Hi,
What is the best (simplest) way to write a byte to a 16 bit port pin with
out disrupting the other MSB pins. In my case I'm trying to write a byte to
the LSByte programmed in C on a dsPIC.

Something like:

unsigned char temp;
temp = 19;
PORTB = temp;

Even though the variable 'temp' is only a byte long, I'm pretty sure it will
toggle the MSByte in the 16 bit PORTB register.

Any ideas?

Thomas Magma

On Thu, 14 Feb 2008 19:04:59 GMT, "Thomas Magma"
<somewhere@overtherainbow.com> wrote:

If you can't do a read-modify-write on the port, something like:
temp = 19;
foo = ((PORTB & 0xFF00) | temp);
PORTB = foo;

then try keeping a shadow register, i.e., *all* changes to PORTB are
first done to ShadowB and then ShadowB is written to the port.

--
Rich Webb Norfolk, VA

Thomas Magma wrote:
I am sure they do it right.

Take a look at the dsPIC30F/33F Programmer’s Reference Manual

http://ww1.microchip.com/downloads/e...Doc/70157B.pdf

Section 4.4, all will be reveled.

donald

Thanks Rich,

What is the 'foo' for? Couldn't I just:
PORTB = ((PORTB & 0xFF00) | temp);

to simplify things?

Thomas

On Thu, 14 Feb 2008 19:31:40 GMT, "Thomas Magma"
<somewhere@overtherainbow.com> wrote:

Probably, although [disclaimer: never worked on dsPICs or with that
specific compiler] it's likely that PORTB is not a simple object and
there could be side effects that I'm not aware of that come into play
when PORTB is referenced. Using a throw-away temp should guarantee
that bad things won't happen.

--
Rich Webb Norfolk, VA

Thanks Donald,

I hadn't seen that programming reference manual before, but it only talks
about byte addressing in assembly language and not in C. I'm kind of looking
for a clean way to do it in C.

Thomas

On Thu, 14 Feb 2008 19:04:59 GMT, "Thomas Magma"
<somewhere@overtherainbow.com> wrote:

Not an answer to your question-- I think that is partly in donald's
suggestion and partly in the C30 compiler User's Guide if you want to
use the byte mode capability-- or use the and/or-- but 99.9% of the
time you should NOT be writing to PORTB in the first place-- rather
use LATB (or B sorry).
Best regards,
Spehro Pefhany
--
"it's the network..." "The Journey is the reward"
speff@interlog.com Info for manufacturers: http://www.trexon.com
Embedded software/hardware/analog Info for designers: http://www.speff.com

I did mean LATB. I've already been through that painful LAT vs PORT learning
curve, but thanks anyways.

On Feb 15, 12:18*am, Rich Webb <bbew...@mapson.nozirev.ten> wrote:
This is the famous trick to be adapted in this scenario

Karthik Balaguru

"Thomas Magma" <somewhere@overtherainbow.com> wrote:

Assuming you are using the Microchip C30 compiler include generic.h and use
an ugly cast like.

((WORD_VAL*)&LATB)->byte.LB = 0x23;

((WORD_VAL*)&LATB)->byte.HB = 0x45;

or define your own structure to map over LATB like LATBbits is done in the
processor header file and add a line to the linker script file to define
its address.

--

donald wrote:
Depends on how PORTB is declared in C.
If it's declared as a 16 bit memory location, the C compiler will
certainly promote the byte value to a 16 bit int and thus write zeros to
the MSB.
If the MSB and LSB of the port can be addressed as two (8 bit) char
locations, you can get away with a simple assignment.

Assumin 16 bit only, I'd go for
PORT = (PORT & 0xff00) | new_lsb_value;
or the "shadow" method if the port can't be read.
The port should probably be declared volatile too.

Anahata