Diode connection.

Maksrus · February 15, 2018, 12:10pm

Help me please!

I do not understand how to connect the LED so that it blinks synchronously with the displayed digit in the com port.
If it is printed 1 it should blink once, 9 times nine times.

int dig1 ;
int dig2;
int dig3;
int dig4;


void setup() 
{
  Serial.begin(9600); // 
   Serial.print( " GO " );
}

void loop() 
{
  int dig1, dig2, dig3, dig4; // 
  int keys[10] = { 0,1, 2, 3, 4, 5, 
  6, 7, 8, 9  }; // 

// 

  for(dig1=1; dig1<10; dig1++-
  {
    for(dig2=1; dig2<10; dig2++-
    {
      for(dig3=1; dig3<10; dig3++-
      {
        for(dig4=1; dig4<10; dig4++=
        {
          Serial.print(keys[dig1]);
          delay(400);
          Serial.print(keys[dig2]);
          delay(400);
          Serial.print(keys[dig3]);
          delay(400);
          Serial.print(keys[dig4]);
          delay(400);
          Serial.println();
          delay(1000);
        }
      }
    }
  }
}

knut_ny · February 15, 2018, 1:00pm

Your syntax is far off.
It is unclear to me what you want your code to do.

Maksrus · February 15, 2018, 1:18pm

I want the diode to flash the corresponding digit

digit 1 flashes once

digit 2 flashes twice
and so on 1-9

Maksrus · February 15, 2018, 1:31pm

above is the code it prints the numbers in order
1111 diode flashes four times with a pause of 1 minute
1112 the diode flashes three times with a pause of one minute and twice with a pause of five seconds
and so on
this is a kind of sorting through the digits with a diode

slipstick · February 15, 2018, 1:52pm

Your code does nothing with any LEDs. You have four "dig" variables (defined twice) but nothing ever gives them any values. Your for loops use invalid syntax. Have another try.

And the specification is very unclear. Can the digits be 0 to 9? When do you have no gap between flashes, when 5 second gap and when 1 minute? E.g. what does 3192 do? How do you tell the difference between 4000 and 1111?

Steve

IanCrowe · February 15, 2018, 1:54pm

Try the following:-

In your setup() function of your code add the following lines:-

pinmode(LED_BUILTIN, OUTPUT);
digitalWrite(LED_BUILTIN, LOW);

This will use the built-in led (connected to pin 13 on Uno and Nano boards and to other pins on other variants).

Now create a simple function called something like ledFlash:-

void ledFlash(int flashes) {
   for(int i = 1; i == flashes; i++) {
      digitalWrite(LED_BUILTIN, HIGH);
      delay(100);
      digitalWrite(LED_BUILTIN, LOW);
      delay(100);
   }

Then every time you use Serial.print to print a digit call ledFlash with the same digit e.g. :-

Serial.print(keys[dig1]);
ledFlash(keys[dig1]);

BTW your syntax for your for statements is wrong. I leave it as an exercise for the reader to interpret the compiler error messages.

Ian

system · February 15, 2018, 6:40pm

How do you blink a "0"? Here's an old program called "poor man's GUI" that might give you some ideas, type a number (1 to 999) in the top of serial monitor and press [ENTER].

// Poor man's GUI


byte led = 13;

void pmGUI(byte led, int num)
{
  int interval = 350, i;

  if(num == 0){
    num = 4; 
    interval = 50;
  }
  for(i = 0;i < num;i++){
    digitalWrite(led,HIGH); 
    delay(50); 
    digitalWrite(led,LOW);
    delay(interval);
  }
  delay(600);
}

void setup() {
  Serial.begin(9600);
  pinMode(led,OUTPUT);
  Serial.println("  Type a number from 0 to 999, then <ENTER> ");
  Serial.println("      A \"stutter\" = zero \n");
}
void loop() {
    // if there's any serial available, read it:
  while (Serial.available() > 0) {

    // look for the next valid integer in the incoming serial stream:
    int num = Serial.parseInt(); 
    // look for the newline. That's the end of your
    // sentence:
    if (Serial.read() == '\n') {
    }
    if(num > 99) pmGUI(led,num / 100);
    if(num > 9) pmGUI(led,num % 100  / 10);
    pmGUI(led,num % 10);
  }
}

IanCrowe · February 15, 2018, 6:53pm

outsider:
How do you blink a "0"?

If you look carefully the code never actually prints a zero (possibly a bug?). All the for loops start at i=1 and terminate when i == 10. Hence dig1, dig2, dig3 and dig4 all range between 1 and 9 and therefore only index keys[1] to keys[9].

