DCD command - need some advice

[malc-c]

From a previous thread Steve suggested a simple test program that I would like to embellish on to send a pattern sequence to port B of a 16F873a. The code I've come up with is shown below, but I keep getting the "Bad expression" error when I try to use a varible in the DCD line.

If I enter PORTB=DCD patt3 the code compiles file, However I want to change the pattern depending on the value of a varible (either swcount or N) and I've tried

PORTB=DCD patt,N
PORTB=DCD patt,(N)
PORTB=DCD patt(N)
And replacing N with swcount and it still errors.

Can the DCD command be used in such a way ?

;************* set up pattern data **********
    
Patt1 DATA 16,1,2,4,8,16,32,64,128,128,64,32,16,8,4,2,1 
Patt2 DATA 8,129,66,36,24,24,36,66,129 
Patt3 DATA 16,1,3,2,6,4,12,8,24,16,48,32,96,64,192,128,0 
Patt4 DATA 16,1,128,2,64,4,32,8,16,8,32,4,64,2,128,1,0 
Patt5 DATA 12,24,60,126,255,231,195,129,0,129,195,231,255 
Patt6 DATA 13,1,2,4,8,17,34,68,136,16,32,64,128,0 
Patt7 DATA 8,128,64,32,16,8,4,2,1

;************* set up PIC ********************

ADCON1=$0F
CMCON = 7                        ' Digital inputs
CCP1CON = 0                      ' PWM off
TRISA=%11100111                  'set PORTA as all input apart from A3 & A4
TRISB=%00000000                  'set PORTB as all output
PORTB=0

;************* set up variables ***************

ByteA   var byte
D var byte
scale var byte                      ;used in the POT command
Scale = 254  
steps var byte
swcount var byte
swcount=3                           ; set to 3 for testing only
N var byte

;**************** main program ********************
main:
if swcount = 1 then read Patt1, steps   ;read the first value in data string patt1 and place it in steps
if swcount = 2 then read Patt2, steps
If swcount = 3 then read Patt3, steps
if swcount = 4 then read Patt4, steps
if swcount = 5 then read Patt5, steps
If swcount = 6 then read Patt6, steps
If swcount = 7 then Read patt7, steps

n = swcount

Pot PORTA.1,scale,D                 ;used to read value from 10k pot
    for bytea=0 to steps
        PORTB=DCD patt,n
        PAUSE D
    NEXT
goto main

The other question I have, is on a 16F628A the datasheet shows the program memory (flash words) as 2048, data memory as 224 bytes SRAM and 128 EEPROM bytes. However the Datasheet for the 16F873A shows Program memory as 7.2K bytes / 4096 single word instructions. Can someone advise why / what is meant by the two entries for the '873A's program memory, and if it's different to the '628A's memory ?

[mister_e]

i'm not sure what you expect at the output, but when you said PORTB=DCD patt3 worked but you want to use some kind of text substitution, and use something like PORTB=DCD patt[x] and if x=3 work... you'll like the following

;************* set up pattern data **********
    
Patt1 DATA 16,1,2,4,8,16,32,64,128,128,64,32,16,8,4,2,1 
Patt2 DATA 8,129,66,36,24,24,36,66,129 
Patt3 DATA 16,1,3,2,6,4,12,8,24,16,48,32,96,64,192,128,0 
Patt4 DATA 16,1,128,2,64,4,32,8,16,8,32,4,64,2,128,1,0 
Patt5 DATA 12,24,60,126,255,231,195,129,0,129,195,231,255 
Patt6 DATA 13,1,2,4,8,17,34,68,136,16,32,64,128,0 
Patt7 DATA 8,128,64,32,16,8,4,2,1

;************* set up PIC ********************

ADCON1=$0F
CMCON = 7                        ' Digital inputs
CCP1CON = 0                      ' PWM off
TRISA=%11100111                  'set PORTA as all input apart from A3 & A4
TRISB=%00000000                  'set PORTB as all output
PORTB=0

;************* set up variables ***************

Patt    var byte [8]
asm
    MOVE?CB _Patt1,_Patt+1  ; Patt[1]
    MOVE?CB _Patt2,_Patt+2  ; Patt[2]
    MOVE?CB _Patt3,_Patt+3  ; Patt[3]
    MOVE?CB _Patt4,_Patt+4  ; Patt[4]
    MOVE?CB _Patt5,_Patt+5  ; Patt[5]
    MOVE?CB _Patt6,_Patt+6  ; Patt[6]
    MOVE?CB _Patt7,_Patt+7  ; Patt[7]
endasm


ByteB   var byte
ByteA   var byte
D var byte
scale var byte                      ;used in the POT command
Scale = 254  
steps var byte
swcount var byte
swcount=3                           ; set to 3 for testing only
N var byte

