Digital touch Screen Combination lock

[astanapane]

ok i think i have some how understood the concept.

my question is.

How could we assign a value that can be stored in memory and represent the state of the pin itself. (0 or 1)

but at the same time to not read the actual pin, but the value that is stored in memory.

[richard]

How could we assign a value that can be stored in memory and represent the state of the pin itself. (0 or 1)


but at the same time to not read the actual pin, but the value that is stored in memory.

simple, to read the value assigned to pins designated as output read latx its value is always what you last set it to
the latx value of pins set as input is meaningless in regards to actual pin logic level

to read the actual pin logic levels at a point in time read portx , its always the actual pin value at read time

pin and latx values are not necessarily equal at a point in time, pins set as output are subject to real world physical constraints dictated by the load impedance they are driving

[astanapane]

i did a small test but it is not what i need.

I would like now once again to clear the following.

Henrik said

"One more thing: You're clearly reading PORTB which must mean you're using those five low bits as inputs, yet, at the top of the program you have TRISB = %00000000 which makes all of PORTB outputs..."

As this statement, how could we read the status of the actual pin without using the port as input. Could you please explain to me the method?

ANSELA = 0
    DEFINE OSC 64 
    OSCCON = $70
    OSCTUNE.6 = 1
    TRISA = %11111100 
                             
led1 var porta.0    ; alias led1 to porta.0
led2 var porta.1    ; alias led2 to porta.1
x    var bit            ; tried to use X var as a bit in order to hold the value of the actual status of the pin.

lata = 0               ; based on the schematic, we write to LATA and clear the ports 

x = 0                  ; we write 0 to var X


loopy:
lata.0 = 1           ; we write 1 to lata.0
gosub holdmem  ; gosub to holdmem

pause 500
read 5,led2        ; write the value of the location 5 to led2

pause 500
lata.0 = x          ; write the value of X to lata.0
lata.1 = x          ; write the value of X to lala.1

pause 1000
goto loopy

holdmem:
write 5,led1  ; read the state of the porta.0 and place it to location 5 in memory.
return

[richard]

As this statement, how could we read the status of the actual pin without using the port as input. Could you please explain to me the method?

simply put you cannot , reading portx is the only direct way to read the actual status of the pins.

what henrik means is why would you read a port to read the value of outputs you have already set. the outputs will be at the level you set them to
(after a suitable interval to allow pin to settle) {providing load is acceptable}.

what are trying to do ?

[richard]

do you mean something like this

no need to wear out the eeprom

ANSELA = 0
    DEFINE OSC 64 
    OSCCON = $70
    OSCTUNE.6 = 1
    TRISA = %11111100 
                             
led1 var porta.0    ; alias led1 to porta.0
led2 var porta.1    ; alias led2 to porta.1
x    var bit            ; tried to use X var as a bit in order to hold the value of the actual status of the pin.


 


x = 0                  ; we write 0 to var X




loopy:
lata.0 = 1           ; we write 1 to lata.0
gosub holdmem  ; gosub to holdmem


pause 500
led2  = x      ; write the value of x to led2
pause 500
led1=0
@ nop
led2=0
pause 1000
goto loopy


holdmem:
x = led1  ; read the state of the porta.0 and place it in x
'x = lata.0  ;or  read the state of the lata.0 and place it in x
return
 
goto loopy

[astanapane]

do you mean something like this

no need to wear out the eeprom

ANSELA = 0
    DEFINE OSC 64 
    OSCCON = $70
    OSCTUNE.6 = 1
    TRISA = %11111100 
                             
led1 var porta.0    ; alias led1 to porta.0
led2 var porta.1    ; alias led2 to porta.1
x    var bit            ; tried to use X var as a bit in order to hold the value of the actual status of the pin.


 


x = 0                  ; we write 0 to var X




loopy:
lata.0 = 1           ; we write 1 to lata.0
gosub holdmem  ; gosub to holdmem


pause 500
led2  = x      ; write the value of x to led2
pause 500
led1=0
@ nop
led2=0
pause 1000
goto loopy


holdmem:
x = led1  ; read the state of the porta.0 and place it in x
'x = lata.0  ;or  read the state of the lata.0 and place it in x
return
 
goto loopy

yes i think this is what i wanted to do. But when we use the following command:

x = led1  ; read the state of the porta.0 and place it in x

arent we write to EEPROM? Or this is temprorarly held in memory?

[richard]

arent we write to EEPROM? Or this is temprorarly held in memory?

your lack of understanding exceeds my wildest expectation ,i wasn't prepared for a beginner.
if you like pm with me with your email and i will send a pic tutorial book i have as a pdf . its too big to post on forum


until you comprehend the basic concepts of memory types and uses , binary numbers , basic boolean algebra , logic
operators , i/o concepts ..... its just too hard

[astanapane]

simply put you cannot , reading portx is the only direct way to read the actual status of the pins.

what henrik means is why would you read a port to read the value of outputs you have already set. the outputs will be at the level you set them to
(after a suitable interval to allow pin to settle) {providing load is acceptable}.

what are trying to do ?

The program for the lockcode, at the moment works fine. As mentioned in my first post im not a programmer, i had to find a way to make it work.

So the best suitable solution for me is to light up some LEDs and then read the status of the ports for confirmation.

That's why i made those pins as output and then read the status to confirm if they are 0 or 1.