I won't even mention the fact that the keys array is completely redundant

Ian

Maksrus · February 16, 2018, 8:48am

int dig1 ;
int dig2;
int dig3;
int dig4;
int ledPin = 13;


void setup() 
{
  Serial.begin(9600); 
   Serial.print( " GO " );
  pinMode(13, OUTPUT);
digitalWrite(13, LOW);
}

void loop() 
{
  int dig1, dig2, dig3, dig4; 
  int keys[10] = { 0,1, 2, 3, 4, 5, 
  6, 7, 8, 9  }; 

 
  for(dig1=2; dig1<10; dig1++)
  {
    for(dig2=2; dig2<10; dig2++)
    {
      for(dig3=2; dig3<10; dig3++)
      {
        for(dig4=2; dig4<10; dig4++)
        {
        
          
          Serial.print(keys[dig1]);
          ledFlash(keys[dig1]);
          delay(800);
          Serial.print(keys[dig2]);
          ledFlash(keys[dig2]);
          delay(800);
          Serial.print(keys[dig3]);
          ledFlash(keys[dig3]);
          delay(800);
          Serial.print(keys[dig4]);
          delay(800);
          ledFlash(keys[dig4]);
          Serial.println();
          delay(5000);
         
        
        
       
          
          
        }
      }
    }
  }
}

void ledFlash(int flashes) {
 for(int i = 1; i ==flashes ; i++) 
  {
    
        
      digitalWrite(13, HIGH);
      delay(10);
      digitalWrite(13, LOW);
      delay(10);
     
     
    
   }
 
 }

unfortunately the code does not work
1 blinks times
on other numbers does not react
I'm not a programmer on this, I do not understand much

IanCrowe · February 16, 2018, 11:08am

Maksrus:
int dig1 ;

int dig2;
int dig3;
int dig4;
int ledPin = 13;

void setup()
{
Serial.begin(9600);
Serial.print( " GO " );
pinMode(13, OUTPUT);
digitalWrite(13, LOW);
}

void loop()
{
int dig1, dig2, dig3, dig4;
int keys[10] = { 0,1, 2, 3, 4, 5,
6, 7, 8, 9 };

for(dig1=2; dig1<10; dig1++)
{
for(dig2=2; dig2<10; dig2++)
{
for(dig3=2; dig3<10; dig3++)
{
for(dig4=2; dig4<10; dig4++)
{

Serial.print(keys[dig1]);
ledFlash(keys[dig1]);
delay(800);
Serial.print(keys[dig2]);
ledFlash(keys[dig2]);
delay(800);
Serial.print(keys[dig3]);
ledFlash(keys[dig3]);
delay(800);
Serial.print(keys[dig4]);
delay(800);
ledFlash(keys[dig4]);
Serial.println();
delay(5000);

}
}
}
}
}

void ledFlash(int flashes) {
for(int i = 1; i ==flashes ; i++)
{

digitalWrite(13, HIGH);
delay(10);
digitalWrite(13, LOW);
delay(10);

}

}




unfortunately the code does not work
1 blinks times
on other numbers does not react
I'm not a programmer on this, I do not understand much
:cold_sweat:

Sorry, my bad, there was a typo in my code the line

for(int i = 1; i ==flashes ; i++)

should read

for(int i = 1; i <=flashes ; i++)

By the way you have now limited your for loops to the range 2 to 9. Was this deliberate?

Note setting the delays in ledFlash to 10ms means you may not be able to see the individual flashes as they are coming at 50Hz which is usually beyond what persistence of vision will allow the human eye to discriminate.

Ian

Maksrus · February 16, 2018, 11:21am

No, this is not a typo, I checked that it really does not blink twice for 2.

Maksrus · February 16, 2018, 11:58am

outsider:
How do you blink a "0"? Here's an old program called "poor man's GUI" that might give you some ideas, type a number (1 to 999) in the top of serial monitor and press [ENTER].
// Poor man's GUI

byte led = 13;

void pmGUI(byte led, int num)
{
int interval = 350, i;

if(num == 0){
num = 4;
interval = 50;
}
for(i = 0;i < num;i++){
digitalWrite(led,HIGH);
delay(50);
digitalWrite(led,LOW);
delay(interval);
}
delay(600);
}

void setup() {
Serial.begin(9600);
pinMode(led,OUTPUT);
Serial.println(" Type a number from 0 to 999, then ");
Serial.println(" A "stutter" = zero \n");
}
void loop() {
// if there's any serial available, read it:
while (Serial.available() > 0) {

// look for the next valid integer in the incoming serial stream:
int num = Serial.parseInt();
// look for the newline. That's the end of your
// sentence:
if (Serial.read() == '\n') {
}
if(num > 99) pmGUI(led,num / 100);
if(num > 9) pmGUI(led,num % 100 / 10);
pmGUI(led,num % 10);
}
}

this code works correctly, but the numbers are entered from the keyboard I need automatically, without my participation.

IanCrowe · February 16, 2018, 12:33pm

Maksrus:
No, this is not a typo, I checked that it really does not blink twice for 2.

Yes it was a typo. For loops continue until the second parameter is false. If you use == instead of <= then the condition is never true and the for loop never executes.

Change the condition to less than or equal to ( <= ) and increase the delays to 100ms and the sketch will function in the way you want. If you keep the delays at 10ms the flashes will merge into one flash of varying length. This is due to the way the human eye works.

Ian

Maksrus · February 16, 2018, 1:11pm

IanCrowe:
Yes it was a typo. For loops continue until the second parameter is false. If you use == instead of <= then the condition is never true and the for loop never executes.

Change the condition to less than or equal to ( <= ) and increase the delays to 100ms and the sketch will function in the way you want. If you keep the delays at 10ms the flashes will merge into one flash of varying length. This is due to the way the human eye works.

Ian

But it does work.
Thank you so much.

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