Bitwise not working

[paul borgmeier]

Steve's does it... but if you do not want to fiddle with the upper nibble latch

For i=0 to $0F
    temp = PORTB & $F0 ' keep upper nibble
    PORTB= temp | i       ' keep upper, set lower to i
next

EDIT - Wow - a lot happens in 10 minutes - I was replying to Steve's #5 and 3 more got in there! :-)
and attachments did not work for me yesterday - I got corrupt error when tried to open

[Ioannis]

Thanks Paul and Steve.

The solution I was expressing at #6 was the one Paul was describing. That is the solution. Steve clever but you do not keep the upper bits.

I have not checked if the interrupt flag for the 4:7 of port b is affected if one is trying to write on input bits of the port. That renains to be checked.

Thanks again,
Ioannis

[sayzer]

Why not using Grandpa's traditional way?

i    VAR BYTE


for i=0 to 15
    PORTB.0 = i.0
    PORTB.1 = i.1
    PORTB.2 = i.2
    PORTB.3 = i.3    
NEXT i

mister_e's #5 code is indeed short. Now it keeps upper ones unchanged.

[mister_e]

As the pin are set to input.... if you write to them, the result stay unchanged... as i posted. Not sure with the interrupt.

Why do you think that PORTB.7=1 or PORTB.7=0 no nothing if the TRIS.7 is not set to 0?

One thing is sure, you will clear the RBIF bit just by reading PORTB, so the above solution using Temp=PORTB will certainely cause some strange interrupt behaviour (at least you will miss some). All in the datasheet in INTCON register description.

[Ioannis]

Steve, yes I know reading the port, RBIF will get reset. But what about writing to the port if the upper nibble is input? Will the write cause Interrupt?

I can tolerate to loose a few while writing to the port. But producing Int's for this reason is not welcome.

Sayzer, nice idea the old recipe. I am just looking too shorten my code. I 'll see which method is shortest.


Thanks,
Ioannis

[paul borgmeier]

Both “Temp = PortB” and “PortB = i” read the Port . Both allow for a chance of missing a change and therefore an interrupt. Steve is right in that preserving the state of the upper four pins does not appear necessary since it does not change the input values when the port is written – just the latched values should the pin directions be changed from inputs to outputs. Go with #5.

[Ioannis]

OK guys. I thought of preservint the interrupts. Anyway, sometimes your thoughts are silly without the ability to see further...

Thanks,
Ioannis