Can't get actual output when, both input logic is high

Question

I want When,

• light=0, ir=0 then output will pin 8 high & pin 12 low.

• light=0, ir=1 then output will pin 8 high & pin 12 high.

• light=1, ir=0 then output will pin 8 low & pin 12 low.

• light=1, ir=1 then output will pin 8 low & pin 12 low (This statement does not work).

Please find the code & suggest me, if any modification is needed.

void setup() {

    pinMode(8,OUTPUT); //logic output.
    pinMode(12,OUTPUT); //logic output.
    pinMode(2, INPUT); //value input.
    pinMode(4, INPUT); //value input.

    Serial.begin(9600);   
}


void loop() {

    int light = digitalRead(2); //value input.
    int ir = digitalRead(4); //value input.

    if (light == LOW && ir == LOW)
    {

      digitalWrite(8, HIGH);
      digitalWrite(12, LOW);
      delay(2000);

    }

    if (light == LOW && ir == HIGH)
    {

      digitalWrite(8, HIGH);
      digitalWrite(12, HIGH);
      delay(2000);

    }

    if (light == HIGH && ir == LOW)
    {

      digitalWrite(8, LOW);
      digitalWrite(12, LOW);
      delay(2000);

    }

    if (light == HIGH && ir == HIGH) //This condition does not work.
    {

      digitalWrite(8, LOW);
      digitalWrite(12, LOW);
      delay(2000);

    }

    digitalWrite(8, LOW);
    digitalWrite(12, LOW);   
    delay(2000);

}

Answer 1

You can simplify your requirements somewhat by splitting your inputs up.

light=0, ir=0 then output will pin 8 high & pin 12 low.

light=0, ir=1 then output will pin 8 high & pin 12 high.

If light is 0 then pin 8 is high. AND ir is echoed to pin 12.

light=1, ir=0 then output will pin 8 low & pin 12 low.

light=1, ir=1 then output will pin 8 low & pin 12 low

If light is 1 then both pins 8 and 12 are low, and you don't care what ir does, since it doesn't have any effect on the outputs.

So your structure could look more like:

if (light == 0) {
    digitalWrite(8, HIGH); // 8 will always be high with light == 0
    digitalWrite(12, ir); // echo the ir value to pin 12
} else {
    digitalWrite(8, LOW); // if light is 1 then everything
    digitalWrite(12, LOW);// is low regardless of ir.
}

Answer 2

Note this is not an answer to the question (I don't see any problem initially), but to clean the code it might help to debug it (using print statements as Sim Son suggests).

Instead of all the if statements, check when pin 8 and pin 12 needs to be HIGH:

 pin 8:  (light == LOW && ir == LOW) 
 pin 12: (light == LOW && ir == LOW)  or (light == LOW && ir == HIGH)

For pin 12, the condition can be rewritten as

 pin 12: (light == LOW) || (ir == LOW or ir == HIGH)

which is equal to

 pin 12: light == LOW

So now you can change the code from the loop function to:

void loop()
{
   digitalWrite(8 , (light == LOW && ir == LOW) ? HIGH : LOW);
   digitalWrite(12, (light == LOW             ) ? HIGH : LOW);
   delay(2000);

   digitalWrite(8, LOW);
   digitalWrite(12, LOW);   
   delay(2000);

}

the x ? : z statements means: if x is true, do y, else z.

The last three statements are fixed (copied from your code).

Now we can even further reduce it, by removing the duplicated code

void loop()
{
   processLeds((light == LOW && ir == LOW), (light == LOW));
   processLeds(False                      , False         );
}

void processLeds(int conditionLed8, int conditionLed12)
{
    digitalWrite( 8, conditionLed8  ? HIGH : LOW);
    digitalWrite(12, conditionLed12 ? HIGH : LOW);
    delay(2000);
}

You pass the condition when led 8 and 12 have to be high or log to a function, which sets the LEDs.

Now you can also use much less print statements for finding your problem.