> Processor is a 18F23K20
> 
> I've never had the requirement to do much math with a PIC, other than simple divide by 2 or 4 (shifts), add and subtract
> 
> So looking for a little help in implementing this 
> 
> temp_F = (temp_C * 9 / 5) + 32;
> 
> The temp_C * 9 can be done with the MULLW instruction.  Just not to sure about the divide by 5 part?

If you don't need decimals:

temp_F = temp_C + 32      Tf=Tc*1.8+32. This line does Tf=Tc*1+32
temp_F = temp_F + HIBYTE(temp_C * 205); Add Tc*0.8 (205/256 is ~0.8)

This works only for temp_C=<0:255> for negative temp_c you could shift 
temp_C to positive with adding 100 and after the conversion subtracting 
  180 from the result. This will give a temp_C range <-100:155>

Djula
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