Loading data into ONLY 4 bits of a port - is it possible?

[Demon]

Just a note, I prefer to use binary instead of hex.

%11110000 is the same as $f0, except it is easy for a guy like me to use %11100000 if I want to refer to top 3 pins only.

There no way I can remember the hex equivalent without using a conversion tool.

Robert

[Art]

You could read portb to a word variable, multiply it, read the high byte into buffera,
zero the high byte, multiply the word var again, and copy the new high byte to bufferb.
This would probably end up more complicated in compiled code than just shifting it.

[Mike, K8LH]

Hi Al (and gang),

Can you better explain what you're trying to do, please? For example, are you trying to take bits b3..b0 in "varA" and copy them to PORTX bits RX3..RX0 and then copy bits b3..b0 in "varB" to PORTX bits RX7..RX4? It so, can you rely on "varA" and "varB" containing only 4 bits of data and simply combine them and write a single byte to PORTX which contains all 8 bits of data?

If you're doing something else, you could always move the 'source' bits into the proper 'target' bit position and then use XOR and AND instructions to copy them to the target. Here's an (assembler) example;

;
;  copy "varA" b3..b0 to PORTX b7..b4
;
        swapf   varA,W          ; move b3..b0 to b7..b4
        xorwf   PORTX,W         ; differences, hi or lo
        andlw   b'11110000'     ; don't change target b3..b0
        xorwf   PORTX,F         ; apply new b7..b4 bits