jan-erik,

 I still hold the belief that 0x08 vs BSR is not wrong.  Since BSR is at
address 0x08, how can it be?
 When the program is assembled, the MOVWF instruction is encoded as
0x008, and has appended to it the=20
 register on which to act, which is the BSR, which corresponds to 0x08.=20
So, in the final machine form,
 it translates to 0x0088 which is "Move w to register 8 which is the
BSR.  Therefore, whether you use=20
 BSR or 0x08, you wind up with the same machine instruction.  If using
0x08 was wrong, the final machine
 code would be different.  It's just a matter of semantics.  One form
isn't more right or wrong than the
 other.  It's just the BSR is the more preferred way.

 Regards,

 Jim

> -------- Original Message --------
> Subject: Re: [PIC]: PIC16F1827 PORTB Problem.
> From: Jan-Erik Soderholm <jan-erik.soderholm@telia.com>
> Date: Wed, June 22, 2011 8:54 am
> To: "Microcontroller discussion list - Public." <piclist@mit.edu>
>=20
>=20
> RussellMc wrote 2011-06-22 15:39:
>=20
> > It may be that JE holds a "different is wrong" mindset for whatever
> > reason, but I hope not.
>=20
> No, not at all, of course.
>=20
> I simply ment that writing "movwf 0x08" is wrong
> when you can write "movwf BSR" instead.
>=20
> *That* is my definition of "wrong". :-)
>=20
> I've not read the rest of your post.
> Sorry for that.
>=20
> Regards,
> Jan-Erik.
> --=20
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