It will affect the output bits, but not the inputs.  All the bits are read in
the first instance, then the high or low value of the bits you are keeping are
re-written to portB, along with the low or high value of the bits your are
writing to portb.

This can cause a problem in this way:
Say you've set bit 7 as output, and are driving it high.
If an external load is driving the pin low (and doing a better job than the PIC)
then when you read in PORTB it will report that that pin is low, even though you
previously told it to output high.  When you write the modified value back to
the port it will set that pin low.

See the instructions BCF and BSF from the data sheet.  In a nutshell, ports do
not exist as registers.  If you only modify a portion of a port you will read
the actual value on the pin, not the intended value.

-Adam

WF wrote:
>
> My question is:
>
> Will affect the remainder bits? The remainders bits are configurated as
> INPUT and OUTPUT...
>
> Miguel
> ----- Original Message -----
> From: Andrew Kunz <akunz@TDIPOWER.COM>
> To: <PICLIST@MITVMA.MIT.EDU>
> Sent: Tuesday, June 13, 2000 4:42 PM
> Subject: Re: [OT]: CCS C COMPILER
>
> > PORTB = (PORTB & 0x0F) | (NEWVALUE & 0xF0);   // Updates high nybble
> >
> > PORTB = (PORTB & 0xF0) | (NEWVALUE & 0x0F);   // Updates low nybble
> >
> > Andy