Using "If then" and "and" together

[Christopher4187]

I have a keypad with four numbers. I'm able to decode the keys just fine but it's the password I'm having trouble with. I'm trying to do something like this:

IF CODE_ONE=13 AND CODE_TW0=13 AND CODE_THREE=16 AND CODE_FOUR=21 AND CODE_FIVE=32 THEN....

but PBP gives me a syntax error. I realize this is probably archaic but I'm wondering two things.

1. Why won't PBP let me compile this code?

2. I realize this code is not efficient. How can I improve it?

[Christopher4187]

Ok, so apparently I'm not intelligent enough to realize the difference between a 0 (zero) and O (the letter O)

If there is a way to make this code better, please let me know.

[HenrikOlsson]

Hi,
You could try a bunch of nested IF statements. It won't be "better" but it might compile to smaller code - worth a try.

IF CODE_ONE = 13 THEN
  IF CODE_TWO = 13 THEN  'With an oh
    IF CODE_THREE = 16 THEN
      IF CODE_FOUR = 21 THEN
        IF CODE_FIVE = 32 THEN
          'Correct code
        ENDIF
      ENDIF
    ENDIF
  ENDIF
ENDIF

Another option would be to use an array instead of "discrete" variables and iterate thru it with a for-next loop. Might be worth a try.

/Henrik.

[Demon]

Using PBP 2.60C, MCSP v2.2.1.1, MPASM v5.46, PIC 18F46K22.

62 bytes:

@   __CONFIG    _CONFIG1H, _FOSC_HSHP_1H & _PLLCFG_OFF_1H & _PRICLKEN_OFF_1H & _FCMEN_OFF_1H & _IESO_OFF_1H
@   __CONFIG    _CONFIG2L, _PWRTEN_ON_2L & _BOREN_SBORDIS_2L & _BORV_285_2L
@   __CONFIG    _CONFIG2H, _WDTEN_OFF_2H
@   __CONFIG    _CONFIG3H, _PBADEN_OFF_3H & _HFOFST_OFF_3H & _MCLRE_EXTMCLR_3H
@   __CONFIG    _CONFIG4L, _STVREN_ON_4L & _LVP_OFF_4L & _XINST_OFF_4L & _DEBUG_OFF_4L
DEFINE OSC 20
ANSELA = 0
ANSELB = 0
ANSELC = 0
ANSELD = 0
ANSELE = 0
TRISA = %00000000
TRISB = %00000000
TRISC = %00000000
TRISD = %10000000
TRISE = %00000000

CODE_ONE        VAR PortB.1
CODE_TWO        VAR PortB.2
CODE_THREE      VAR PortB.3
CODE_FOUR       VAR PortB.4
CODE_FIVE       VAR PortB.5

  IF CODE_ONE = 13 THEN
    IF CODE_TWO = 13 THEN  'With an oh
      IF CODE_THREE = 16 THEN
        IF CODE_FOUR = 21 THEN
          IF CODE_FIVE = 32 THEN
            'Correct code
          ENDIF
        ENDIF
      ENDIF
    ENDIF
  ENDIF
END

138 bytes:

@   __CONFIG    _CONFIG1H, _FOSC_HSHP_1H & _PLLCFG_OFF_1H & _PRICLKEN_OFF_1H & _FCMEN_OFF_1H & _IESO_OFF_1H
@   __CONFIG    _CONFIG2L, _PWRTEN_ON_2L & _BOREN_SBORDIS_2L & _BORV_285_2L
@   __CONFIG    _CONFIG2H, _WDTEN_OFF_2H
@   __CONFIG    _CONFIG3H, _PBADEN_OFF_3H & _HFOFST_OFF_3H & _MCLRE_EXTMCLR_3H
@   __CONFIG    _CONFIG4L, _STVREN_ON_4L & _LVP_OFF_4L & _XINST_OFF_4L & _DEBUG_OFF_4L
DEFINE OSC 20
ANSELA = 0
ANSELB = 0
ANSELC = 0
ANSELD = 0
ANSELE = 0
TRISA = %00000000
TRISB = %00000000
TRISC = %00000000
TRISD = %10000000
TRISE = %00000000

CODE_ONE        VAR PortB.1
CODE_TWO        VAR PortB.2
CODE_THREE      VAR PortB.3
CODE_FOUR       VAR PortB.4
CODE_FIVE       VAR PortB.5

  IF CODE_ONE = 13 and CODE_TWO = 13 and CODE_THREE = 16 and CODE_FOUR = 21 and CODE_FIVE = 32 THEN
    'Correct code
  ENDIF
END

I know, Port be is defined as output and used as input (noticed after test), but that has no impact on test significance.

More than twice as big using single IF, nested IF is definitely more efficient. I remember reading about that somewhere recently too, can't remember where.

Robert

[aerostar]

you could also turn it on its head,

IF CODE_ONE <> 13 THEN EXIT_TO_ERROR

AND SO ON,

Then after the last comparison you have a valid password.

[HenrikOlsson]

Hi,
I tried four examples here, compiled for a 16F628.
First test:

  IF CODE_ONE = 13 and CODE_TWO = 13 and CODE_THREE = 16 and CODE_FOUR = 21 and CODE_FIVE = 32 THEN
   'Correct code
  ENDIF

Result: 75 WORDS

Second test:

If CODE_ONE = 13 THEN
 IF CODE_TWO = 13 then
   if CODE_THREE = 16 then
     if CODE_FOUR = 21 then
       if CODE_FIVE = 32 THEN
       'correct code
       Endif
     endif
   endif
 endif
endif

Result: 26 WORDS

Third test:

IF CODE_ONE <> 13 THEN EXIT_TO_ERROR
IF CODE_TWO <> 13 THEN EXIT_TO_ERROR
IF CODE_THREE <> 13 THEN EXIT_TO_ERROR
IF CODE_FOUR <> 13 THEN EXIT_TO_ERROR
IF CODE_FIVE <> 13 THEN EXIT_TO_ERROR
'correct code

EXIT_TO_ERROR:

RESULT: 26 WORDS (same as test 2)

Forth test:
This one uses an array to hold the entered code and assumes that the correct password sequence is stored in EEPROM, starting at location 0

Correct = 1
For i = 0 to 4
 Read i, Password
  if Code[i] <> Password THEN
    Correct = 0
  ENDIF
NEXT
If Correct THEN
'correct code
endif

RESULT: 41 WORDS

So between these four method 2 or 3 produces the smallest code - by far. Personally I think Aerostars suggestion is the easiest to read but that might just be me...

/Henrik.

[longpole001]

Henrik , can you advise why the big dif in words count between the samples ?