help with variable?

Projects General Guidance
3

hi ray thanks for the help

it didnt take 20 secs in the end

will this work instead?

int led1 = 4;
int led2= 5;
int led3 = 6;
int led4 = 7;
int led5 = 8;
int led6 = 9;
int led7 = 10;
int led8 = 11;
int led9 = 12;
int led10 = 13;
void setup() { 
  Serial.begin(9600);
  pinMode(led1, OUTPUT);
  pinMode(led2, OUTPUT);
  pinMode(led3, OUTPUT);
  pinMode(led4, OUTPUT);
  pinMode(led5, OUTPUT);
  pinMode(led6, OUTPUT);
  pinMode(led7, OUTPUT);
  pinMode(led8, OUTPUT);
  pinMode(led9, OUTPUT);
  pinMode(led10, OUTPUT);
  unsigned long secondsCounter = 0;}
void loop() {
  digitalWrite(led1, HIGH);
  delay(1000);
  secondsCounter += 1;
Serial.println(secondsCounter);
  digitalWrite(led2, HIGH);
  delay(1000);
  secondsCounter += 1;
Serial.println(secondsCounter);
  digitalWrite(led3, HIGH);
  delay(1000);
  secondsCounter += 1;
Serial.println(secondsCounter);
  digitalWrite(led4, HIGH);
  delay(1000);
  secondsCounter += 1;
Serial.println(secondsCounter);
  digitalWrite(led5, HIGH);
  delay(1000);
  secondsCounter += 1;
Serial.println(secondsCounter);
  digitalWrite(led6, HIGH);
  delay(1000);
  secondsCounter += 1;
Serial.println(secondsCounter);
  digitalWrite(led7, HIGH);
  delay(1000);
  secondsCounter += 1;
Serial.println(secondsCounter);
  digitalWrite(led8, HIGH);
  delay(1000);
  secondsCounter += 1;
Serial.println(secondsCounter);
  digitalWrite(led9, HIGH);
  delay(1000);
  secondsCounter += 1;
Serial.println(secondsCounter);
  digitalWrite(led10, HIGH);
  delay(1000);
  secondsCounter += 1;
Serial.println(secondsCounter);
  digitalWrite(led1, LOW);
  digitalWrite(led2, LOW);
  digitalWrite(led3, LOW);
  digitalWrite(led4, LOW);
  digitalWrite(led5, LOW);
  delay(1000);
  secondsCounter += 1;
Serial.println(secondsCounter);
  digitalWrite(led6, LOW);
  digitalWrite(led7, LOW);
  digitalWrite(led8, LOW);
  digitalWrite(led9, LOW);
  digitalWrite(led10, LOW);
  delay(1000);
  secondsCounter += 1;
Serial.println(secondsCounter);
  digitalWrite(led1, HIGH;
  digitalWrite(led2, HIGH);
  digitalWrite(led3, HIGH);
  digitalWrite(led4, HIGH);
  digitalWrite(led5, HIGH);
  delay(1000);
  secondsCounter += 1;
Serial.println(secondsCounter);
  digitalWrite(led6, HIGH);
  digitalWrite(led7, HIGH);
  digitalWrite(led8, HIGH);
  digitalWrite(led9, HIGH);
  digitalWrite(led10, HIGH);
  delay(1000);
  secondsCounter += 1;
Serial.println(secondsCounter); 
  digitalWrite(led1, LOW);
  digitalWrite(led2, LOW);
  digitalWrite(led3, LOW);
  digitalWrite(led4, LOW);
  digitalWrite(led5, LOW);
  delay(1000);
  secondsCounter += 1;
Serial.println(secondsCounter);
  digitalWrite(led6, LOW);
  digitalWrite(led7, LOW);
  digitalWrite(led8, LOW);
  digitalWrite(led9, LOW);
  digitalWrite(led10, LOW);
  delay(1000);
  secondsCounter += 1;
Serial.println(secondsCounter);
}

Hackscribble:
Hi there

An answer to the specific question you have asked is ...

Declare a variable and initialise it to 0:

unsigned long secondsCounter = 0;

At the end of your loop, add 20 and print it:

secondsCounter += 20;

Serial.println(secondsCounter);




BUT ...

How do you know your loop will take **exactly** 20 seconds to run? It would be better to use the millis() function to check how much time has passed since the last time round the loop and print that.

For example:

unsigned long lastMillis = 0;
unsigned long nowMillis = 0;

nowMillis = millis();
Serial.println((nowMillis - lastMillis) / 1000.0, 1);  // Forced to float to display decimal part of time, and set to 1 decimal place
lastMillis = nowMillis;




Regards

Ray

thanks,
Jason