4

I made some arduino code for leds that blink down a row and then back again, but when the program is run, there is an approximately 2 second delay in between the first for loop and the second. This is my code in the loop function:

int leds[] = {13, 12, 11};
int i;

void setup() {
  Serial.begin(9600);

  for (int i = 0; i < sizeof(leds); i ++) {
     pinMode(leds[i], OUTPUT);
  }
}

void loop() {
    for (i = 0; i < sizeof(leds); i ++) {
       digitalWrite(leds[i], HIGH);
       delay(250);
       digitalWrite(leds[i], LOW);
    }
    for (i; i > 0; i --) {
       digitalWrite(leds[i], HIGH);
       delay(250);
       digitalWrite(leds[i], LOW);
    }
}

It really makes the program look weird when the LEDs are blinking on a breadboard. Keep in mind that this only happens after the first for loop, the second is continuous back to the first. Any idea why this might be and how to fix it?

2
  • Please include the FULL code. Yes, including the setup even though it is not really relevant for you.
    – Dat Ha
    Commented Nov 10, 2016 at 2:01
  • Alright, I edited it. Does that help?
    – Minecat40
    Commented Nov 10, 2016 at 2:04

3 Answers 3

-1

The cause it that your 3rd led (2nd element in the array) is repeated twice.

Try this instead:

int leds[3] = {13, 12, 11};

void setup() {   
  for (int i = 0; i < 3; i++) {
    pinMode(leds[i], OUTPUT);
  }
}

void loop() {
  for (int i = 0; i < 2; i++) {
    digitalWrite(leds[i], HIGH);
    delay(250);
    digitalWrite(leds[i], LOW);
  }
  for (int i = 2; i > 0; i--) {
    digitalWrite(leds[i], HIGH);
    delay(250);
    digitalWrite(leds[i], LOW);
  }
}
1
  • Thank you! I thought that I wouldn't have to decrement the length of the array, guess I was wrong.
    – Minecat40
    Commented Nov 10, 2016 at 2:20
5

The claim “The cause it that your 3rd led (2nd element in the array) is reapeted twice” is wrong.

Apparently that sentence is supposed to mean that the long delay is due to i being one too large, coming out of the first loop and entering the second one. Indeed, i is too large, but that's because each element of the array is a 2-byte int, so that instead of for (int i = 0; i < sizeof(leds); i ++) being equivalent to for (int i = 0; i < 3; i ++), it is equivalent to for (int i = 0; i < 6; i ++).

Thus, one second goes by with an extra 4 occurrences of delay(250); in the first loop, and another one second with 4 extra occurrences of delay(250); in the other loop.

To correct the problem, write sizeof leds/sizeof leds[0] instead of sizeof(leds), and write for (--i; i > 0; i--) instead of for (i; i > 0; i --).

0
1

The problem is on "for loops" condition,
Try this:

int leds[] = {13, 12, 11};
int i;

void setup() {
  Serial.begin(9600);

  for (int i = 0; i < (sizeof(leds)/sizeof(leds[0])); i ++) {
     pinMode(leds[i], OUTPUT);
  }
}

void loop() {
    for (i = 0; i < (sizeof(leds)/sizeof(leds[0]); i ++) {
       digitalWrite(leds[i], HIGH);
       delay(250);
       digitalWrite(leds[i], LOW);
    }
    for (i=((sizeof(leds)/sizeof(leds[0])-1); i>=0; i --) {
       digitalWrite(leds[i], HIGH);
       delay(250);
       digitalWrite(leds[i], LOW);
    }
}
9
  • This is not true. It depends on the persons intention.
    – Dat Ha
    Commented Nov 10, 2016 at 3:10
  • Well my friend, the "intention" is to blink a LED. (1) You turn on a LED, (2) put it on hold by using delay, (3) You turn the LED off, (4) loops, the LED is turned on again. Well, if by "blinking" you mean its turn off for less than 10 clocks instruction, I accept.
    – duck
    Commented Nov 10, 2016 at 3:18
  • 1
    You turn it off then change leds so there is a delay anyways.
    – Dat Ha
    Commented Nov 10, 2016 at 3:21
  • ah you are right. Sorry my bad.
    – duck
    Commented Nov 10, 2016 at 3:22
  • 2
    You mean sizeof(leds)/sizeof(leds[0]). Commented Nov 10, 2016 at 8:58

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