Hi,

Gavin Jackson Wrote:

> Hi there
>
> I would just like some opinions on this. Does this
> solve the read-modify-write problem.
>
> All PIC pins have external pull-ups.  When the program
> starts the instruction CLRF PORTX is issued to write a
> logic '0' to the data latch.  If you want a logic '1' on the
> port you simply make it an input, by writing a '1' in the
> TRISX register corresponding to the bit position.
> If you want a logic '0' you make the port pin an output and as
> the data latch already contains a '0' you know that the pin
> will pull the external pull-up to ground.
>
> You no longer read and modify the PORT, but rather the
> TRIS register for that port.
>
> Any thoughts on this method?
>
> Regards
>
> Gavin

No, it doesn't.  Suppose you want RB0 to be a 0 and RB1 to be a 1.
According to your scheme, you will make RB0 an output and RB1 an input.
This will have the desired effect.  Now, lets say you do a bsf RB0.  This
instruction will (1) Read PORTB, (2) set bit 0 and (3) Write the result
back to PORTB.  When PORTB is read, RB1 will read as a high because of
the external pull-up.  Thus RB0's output latch value will be set to 1 in
step 3.  If you now make RB0 low (set it to an output), it will thus stay
high.

The only way (I know of) to beat this, is to use a ram register (shadow
register) to hold the contents of the output latches of the port.  If you
want to perform a bit modify, perform it on the shadow register and
transfer the register to the port:
bsf   Shadow_PORTB
movf  Shadow_PORTB, W
movwf PORTB

Hope this helps,
Niki