Re: [avr-gcc-list] efficiency of assigning bits

[E. Weddington]

Jamie Morken wrote:

> Hi,
>
> ----- Original Message -----
> From: "E. Weddington" <address@hidden>
> Date: Monday, March 14, 2005 8:52 am
> Subject: Re: [avr-gcc-list] efficiency of assigning bits
>
>  
>
> > Jamie Morken wrote:
> >
> >    
>
>  
>
> > Not quite. Do this:
> >
> > PORTD &= ~(_BV(PD1) | _BV(PD4));
> >
> > That will clear the two bits at the same time.
> >    
>
> How can you set 3 bits and clear 1 bit on the same port with one line like 
> above?  I read through your tutorial below but I couldn't figure out this still 
> :)
>
>  
Forgive me if I don't have the correct bits that you want; this is just 
an example.

To clear bit 0 and set bits 1,2,3 without disturbing the other bits:

PORTD = (PORTD & ~(_BV(0))) | (_BV(1)|_BV(2)|_BV(3));

Use the bit operator truth tables in the tutorial and work out each 
operation.

Eric

PS: Please reply to the list and not me personally. Thanks.