if statements, masking, outputs, problems!

Forum 2005-2010 (read only) Hardware Troubleshooting
1

hey guys,

I am new to this site (and relatively new to arduino, I have only used a basic stamp before :o )

anyway, I currently have a ardunio duemilanove 328 hooked up to a tri-colour LED 8x8 matrix via 3 CMOS 4514 (one per LED colour) and one CMOS 4515 (for common ground) (most of this is irrelevant, just a bit of background, but stay with me!)

I have written some code which should hopefully light each (blue) LED one after the other, stating from the first led and moving up to the 64th.
I have created a FOR loop which should count from 0 to 63, inside which each bit of the variable n is masked off and then a pin output is writen to the latch (BCD value between 0 - 63, after which the FOR loop should reset).

but nothing happens when I upload it, void blue, void red and void latch are all tested as working.

pins are connected as follows

3 latch

5 red led enable

6 blue led enable

(the latch latches the output from the data pins, into a BCD decoder, one decoder is used per LED colour, when the enable goes LOW, the decoder is allowed to output onto the LED matrix).

pins:

8 data 0
9 -
10 -
11 -
12 -
13 data 5

int b0 = B0;
int b1 = B0;
int b2 = B0;
int b3 = B0;
int b4 = B0;
int b5 = B0;

void setup(){
  pinMode(3,OUTPUT);
  pinMode(5,OUTPUT);
  pinMode(6,OUTPUT);
  pinMode(8,OUTPUT);
  pinMode(9, INPUT);
  pinMode(10,OUTPUT);
  pinMode(11,OUTPUT);
  pinMode(12,OUTPUT);
  pinMode(13,OUTPUT); 
  
  digitalWrite(3,LOW);
  digitalWrite(5, HIGH);
  digitalWrite(6,HIGH);
  

}

void latch (){
  digitalWrite(3, HIGH);
  delay(2);
  digitalWrite(3, LOW);
}

//red is on pin 5

void red () {  
  digitalWrite(5,LOW);
  delay(1);
  digitalWrite(5, HIGH);
}
  
//blue is on pin 6

void blue (){
  digitalWrite(6,LOW);
  delay(2);
  digitalWrite(6, HIGH);
}

//all code up to this point works on other sektches

void loop(){
  
  for(int n = B0; n == B111111; n + B1){
    //mask each bit of n to create 6 bit output
    b0 = n & B000001;
    
    if(b0 == B1) {
       digitalWrite(8, HIGH);
    }
    else{ 
      digitalWrite(8, LOW);
    }
 
     b1 = n & B000010;
    
    if(b1 == B1) {
       digitalWrite(9, HIGH);
    }
    else{ 
      digitalWrite(9, LOW);
    }

    b2 = n & B000100;
    
    if(b2 == B1) {
       digitalWrite(10, HIGH);
    }
    else{ 
      digitalWrite(10, LOW);
    }
    
        b3 = n & B001000;
    
    if(b3 == B1) {
       digitalWrite(11, HIGH);
    }
    else{ 
      digitalWrite(11, LOW);
    }  

    b4 = n & B010000;
    
    if(b4 == B1) {
       digitalWrite(12, HIGH);
    }
    else{ 
      digitalWrite(12, LOW);
    }
    
     b5 == n & B100000;
    
    if(b5 == B1) {
       digitalWrite(13, HIGH);
    }
    else{ 
      digitalWrite(13, LOW);
    }
    
    latch();
    blue();
  }

}

any ideas why this fails to work?

2

What's this supposed to do ?

for(int n = B0; n == B111111; n + B1){
3

create variable n and assigns it a value (binary) 0, tells it to terminate the loop when n is equal to 63, and then add (binary) 1 to the value of n each time the loop loops.

4

create variable n and assigns it a value (binary) 0, tells it to terminate the loop when n is equal to 63, and then add (binary) 1 to the value of n each time the loop loops.

That code doesn't do what you want to. Let's dissect a for loop really quickly.

for( initialization ; condition ; upkeep ) {}

Your initialization is OK.

Your condition is backwards though. The loop will continue to execute as long as condition is true. So, in your case, it'll do the initialization, evaluate the condition, see that condition is false, and never execute the body of the loop. Invert that by using != and it will run until n is equal to your target value.

Your upkeep is not going to do what you want. Yes, the expression is going to be evaluated but consider that there is a world of difference between the following two statements:

n+B1   // Wrong. Adds B1 to n but does nothing with the result
n+=B1  // Right. Adds B1 to n and then saves it in n