help me understand this AVR C code

Projects Programming
1

this is a working code that is splitting the 8 bit data on two different ports and I am not understanding how it is working.
if we AND PORTB to ~0x3F which is 0xC0 , it will make bits DB0-DB5 all zeros so how is the PORTB and PORTC
assignments working ?

/*-------------------------------------------------
DATA:
    DB0 DB1 DB2 DB3 DB4 DB5 DB6 DB7 
    
    PB0 PB1 PB2	PB3 PB4 PB5 PC6 PC7

CONTROL:
    RES  RD   WR  RS
    PC2  PC3  PB4 PB5
    
---------------------------------------------------*/
 #define DMSK				0x3F
 

void write_8(unsigned char c)
{
	PORTB = (PORTB & ~DMSK)|(c & DMSK); 
	PORTC = (PORTC & DMSK)|(c & ~DMSK);
     	.
	.
}

void TFT_setup(void)
{
	DDRB |= DMSK;               
	DDRC |= ~DMSK;			
	DDRC |= 0x3C;              
        .
        .
}
2

This tutorial on the Playground might help you figure that code out

http://playground.arduino.cc/Code/BitMath

Just break each part of the operation down, and use binary values. The playground article will explain the operations. Then it is just a matter of working them out. Tough to get thinking this way, but practice will make it easier.

3 
	PORTB = (PORTB & ~DMSK)|(c & DMSK); 
	PORTC = (PORTC & DMSK)|(c & ~DMSK);

The mask is 0x3F which has the lower 6 bits set. c & DMSK gives a byte with the lower 6 bits from c and the upper 2 bits set to zero. PORTB & ~DMSK gives a byte with the upper 2 bits from PORTB and the lower 6 bits set to zero. Or-ing these two bytes gives a byte with the upper 2 bits from PORTB and the lower 6 bits from c. Similar analysis can be applied to the PORTC assignment.

4

See comments in the code. What AVR are you going to run this on as it won't work on the UNO as is.

/*-------------------------------------------------
  DATA:
  DB0 DB1 DB2 DB3 DB4 DB5 DB6 DB7 
  
  PB0 PB1 PB2	PB3 PB4 PB5 PC6 PC7
  
  CONTROL:
  RES  RD   WR  RS
  PC2  PC3  PB4 PB5
  
---------------------------------------------------*/
#define DMSK  0x3F // = Binary 00111111


void write_8(unsigned char c)
{
	PORTB = (PORTB & ~DMSK)|(c & DMSK); // Read PORTB and mask to keep upper 2 bits, Mask c to keep lower 6 bits, OR both together and write back to PORTB
	PORTC = (PORTC & DMSK)|(c & ~DMSK); // Read PORTC and mask to keep lower 6 bits, mask c to keep upper 2 bits, OR both and write back to PORTC
}

void TFT_setup(void)
{
	DDRB |= DMSK; // Set lower 6 bits of PORTB as output (upper 2 bits remain the same as the value is OR'ed)
	DDRC |= ~DMSK; // Set upper 6 bits of PORTC as output
	DDRC |= 0x3C; // Set bits 2-5 of PORTC as output
}
5

Riva:
See comments in the code. What AVR are you going to run this on as it won't work on the UNO as is.

/*-------------------------------------------------

DATA:
  DB0 DB1 DB2 DB3 DB4 DB5 DB6 DB7
 
  PB0 PB1 PB2 PB3 PB4 PB5 PC6 PC7
 
  CONTROL:
  RES  RD   WR  RS
  PC2  PC3  PB4 PB5
 
---------------------------------------------------*/
#define DMSK  0x3F // = Binary 00111111

void write_8(unsigned char c)
{
PORTB = (PORTB & ~DMSK)|(c & DMSK); // Read PORTB and mask to keep upper 2 bits, Mask c to keep lower 6 bits, OR both together and write back to PORTB
PORTC = (PORTC & DMSK)|(c & ~DMSK); // Read PORTC and mask to keep lower 6 bits, mask c to keep upper 2 bits, OR both and write back to PORTC
}

void TFT_setup(void)
{
DDRB |= DMSK; // Set lower 6 bits of PORTB as output (upper 2 bits remain the same as the value is OR'ed)
DDRC |= ~DMSK; // Set upper 6 bits of PORTC as output
DDRC |= 0x3C; // Set bits 2-5 of PORTC as output
}

oh I see it now ,, PORT B bits 0-5 are made zero but when OR'ed with data gives the pins correct values.
I am running this code on atmega1284 .

regards

6

so if i change the pin mapping to follows :

DATA:
    DB0   DB1     DB2    DB3    DB4    DB5     DB6   DB7 
    PB0    PB1     PB2    PC3     PC4     PC5     PC6    PC7
CONTROL:
    RES   RD     WR      RS
    PC2   PB5    PD3    PD2

would the following assignments be correct ?

void write_8(unsigned char c)
{
 	PORTB = (PORTB & 0xF8)|(c & 07);  // Read PORTB and mask to keep upper 5 bits, Mask c to keep lower 3 bits,  OR both together and write back to PORTB
 	PORTC = (PORTC & 07)|(c & F8);    // Read PORTC and mask to keep lower 3 bits, mask c to keep upper 5 bits,  OR both and write back to PORTC
}

void TFT_setup(void)
{
DDRB |= ((1<<PORTB0)|(1<<PORTB1)|(1<<PORTB2)|(1<<PORTB5)) 
DDRC |= ((1<<PORTC2)|(1<<PORTC3)|(1<<PORTC4)|(1<<PORTC5)|(1<<PORTC6)|(1<<PORTC7))
DDRD |= ((1<<PORTD2)|(1<<PORTD3))
}

please note I am using one DDRC command to set the data and control bits as output and not two DDRC commands as in original post.

7

I obviously have no idea of your hardware layout/limitations but if your designing it yourself then life would be so much easier if you could dedicate all 8 pins on a port to the 8 bit data bus. It should also speed up display refresh quite a bit as less instructions needed to just write a value to a single port instead or Reading, AND'ing, OR'ing & writing to two separate ports.

8

my hardware requirment does not leave me any contigious sets of pins to do that.
can someone tell me if the following two are equivalent ? :

DDRB |= 0x27;
DDRC |= 0xFC;
DDRD |= 0x0C;

DDRB |= ((1<<PORTB0)|(1<<PORTB1)|(1<<PORTB2)|(1<<PORTB5))
DDRC |= ((1<<PORTC2)|(1<<PORTC3)|(1<<PORTC4)|(1<<PORTC5)|(1<<PORTC6)|(1<<PORTC7))
DDRD |= ((1<<PORTD2)|(1<<PORTD3))

9

It is easy to determine equivalence. Just do something like this and look at the Serial monitor

#include <Arduino.h>
#include <avr/io.h>

void setup()
{
  Serial.begin(9600);
  uint8_t tmp = 0;
  tmp |= ((1<<PORTB0)|(1<<PORTB1)|(1<<PORTB2)|(1<<PORTB5));
  Serial.println(tmp, HEX);

  tmp = 0;
  tmp |= ((1<<PORTC2)|(1<<PORTC3)|(1<<PORTC4)|(1<<PORTC5)|(1<<PORTC6)|(1<<PORTC7));
  Serial.println(tmp, HEX);

  tmp = 0;
  tmp |= ((1<<PORTD2)|(1<<PORTD3));
  Serial.println(tmp, HEX);
}

void loop()
{
}

produces:

27
FC
C

However, the original code is more readable. It is much clearer what the line of code intends.