;**************** main program ********************
main:
if swcount = 1 then read Patt1, steps   ;read the first value in data string patt1 and place it in steps
if swcount = 2 then read Patt2, steps
If swcount = 3 then read Patt3, steps
if swcount = 4 then read Patt4, steps
if swcount = 5 then read Patt5, steps
If swcount = 6 then read Patt6, steps
If swcount = 7 then Read patt7, steps

n = swcount

Pot PORTA.1,scale,D                 ;used to read value from 10k pot
    FOR ByteA=0 to steps
        PORTB=DCD Patt[n]
        PAUSE D
    NEXT
PAUSE 500
GOTO main

As you see, i didn't touch too much your previous code.. just add some spices in.

Maybe Darrel or someone else have a better solution?

[SteveB]

The other question I have, is on a 16F628A the datasheet shows the program memory (flash words) as 2048, data memory as 224 bytes SRAM and 128 EEPROM bytes. However the Datasheet for the 16F873A shows Program memory as 7.2K bytes / 4096 single word instructions. Can someone advise why / what is meant by the two entries for the '873A's program memory, and if it's different to the '628A's memory ?

In the Mid-Range PICs (16Fxxxx), each instruction is 14 bits long, and is considered a "single word instruction." So, 2048 words in the 628A is half the instructions as 4096 words in the 873A. The 7.2K bytes is just (4096 words * 14 bits)/8 bits = total actual bytes. Not a very useful number (IMO) when using these devices.

HTH,
SteveB

[malc-c]

Thanks guys for your replies.

Steve,
When I said it worked, I mean it compiles without error, not that it actually produced the correct pattern sequence on the LEDs attached to PORT B. I based this on the example you provided in the post on my troubles with programming the 2550 chip. Your code below produced a nice chase sequence whe programmed into the 16F873A

ByteA   var byte
 
main:
    for bytea=0 to 7
        PORTB=DCD BYTEA
        PAUSE 250
        NEXT
    PORTB=255
    PAUSE 250
    goto main

If I read your code correctly DCD is converting the value of bytea (0 to 7) to binary and then displaying this vaule on port B. ie 1 = 00000001, 2 = 00000010 etc. I was hoping that I could read the pattern string, take the values of each step (0 to 255) and use DCD to display the pattern on port B. Hence

for bytea=0 to steps
        PORTB=DCD patt3
        PAUSE D
    NEXT

Funny thing is that as I'm typing this I think I know where 'm going wrong. I need to read each value in the sequence. My old code (for the 16F628A) used

if swcount = 1 then READ (Patt1+counts), PORTB  ;read the next value in patt1 and display it on PORTB
if swcount = 2 then READ (Patt2+counts), PORTB
if swcount = 3 then READ (Patt3+counts), PORTB
if swcount = 4 then READ (Patt4+counts), PORTB
if swcount = 5 then READ (Patt5+counts), PORTB
if swcount = 6 then READ (Patt6+counts), PORTB
if swcount = 7 then READ (Patt7+counts), PORTB

So I tried the following (hopefully you will see my logic

n = swcount

Pot PORTA.1,scale,D                 ;used to read value from 10k pot
    for bytea=0 to steps
       READ (Pattn+bytea), PORTB 
        PAUSE D
    NEXT
goto main

Thinking that it would READ pattern 3 (as I set SWCOUNT to 3 for testing) + the value for bytea (0, to STEPS) and then output it to PORT B. But I get an error "Bad Expression" with the READ (PatN+bytea), PORTB line.

SteveB (and Steve if you can help )

Would the differences between the two chips that you stated cause the use of the DATA statement be handled differently between the two PICs. The reason I ask was then If I use the code I originally wrote for the 16F628a with the 16F873a, the patterns are not displayed in the same way on the '873a.

Thanks for your help

[mister_e]

how about this one?

;************* set up pattern data **********
    
Patt1 DATA 16,1,2,4,8,16,32,64,128,128,64,32,16,8,4,2,1 
Patt2 DATA 8,129,66,36,24,24,36,66,129 
Patt3 DATA 16,1,3,2,6,4,12,8,24,16,48,32,96,64,192,128,0 
Patt4 DATA 16,1,128,2,64,4,32,8,16,8,32,4,64,2,128,1,0 
Patt5 DATA 12,24,60,126,255,231,195,129,0,129,195,231,255 
Patt6 DATA 13,1,2,4,8,17,34,68,136,16,32,64,128,0 
Patt7 DATA 8,128,64,32,16,8,4,2,1

