// declareer de PINOUT van de lampen buitenom de scope van Void Setup en de Void Loop

// opdracht 9 en 13 alleen dan voor 3 ledjes
int LampPin[] = {0, 1, 2,3};
int aanDelay = 25;
int uitDelay = 1000;
int lampDelay = 1000;

byte tempNum = 0;

void setup() {
  // put your setup code here, to run once:
Serial.begin(9600);


// Setup Pins 

// Opdracht 1 alleen dan voor 3 lampjes
for (int idx = 0; idx <= 3; idx++){
pinMode(LampPin[idx],OUTPUT); // Lampjes initialiseren.
digitalWrite(LampPin[idx], LOW);
}
}

void loop() {
  // put your main code here, to run repeatedly:
  // Opdracht 2 alleen dan voor 3 lampjes
  // Student nummer in code van lampjes
  char studentNr[] = "1072654";
  //studentNr.concat(1072654); // studentNr.length()
for (int idx = 0; idx <= 7; idx++){
tempNum = studentNr[idx];
Serial.println(studentNr);
Serial.println(tempNum);

INT2BIN(7,4);

delay(uitDelay);
for (int idx = 0; idx <= 3; idx++){
digitalWrite(LampPin[idx], LOW);
}
}




}
int INT2BIN(int nummer,int led){
  char binary[4] = {0};
  byte someValue = nummer; //For this example, lets convert the number 20  
  someValue += 16; //(led^2); //Voeg 16 (4^2 toe) zodat er altijd 4 getallen staan in de string
  itoa(someValue,binary,2); //Conver someValue to a string using base 2 and save it in the array named binary
  char* string = binary + 1; //get rid of the most significant digit as you only want 4 bits bij nul heb je anders nog 5 bits
  // = binary + 1; //get rid of the most significant digit as you only want 4 bits bij nul heb je anders nog 5 bits
  
   Serial.println(string);
   Serial.println(binary);
   
  for (byte idx = 0; idx <= (led-1); idx++){
  //if (string[idx] = 1){ 
  Serial.println(string);  
  digitalWrite(LampPin[idx], string[idx] -'0');
  //}
  //else{
  //digitalWrite(LampPin[idx], LOW);
  //}
  delay(lampDelay);
  
}
return;
}