Save some bytes in program memory

[pedja089]

You can save some code space if you use multiple If condition instead one with multiple AND.
Example:

IF portb.0 AND portb.1 AND portb.2 AND portb.3 THEN
ENDIF '46words

But if you break that in multiple if's

IF portb.0 THEN
 IF portb.1 THEN
  IF portb.2 THEN
   IF portb.3 then
   ENDIF
  ENDIF
 ENDIF
ENDIF

Code is just 13 words.

[Mike, K8LH]

You can save some code space if you use multiple If condition instead one with multiple AND.
Example:

IF portb.0 AND portb.1 AND portb.2 AND portb.3 THEN
ENDIF '46words

But if you break that in multiple if's

IF portb.0 THEN
 IF portb.1 THEN
  IF portb.2 THEN
   IF portb.3 then
   ENDIF
  ENDIF
 ENDIF
ENDIF

Code is just 13 words.

Wow! Big difference. Here's your first example (46 words) in BoostC -> 16 words;

    if(portb.0 & portb.1 & portb.2 & portb.3)
      nop();

     if(portb.0 & portb.1 & portb.2 & portb.3)
00ED  01FD      CLRF CompTempVar20
00EE  1C86      BTFSS gbl_portb,1
00EF  28F3      GOTO    label14
00F0  1C06      BTFSS gbl_portb,0
00F1  28F3      GOTO    label14
00F2  0AFD      INCF CompTempVar20, F
00F3        label14
00F3  01FE      CLRF CompTempVar21
00F4  1C7D      BTFSS CompTempVar20,0
00F5  28F9      GOTO    label15
00F6  1D06      BTFSS gbl_portb,2
00F7  28F9      GOTO    label15
00F8  0AFE      INCF CompTempVar21, F
00F9        label15
00F9  1D86      BTFSS gbl_portb,3
00FA  28FE      GOTO    label16
00FB  1C7E      BTFSS CompTempVar21,0
00FC  28FE      GOTO    label16


       nop();         
00FD  0000      NOP
00FE        label16

Your second example (13 words) in BoostC -> 8 words;

     if(portb.0)
       if(portb.1)
         if(portb.2)
           if(portb.3)
             nop();

     if(portb.0)
00DD  1C06      BTFSS gbl_portb,0
00DE  28E6      GOTO    label12
        if(portb.1)
00DF  1C86      BTFSS gbl_portb,1
00E0  28E6      GOTO    label12
          if(portb.2)
00E1  1D06      BTFSS gbl_portb,2
00E2  28E6      GOTO    label12
            if(portb.3)
00E3  1D86      BTFSS gbl_portb,3
00E4  28E6      GOTO    label12

              nop();
00E5  0000      NOP
00E6        label12

This variation only uses four words, which is pretty much how you might code it in assembler;

     if(portb ^ 15)
       nop();

     if(portb ^ 15)
00E6  300F      MOVLW 0x0F
00E7  0606      XORWF gbl_portb, W
00E8  1903      BTFSC STATUS,Z
00E9  28EB      GOTO    label13


       nop();
00EA  0000      NOP
00EB        label13

It looks like the moral is to try and avoid multiple operations within a single instruction statement (line).

[Mike, K8LH]

I was too quick on the trigger... That last example should be;

    if((portb ^ 15) == 0)
      nop();

     if((portb ^ 15) == 0)
00E8  300F      MOVLW 0x0F
00E9  0606      XORWF gbl_portb, W
00EA  1D03      BTFSS STATUS,Z
00EB  28ED      GOTO    label13


       nop();
00EC  0000      NOP
00ED        label13

[Art]

This variation only uses four words, which is pretty much how you might code it in assembler;

You could load portb into the temp variable, rotate it four times left, and check the value is b11110000,
and it wouldn't matter what was happening on the high bits of portb.

[Mike, K8LH]

This variation only uses four words, which is pretty much how you might code it in assembler;

You could load portb into the temp variable, rotate it four times left, and check the value is b11110000, and it wouldn't matter what was happening on the high bits of portb.

Is there some advantage doing it that way, Art?

[Art]

Only if there was something else going on with the 4 high bits of portb,
Otherwise any input on the high bits breaks the condition of portb ^ 15 = 0. or am I mistaken there?

[Mike, K8LH]

You're correct. Sorry I misunderstood, Art...

[rsocor01]

It looks like the moral is to try and avoid multiple operations within a single instruction statement (line).

Hmmm, I've never looked at it that way, but you might be right .