Motorola 68Hc08 programming help

If you are programming in C, you don't have to worry about what is in the accumulator or the registers. That is one of the benefits of using a higher level language: you don't have to be aware of the CPU architecture, although it often makes your C code more efficient if you are, especially with simple, 8-bit CPUs. What is it you are trying to do in C that requires you to manipulate the CPU registers?

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How you declare the I/O registers depends on the C compiler you are using. For the open-source SDCC compiler, you do it like this:

volatile unsigned char __at 0x00 PORTA; volatile unsigned char __at 0x04 DDRA;

For the Metrowerks CodeWarrior compiler, it's like this:

volatile unsigned char PORTA _IO_AT(0x00); volatile unsigned char DDRA _IO_AT(0x04);

SDCC is free. There is a demo version of CodeWarrior available for free that will do up to 4K of object code and a maximum of 16 sources files, IIRC. If you're using another C compiler, RTFM. Both SDCC and CodeWarrior have header files that define all the registers, so all you need to do is something like:

#include "mc68hc908gp32.h"

to define all the I/O registers for the GP32. Once you have declared your I/O registers, it's as easy as:

unsigned char somevar; /* declare an unsigned char */ DDRA = 0x00; /* make Port A an input port */ somevar = PORTA; /* read the value on Port A */ DDRA = 0xFF; /* make Port A an output port */ PORTA = 0xAA; /* output alternating 1s & 0s to Port A */

Hope this helps.

NM

Freescale got it backwards: 1 = Output and 0 = Input. So putting 0xAA in DDRA makes the even bits inputs and the odd bits outputs.

How many bits a variable (or register) contains depends on how it is declared.

unsigned char var1; /* var1 is 8 bits */ unsigned short int var2; /* var2 is probably 16 bits, RTFM */ unsigned int var3; /* var3 might be 16 bits, RTFM */ unsigned long int var4; /* var4 is probably 32 bits, RTFM */

With standard C, shorts are 16 bits, ints are 32 bits, and longs are 64 bits, but C compilers for smaller 8-bit and 16-bit CPUs tend to deviate from standard C making shorts and ints 16 bits and longs 32 bits, due to smaller CPUs having to work harder with larger bit sizes. As always, Read The Fine Manual for your C compiler.

With constants, as long as the value fits into the variable, things will be fine:

var4 = 0xAA; /* sets var4 to 0x000000AA */ var2 = 0x55; /* sets var2 to 0x0055 */ var1 = 0xAAAA; /* generates a compile-time error, as you can't fit a 16-bit value into a 8-bit variable. */

Or you could make the constant the size of the variable as a reminder as to the size of the variable:

var4 = 0x000000AA; var2 = 0x0055;

NM

PS: You should look at the header files for the C compiler you are using. They are probably in a directory like \lib\include. These will show you how to declare registers and even the individual bits in them. If you don't already have a compiler, download SDCC from

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