Multiplexer address control..

[pedja089]

If you are using PIC 18F, then use simplest solution is
LATB.4=Control.0
LATB.5=Control.1
LATB.6=Control.2
LATB.7=Control.3

EDIT:
For PIC16

PortBCopy var byte

PortBCopy = PORTB
PortBCopy .4=Control.0
PortBCopy .5=Control.1
PortBCopy .6=Control.2
PortBCopy .7=Control.3
PORTB=PortBCopy

[retepsnikrep]

Yes thanks, but how to step through the options in the most efficient way without disturbing ports B0-3

[pedja089]

for control = 0 to 15
LATB.4=Control.0
LATB.5=Control.1
LATB.6=Control.2
LATB.7=Control.3
next control

[Art]

It would be quicker if you used the port bits 0-3 instead, so you could just increment the port b value.
The quickest way also depends where the enable bit is connected.

[Dave]

Pedja089, In your suggested code you are treating "Control" as an array. If it is not then you are just resetting the LATx bits 16 times.

[pedja089]

I'm using as single bit, not array. And changing only upper 4 bit.
For this PBP will generate 4 asm instruction per PBP line, and execute 2 or 3 of them.
In worst case scenario 12 asm instruction are executed.
Compared to any PBP math function size, and speed are better in this case.
Maybe using logic you can get same result. You can try it if you want.
Let me explain little more...
Control=2
LATB.4=Control.0
LATB.5=Control.1
LATB.6=Control.2
LATB.7=Control.3
Bit 4 of LAT register is set to 0, bit 5 of LAT register is set to 1, bits 6,7 of LAT register is set to 0.
Bits 0,1,2,3, are unaffected.


PS
Array would be Control[i]...

[Dave]

Pedja089, Why not just say:

for control = 0 to 15
LATB = LATB & $0F 'strip upper 4 bits
LATB = LATB | Control << 4 'add new control bits
next control