HIGH || digitalRead(s1ArrivalPinB) == HIGH))
        {
          Serial.println(F("Train on siding 1 status IN TRANSIT"));
          delay(1);
        }
      else if ((digitalRead(s1DeparturePinA) == HIGH || digitalRead(s1DeparturePinB) == HIGH) &&
               (digitalRead(s1ArrivalPinA) == LOW || digitalRead(s1ArrivalPinB) == LOW))
        {
          Serial.println(F("Train on siding 1 status ARRIVAL"));
          delay(1);
          s1ArrivalDetect = LOW;
        }
      else if ((digitalRead(s1DeparturePinA) == LOW || digitalRead(s1DeparturePinB) == LOW) && 
               (digitalRead(s1ArrivalPinA) == LOW || digitalRead(s1ArrivalPinB) == LOW) &&
                s1ArrivalDetect == LOW)
        {
          Serial.println(F("Train on siding 1 status STOP"));
          delay(1);       
        trainOneOut = LOW; //train selection flip_flop L=train on siding 2
        s1ArrivalDetect = HIGH;
        digitalWrite(l1_TurnoutL, LOW);  //unset departure and arrival turnouts for train1
        digitalWrite(s1_Power, LOW);        //unset power on for train1
        digitalWrite(l1_TurnoutR, HIGH);  //set departure and arrival turnouts for train2
        digitalWrite(s2_Power, HIGH);        //set power on for train2
        }
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  • /members/colporteur ok I use a lot of switches and have found the cheapest and easiest is to use a Schmitt Trigger (.64$ @digikey)/6 = 0.166$ for each section with an R/C filter. take a look at this blog of mine.

    imageimageThis is cheaper than the 555 solution, as for six debouncers but then you need a 555 (0.40$@digikey) * 2 = 0.80$) for your project. This gets pricey as they will require 2x resistors & capacitors, and a lot more soldering.

    Normally I put the R/C networks on an IC header if it is a 16 pin header I can put 4 networks on it. and this is on a Wire-Wrap socket. 

    image

  • in reply to phoenixcomm

    when you already have a microprocessor involved in the design I think the cheapest thing is to do it by software, but surely something escapes me, hardware is not my specialty.

  • in reply to javagoza

    /members/javagoza yes you are right, but It depends of what's going on with the software. I personally don't like to waste my CPU doing waits. with my Fuel Indicator I  have two switches both being an SPDT which means I have 5 states. 0 off,  Fill, Test, Set UP, Set Down, so the output of the Schmitt triggers go to a priority encoder inputs. whose output goes to an interrupt.  Here I have an ISR to determine what the code is 000, 001, 010, 011 this takes 3 bits as you don't have to read 0 as it is assumed. then you need to have 4 reads and a little hex math to get switch code.  and away you go.

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  • in reply to javagoza

    /members/javagoza yes you are right, but It depends of what's going on with the software. I personally don't like to waste my CPU doing waits. with my Fuel Indicator I  have two switches both being an SPDT which means I have 5 states. 0 off,  Fill, Test, Set UP, Set Down, so the output of the Schmitt triggers go to a priority encoder inputs. whose output goes to an interrupt.  Here I have an ISR to determine what the code is 000, 001, 010, 011 this takes 3 bits as you don't have to read 0 as it is assumed. then you need to have 4 reads and a little hex math to get switch code.  and away you go.

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