Trying to understand bit ram address

Question

I'm trying to understand this piece of code to address bits:

/*  GPIO bits   */
static bit  GP5 @ (unsigned)&GPIO*8+5;
static bit  GP4 @ (unsigned)&GPIO*8+4;
static bit  GP3 @ (unsigned)&GPIO*8+3;
static bit  GP2 @ (unsigned)&GPIO*8+2;
static bit  GP1 @ (unsigned)&GPIO*8+1;
static bit  GP0 @ (unsigned)&GPIO*8+0;

The GPIO is defined in this way:

static volatile unsigned char   GPIO    @ 0x06;

Why GPIO address is multiplied by 8 and then added by the number of the bit? What the result of this macro and how can I address the bit?

The code above is for XC8 compiler for PIC Microcontrollers. Atmel uses the same when they use the macro IOPORT_CREATE_PIN. This macro is defined as below:

#define IOPORT_CREATE_PIN(port, pin)    ((IOPORT_##port)*8 + (pin))

look at the documentation for the part, it should be pretty obvious. Examine the size of the I/O registers for reading a port and/or pin within a port. That should explain the address calculation. I would not expect this to be consistent among vendors, esp from microchip to atmel, that is simply a coincidence in this one case. The compiler is interesting, but the more important detail we need is the part number. Your question is extremely specific to that part or family. — old_timer, Sep 18 '13 at 13:55
This syntax is not C so much as it is XC8, you may want to adjust the tags accordingly. — Brian Cain, Sep 18 '13 at 13:58
@BrianCain, this sintax can be explained here: http://stackoverflow.com/questions/15955856/sign-in-c-variable-declaration — Daniel Grillo, Sep 18 '13 at 14:01
@DanielGrillo, I understand the syntax, but it's not legal C, so you may find that SO frowns on the use of the 'C' tag for stuff that's not *quite* C. — Brian Cain, Sep 18 '13 at 14:03

score 2 · Accepted Answer · answered Sep 18 '13 at 20:01

"Why GPIO address is multiplied by 8 and then added by the number of the bit? What the result of this macro and how can I address the bit?"

It's the count of bits from the lowest address: 8 bits per byte plus offset into the byte.

You can address the bit by that name, e.g., GP3 = 1;. The compiler knows it is a single bit. As pointed out, this is a particular compiler extension for the PIC.

score 1 · Answer 2 · answered Sep 18 '13 at 14:56

1

Because each special function register has 8 bits (depends on the microcontroller). The address of GP0 is 0x06*8+0 = 0x30. Ways to address a bit also depends on the microcontroller. Sorry, I don't familiar with PIC. You may figure it out on your own.

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answered Sep 18 '13 at 14:56

This is the problem. There's nothing saying about reference a bit neither PIC nor AVR. – Daniel Grillo Sep 18 '13 at 16:46
@DanielGrillo Since GPx(e.g. GP0) has been defined, it can be addressed directly, I think. Like this: `GP0 = 1; // set GP0` – Sep 18 '13 at 17:46
But why it is accessed by the address 0x30 instead of 0x06? – Daniel Grillo Sep 18 '13 at 18:06
@DanielGrillo It is not a byte address. The PIC compiler also employs bit addresses. – UncleO Sep 18 '13 at 20:02

Trying to understand bit ram address

2 Answers2