3x4 matrix iluminated from left to right.

Projects Programming
1

I have a very simple task of running a 3 row (cathodes) by 4 Column (anodes) from top left to bottom right, only illuminating one LED at a time.

I do apologise about the length of my code and how simple this really is. I will appreciate anyone taking a look at my code for me. many thanks.

/* The aim is to make an LED MATRIX of 3 Rows by 4 Columns be illuminated from top left to bottom right, only illuminating one LED at a time. Like an LED test run.
 */

int rowPin1 = 9;
int rowPin2 = 8;
int rowPin3 = 7;
int timer = 100;      // the time the LED will be on for.
int timer2 = 400;     // the time a row takes to finnish a loop from left to right.
int timer3 = 800;     // the time 2 rows take to finnish a loop from left to right.

void setup() {

//the columns (Anodes) will loop and switch from high to low at t = (timer), from left to right
(int thisPin = 2; thisPin < 6; thisPin++) {
    pinMode(thisPin, OUTPUT);}

{ 
 //set row pins as outputs and change whether they go high or low. The row that must be illuminated will have a low out put while the other 2 rows must be high.
  pinMode(rowPin1, OUTPUT);
  pinMode(rowPin2, OUTPUT);
  pinMode(rowPin3, OUTPUT);
  }
}   

void loop()
{// colomn loop.
  for (int thisPin = 2; thisPin < 6; thisPin++) {
    pinMode(thisPin, OUTPUT); 
    digitalWrite(thisPin, HIGH);
    delay(timer);
    // turn the pin off:
    digitalWrite(thisPin, LOW);


  digitalWrite (rowPin1,LOW);  
  //row1(top row) must be low for the time it takes for the column loop to loop once.
  delay (timer2);
  //turn back high for the time it takes the other two rows to loop once.
  digitalWrite (rowPin1,HIGH);
  delay (timer3);
  

  digitalWrite(rowPin2, HIGH);
 //turn row2 high for the time it takes row1 to loop once.
  delay (timer2);
  digitalWrite(rowPin2, LOW);
  //turn row2 low for the time it takes the columns to loop once.
  delay (timer2);
  digitalWrite(rowPin2, HIGH);
 //turn row2 back high for the time it takes the other two rows to loop once.
  delay (timer3);
  


  digitalWrite(rowPin3, HIGH); 
  delay (timer3);
  digitalWrite(rowPin3, LOW);
  delay (timer2);
  digitalWrite(rowPin3, HIGH);
  delay (timer3);
  
}

my error message is always :

Arduino: 1.6.12 (Windows 10), Board: "Arduino/Genuino Uno"

expected primary-expression before 'int'

(int thisPin = 2; thisPin < 6; thisPin++) {

exit status 1
expected primary-expression before 'int' .

I have trouble knowing how many loops the 'void loop' heading should have under it before I rather run a separate loop.

2

You are missing the closing } for both setup() and loop().
Try fixing that, see it helps.
Use CTRL-T, will help you check that.

3

No 'for' in your setup() for statement

4

much appreciated

5

under the 'void loop' will the Arduino carry out each instruction it sees, between these '{}' two parenthesis, by going down line by line until the very last '}' and then repeat the loop?

If this is so, how can I run two loops simultaneously?