;************* set up PIC ********************

ADCON1=$0F
CMCON = 7                        ' Digital inputs
CCP1CON = 0                      ' PWM off
TRISA=%11100111                  'set PORTA as all input apart from A3 & A4
TRISB=%00000000                  'set PORTB as all output
PORTB=0

;************* set up variables ***************

Patt    var byte [8]
asm
    MOVE?CB _Patt1,_Patt+1 ; Patt[1]
    MOVE?CB _Patt2,_Patt+2 ; Patt[2]
    MOVE?CB _Patt3,_Patt+3 ; Patt[3]
    MOVE?CB _Patt4,_Patt+4 ; Patt[4]
    MOVE?CB _Patt5,_Patt+5 ; Patt[5]
    MOVE?CB _Patt6,_Patt+6 ; Patt[6]
    MOVE?CB _Patt7,_Patt+7 ; Patt[7]
endasm

ByteA   var byte
D var byte
scale var byte                      ;used in the POT command
Scale = 254  
steps var byte
swcount var byte
swcount=7                           ; set to 3 for testing only
D=250
;**************** main program ********************
main:
    for swcount=1 to 7
        read patt[swcount],steps
        for bytea=1 to steps
           READ PATT[SWCOUNT]+BYTEA,PORTB
           PAUSE D
           NEXT
        next
    GOTO MAIN

noticed that i skiped the POT command but use a fixed 250mSec delay. It's working here. if you want to use DCD, you'll need to modify the DATA lines and use bit# instead of Decimal value. No big deal

It shouldn't be different from one to another. DATA remain DATA. PORTB remain PORTB. Well i can't explain why if it doesn't create the expected results with the above suggestion

Too bad, MPASM text substitution doesn't seems to work with Bytes variable in the #V(Variable) area.. it would be too convenient...

EDIT: Note that MOVE?CB _Patt1,_Patt+1 can be replace by Patt[1]=Patt1 etc...

The whole MOVE?CB block (or Patt[x]=Pattx) could also be replace using MPASM text substitution like this

asm
    local a=1
    while a<7
        MOVE?CB _Patt#v(a),_Patt+a
a+=1
        endw
endasm

[malc-c]

how about this one?

Yup - works a treat here too... the LEDs are cycling through all the pattern sequences quite nicely

Now sorry to be a pain in the a$$, but you lost me right at the point you went asm
MOVE?CB _Patt1,_Patt+1 ; Patt[1]


I assume that this could be done in PBP without dropping out to assembly code, hence the Note that MOVE?CB _Patt1,_Patt+1 can be replace by Patt[1]=Patt1 etc... statement ????

Sorry if this seems dumb questions, but now that 'm moving away from the simple basic commands and the fact I was (and still am) a bit rusty with my coding, all help to expand the learning curve with PBP is greatly apreciated

[mister_e]

So, i think i lost you pretty much more with the second asm code using the text substitution?

Sorry sometimes i'm getting sick with those MOVE?CB, PAUSE?C, macro and whatever else... It's the Darrel's fault!

Yup indeed the whole MOVE?CB ... block could be replace by using...

Patt[1]=Patt1
Patt[2]=Patt2
Patt[3]=Patt3
Patt[4]=Patt4
Patt[5]=Patt5
Patt[6]=Patt6
Patt[7]=Patt7

Now compile those line and open the .asm file... you'll discover somewhere those lines...

; C:\PBP_PROG\A.BAS        	00040	Patt[1]=Patt1
	MOVE?CB	_Patt1, _Patt + 00001h

; C:\PBP_PROG\A.BAS        	00041	Patt[2]=Patt2
	MOVE?CB	_Patt2, _Patt + 00002h

; C:\PBP_PROG\A.BAS        	00042	Patt[3]=Patt3
	MOVE?CB	_Patt3, _Patt + 00003h

; C:\PBP_PROG\A.BAS        	00043	Patt[4]=Patt4
	MOVE?CB	_Patt4, _Patt + 00004h

; C:\PBP_PROG\A.BAS        	00044	Patt[5]=Patt5
	MOVE?CB	_Patt5, _Patt + 00005h

; C:\PBP_PROG\A.BAS        	00045	Patt[6]=Patt6
	MOVE?CB	_Patt6, _Patt + 00006h

; C:\PBP_PROG\A.BAS        	00046	Patt[7]=Patt7
	MOVE?CB	_Patt7, _Patt + 00007h

Interesting eh?

NOW, why i use the MOVE?CB first? Well didn't thought Patt[1]=Patt1 would worked... later i decide to try and... Woohhooo

So this is why i edited my previous post. I'm still learning as well. I guess we never stop anyways!

MOVE?CB: Is a PBP macro wich copy a Constant value to a Byte variable. Not much.

Glad to hear it's working as expected
Have fun!