Turning LED ON/OFF

Projects Programming
1

If I have 5 LEDs and I use GPIOs 5-9, I can turn them on and off just like a Blink sketch, but with five of them. Or I can use for loop such as:


for(int i=5; i<10; i++){
digitalWrite(i, HIGH);
}

But what if I use pins 4, 5, 9, 10, 13, or 10, 6, 7, 13, 2? My guess is it must be in an array. Not sure how to achieve it.

2 
const byte ledPins[5] = {4, 5, 9, 10, 13};
...
for(int i=0; i<sizeof(ledPins)/sizeof(ledPins[0]); i++){
  digitalWrite(ledPins[i], HIGH);
}
3

Hello who_took_my_nick

either

constexpr byte LedPins[]             {4,5,9,10,13};

or

constexpr byte LedPins[]             {10,6,7,13,2};

You can access the array by using an index from 0 to 4.

Or using the range based loop

for (auto Led:LedPins) digitalWrite(Led,HIGH);
4

What's the difference?

5

port pin addresses ?

6

I'm embarrassed. If I had scrolled to the right I would have seen the list of pin numbers, which is different!

7

I just tried this:

const byte rlyPins[8] = {2, 3, 4, 5, 13, 8, 7, 6};
void setup() {
  for(int i=0; i<sizeof(rlyPins)/sizeof(rlyPins[0]); i++){
    pinMode(rlyPins[i], OUTPUT);
    digitalWrite(rlyPins[i], LOW);
    }
}

void loop() {
  for(int i=0; i<sizeof(rlyPins)/sizeof(rlyPins[0]); i++){
  digitalWrite(rlyPins[i], HIGH);
  delay(100);
  }
  for(int i=0; i<sizeof(rlyPins)/sizeof(rlyPins[0]); i++){
  digitalWrite(rlyPins[i], LOW);
  delay(100);
  }
}

and it worked just fine.

Another one.
Say, I want to limit the first 6 LEDs to blink. I should add -2 in for loop?

for(int i=0; i<sizeof(rlyPins)/sizeof(rlyPins[0])-2; i++){
  digitalWrite(rlyPins[i], LOW);
  delay(100);
  }
8

try it

9

Actually, it is:

for(int i=0; i<(sizeof(rlyPins)/sizeof(rlyPins[0]))-2; i++){
  digitalWrite(rlyPins[i], LOW);
  delay(100);
  }

And it works.

Thank you, guys.

10

Have a nice day and enjoy.