What am I doing wrong?

[mackrackit]

Not sure but

key = (row * 4) + (NCD (col ^ $f))
'NOTE: for 12-key keypad, change to key = (row * 3)

And please use code tags, makes it easier to read...

[tazntex]

Thanks Mackrackit, I tried that but that doesn't work right either. I first thought since the orginal code had "row = 0 to 3, and I am using TRISB = (DCD row) ^ $fe that it would seem that I would have to state "row = 1 to 3" the $fe in TRISB = (DCD row) ^ $fe allows for when I flip the bits to set only one to output bit 0 becomes an input, I am using bit 0 for my interrupt in which when no key is pressed it goes to sleep and right before I go back to the main program I do this:
CMCON=%00000111
vrcon = 0
pauseus 10
TRISB = $fe ' B7-B1 in, B0-out
PORTB = %00000001 ' All output-pins low except bit 0
pauseus 10
intcon.1 = 0 'clears flag
resume
then my main program is this:

enable
cycle:
pause 1
@ Sleep
@ nop
@ nop
goto cycle

I must admit I am confused about row = 1 to 3, that does mean bit 1-3 in the variable Row doesn't it? if so I thought perhaps I could shift by one bit and account for the change, key = ((row >> 1) * 3) + (NCD (col ^ $f) but that doesn't work either. Anyhow I haven't abandoned the ship yet

Again, Many thanks for your reply

[mackrackit]

I am still probably not seeing the problem

FOR row = 0 TO 3        ' 4 rows in keypad

In your case

FOR row = 0 TO 2        ' 3 rows in keypad

Or maybe it is wired wrong...

[tazntex]

I double checked everything with my Fluke my rows are on Portb. 1,Portb.2,and Portb.3 my columns are on Portb.4 - 7. I had everything setup like the orginal example on Bruce's site and it worked great but I wanted to use Portb.0 for my interrupt, thats when I moved everthing over one bit. It works, about eight functions work correctly but now I can send on two different buttons things like two decimal 7's or 5's , etc. instead of decimal 1-12. When I break the code down on paper to calculate what the output would be it concurs the same thing so thats why I was wondering how I could adjust the code, as far as I can see it has to do with the line "for row = 1 to 3" thats why I asked if 1 to 3 does refer to bit 1-3 in the variable row, and if so how can I compensate for three rows that are now shifted by one bit. I tried to see if ((row >> 1) *3) + (ncd (col ^$fe)) but about the same results.

Thanks again.

[mackrackit]

OK, now I think I see what you are doing. Me a little slower than normal today...

row = PORTB & %00001110

Should isolate bit 1 ,2, and 3.

[tazntex]

Well I tried that also, but that doesn't work either. I believe I mentioned earlier that I in my mind I was thinking since I moved the rows and columns over by one bit why don't I just create another variable such as:

row1 = row >> 1

then I could still use the same formula:

key = (row * 3) + (NCD (col ^ $f)) ' (row * 3) because I have three rows and when I return the priority bit
'of the columns after I flip the state of the bits ^$f that will tell me which
' column is actually the one being used at that moment of time as an output.

But I tried it and still have not gained any ground.

hmmm, back to the drawing board.

Thanks again...

[tazntex]

If I substitute this:

if row = 1 && col.0 = 0 then key1 = 1
if row = 1 && col.1 = 0 then key1 = 2
if row = 1 && col.2 = 0 then key1 = 6
if row = 1 && col.3 = 0 then key1 = 7
if row = 2 && col.0 = 0 then key1 = 5
if row = 2 && col.2 = 0 then key1 = 10
if row = 2 && col.3 = 0 then key1 = 11
if row = 3 && col.0 = 0 then key1 = 3
if row = 3 && col.1 = 0 then key1 = 4
if row = 3 && col.2 = 0 then key1 = 8
if row = 3 && col.3 = 0 then key1 = 9

for this:
key = (row1 * 3) + (NCD (col ^ $f))

everything works great but isn't there a simpler way?

Thanks