>>         movf    AD_LOW,w
>>         subwf   REF_LOW,w       ; C set if REF_LO >= AD_LO
>>         movf    AD_HIGH,w
>>         btfss   C
>>          incf   AD_HIGH,w
>>         subwf   REF_HIGH,w      ; C set if REF_HIGH >= AD_HIGH...
>>                                 ;    ...and REF_LOW >= AD_LO,
 or...
>>                                 ; ...if REF_HIGH >= (AD_HIGH+1)
>>         bcf     STAT,2
>>         btfss   C
>>          bsf    STAT,2
>>
>> Note that the above routine is 9 cycles isochronous (slightly better than
>> HiTech's) but that it will not work if AD >= 65280.  Given that the
>> problem stipulated that AD < 4096, this should not pose a problem.

>Hello John.

>To overcome H'FFxx' "barrier" i offer slightly modified variant of your
>code:

>        movf    AD_LOW,w
>        subwf   REF_LOW,w

>        movf    AD_HIGH,w
>        btfss   C
>        incfz   AD_HIGH,w       ;this point was modified
>        subwf   REF_HIGH,w

>        bcf     STAT,2
>        btfss   C
>        bsf     STAT,2

>This code has no limitation.

>WBR Dmitry.


Since I'm the one that initiated this "challenge", I'll be the one
that ends it.

Thanks to all that contributed.

John's and Dmitry's solutions are equally good, since the AD-value
will be less than 4096.

But as a general 16-bit compare Dmitry's solution is the best.

It's nice to see that hand-optimized code still beats what a compiler
produces. (But the Hi-Tech code impressed me....)


-Oyvind