On Tue, 2006-11-14 at 18:07 -0800, Shawn Wilton wrote:


> Does anyone see anything wrong with this code:
>
> //Set page 103 for details regarding the +2
>     tmr0l = (unsigned char)65000;
>     tmr0h = (unsigned char)(65000 >> 8);  //Right shift so the top byte
> becomes the bottom byte

Others may or may not be correct about the default signedness of
chars, but I believe your problem is even more fundamental. When
writing to a 16-bit timer, you must write the high byte first. The
write to the low byte will initiate a full 16-bit to real
hardware. Similarly, to read a 16-bit timer, read the low byte
first. This will initiate an access of the high byte too which you may
read on the subsequent instruction.

Scott
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