Have you looked at the Software UART library routines in C18. In Mplab,=20
Help > Topics > MPLAB C18 Libraries > Index >Uart, software will get you=20
close. It is necessary to program 3 delays in "C" using the delay=20
functions, but there is a good explanation there, a little calculation,=20
and add the delay functions. Using an 18F1320 @ 8M chrystal, and 9600=20
baud it looks like this: (Caution, I have the transmit delay working=20
good, using this same method, just haven't needed the receive function=20
yet. If someone wants to critique this, that's OK.


DelayTXBitUART
      =20

=09
Delay for:
((((2*Fosc) / (4*baud)) + 1) / 2) - 12 cycles


the numbers look like:
((((2 * 8,000,000)/(4 * 9600)) + 1) / 2 ) - 12 =3D 196.83 say 197

and the code:

  void DelayTXBitUART (void)
{
Delay10TCYx(19);
Delay1TCY();
Delay1TCY();
Delay1TCY();
Delay1TCY();
Delay1TCY();
Delay1TCY();
Delay1TCY();
}

On 11/7/2010 6:53 PM, Dario Greggio wrote:
> Il 13/09/2010 8.42, PICdude ha scritto:
>   =20
>> Not the answer I wanted to hear.  I was expecting some way to switch
>> off optimizations perhaps, and a way (such as a profiler) to figure
>> out how many instructions or clock-cycles a particular code sequence
>> takes.
>>     =20
>
> Using inline (asm/endasm blocks) assembler has the advantage to ALSO
> drop off optimization for that block of code (depending upon the
> compiler, C18 disables for the whole function, other may do somewhat
> differently) so it's a good choice.
>
> Dario
>   =20
--=20
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