Don Holtz wrote:

> I understand that:
>
>  movlw 1
>  iorwf STATUS,F
>
> will not set C.

Yes. Due to IORWF instruction change the _Z bit all _Z & _DC and _C
group
of bits are protected from direct changing and it changed only according
results of operation. In case of IORWF _DC & _C are simply locked while
_Z will set according result.

But if you want to change _C in this manner you may add one more
operation
and write your example as following: (myke predko was offer this
variant)

movlw 1
iorwf STATUS,W
movwf STATUS    ;this operation do not change any flags
                ;due to that we have possibility write directly to
STATUS

> Does:
>
> bsf STATUS,C
>
> set the C?

Yes it does. This operation directly change only _C bit and not fluent
to others.

> (bsf/bcf are READ-MODIFY-WRITE instructions... so I hope the data would
> be treated in exactly the same way as a READ instruction follow byte a
> MODIFY-WRITE instruction?)

You may easyly apply bcf/bsf for playing with _Z or _DC or _C bits in
status
register.

> In any case:
>
>  addlw 0 ; will always clear C
Yes.

>  sublw 0 ; will always set C
No. PIC substracting logic are based literally on the adding core. So if
no
borrow after operation then _C=1 (result is positive) Other case when
_C=0
(result is negative and borrow take a place). The only situation when
sublw_0
will result to _C=1 when W=0 before operation, so 0-0=0 and result is
positive
and no borrow was taken.

WBR Dmitry.