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Assembly question..

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rewguy

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Is there a way in PIC assembler to set all bits of a register to 1 without using the W register? Basically, I want to do this:

Code:
movlw    0xFF
moxwf    PORTB

Except I want to preserve what was in the w register. It would simply be a reversal of

Code:
CLRF     PORTB

The reason I want to do this is in displaying a 4 digit number on 7 seg LEDs with common Anode. I want to turn on a number and call the delay. Then after the delay, I want to call the lookup table for the next number, Set PortA to all 1s (turn off the display), do RLF to get to the next digit, then finally put the contents of w back on PortB. The code looks like this:

Code:
<Some digits is being displayed> ;when both PortA and PortB are low, digit is displayed

         call           Delay

         movlw          h'06'          ;Put some number (6) in the W register
         call           LUTable        ;Call the lookup Table with 6
        __?__         PORTB          ;Turn off all LEDS momentarily
         RLF            PORTA          ;Move to the next digit
                                       ;NOTE: W is still value of 6 in LED format
         movwf          PORTB          ;Load PORTB with new number
                                             ;digit is displayed
I know that people normally just use the lookup table afterthey have turned off the LED, but doesn't this mean you have to make your delay faster to reduce flicker? Wouldn't it be much clearer if the lookup table was called while the LED was on?
 
There isn't such a instruction...
Save your W register to RAM temporarily or just call the table while leds are off. Even at 4Mhz pic does this so fast you won't even notice
 
Code:
    MOVWF    PORTB       ; Set some of the bits
    XORLW    H'FF'          ; Invert W reg
    IORWF    PORTB,F    ;Set the rest of the bits
    XORLW    H'FF'          ; Reverse inversion
 
motion said:
Code:
    MOVWF    PORTB       ; Set some of the bits
    XORLW    H'FF'          ; Invert W reg
    IORWF    PORTB,F    ;Set the rest of the bits
    XORLW    H'FF'          ; Reverse inversion

problem is all the bits won't be set at once... if the loop is reapeated fast enough it might be visible :?
 
Get real! Even you said that the the PIC is fast enough that the difference would not be noticable. A 2usec difference in timing even for a delay loop of 1msec would not be detected even if you have eyes of a fly. Besides, setting the bits turns off the LEDs.

You would be getting greater variations in timing through the delay loop.
 
i said turning it off - loading the lookup table - turning it on. will almost not be visible. (maybe a little bit dimmer)

because you have nothing - a value - nothing -... you won't see anything mixed

But displaying a certain value for a time. Another for a shorter time in a tight loop (he might be running that loop 50000 times a second for all we know) i think it's possible the other value shimmers trough a little.

I'm not saying it is so. I'm just saying such things are ofthen much diffirent in practice then in theory. The only way to know for sure is to test it :)
 
In practical applications you don't multiplex displays at such a tight loop. Even a 1 msec loop is very tight. So in the real world, the kind of problems you are suggesting is non-existent.

Another thing, this code segment is normally part of a bigger program where other tasks are being performed. So the delay through the loop will vary far greater than 2usec.

Thirdly, each LED is current limited by a resistor of up to 5% tolerance. That should produce a much wider variation in light intensity.

I suggest you try it in the real world and see for yourself.

BTW, the code needs a slight change.

Code:
    IORWF    PORTB,F       ; Set some of the bits 
    XORLW    H'FF'          ; Invert W reg 
    IORWF    PORTB,F    ;Set the rest of the bits 
    XORLW    H'FF'          ; Reverse inversion

This will prevent LED segments that are already turned off from being turned ON momentarily.
 
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