----- Original Message -----
From: Dave Bell <dbell@THEBELLS.NET>
To: <PICLIST@MITVMA.MIT.EDU>
Sent: Wednesday, January 10, 2001 5:16 PM
Subject: Re: [PIC]: yet another algorithm challange


> Thanks to Andrew Vassallo and Bob Ammerman for some quick and detailed
> help on my "bit stretcher" algorithms!
>
> Here's what I ended up with for the 6X routine:
>
> //
> int   OUT[6];
>
> #byte Out1 = OUT
> #byte Out2 = OUT+1
> #byte Out3 = OUT+2
> #byte Out4 = OUT+3
> #byte Out5 = OUT+4
> #byte Out6 = OUT+5
>
> int   inval;
>
> void X6(int v)
> {
> #asm
>    movwf    inval
>
>    movlw    0                 // start output byte 6
>    btfsc    inval,0
>    iorlw    0x3F
>
>    btfsc    inval,1
>    iorlw    0xC0
>    movwf    Out6
>
>    movlw    0                 // start output byte 5
>    btfsc    inval,1
>    iorlw    0x0F
>
>    btfsc    inval,2
>    iorlw    0xF0
>    movwf    Out5
>
>    movlw    0                 // start output byte 4
>    btfsc    inval,2
>    iorlw    0x03
>
>    btfsc    inval,3
>    iorlw    0xFC
>    movwf    Out4
>
> // now, repeat for (bit+4) and (byte-3)
>    movlw    0                 // start output byte 3
>    btfsc    inval,4
>    iorlw    0x3F
>
>    btfsc    inval,5
>    iorlw    0xC0
>    movwf    Out3
>
>    movlw    0                 // start output byte 2
>    btfsc    inval,5
>    iorlw    0x0F
>
>    btfsc    inval,6
>    iorlw    0xF0
>    movwf    Out2
>
>    movlw    0                 // start output byte 1
>    btfsc    inval,6
>    iorlw    0x03
>
>    btfsc    inval,7
>    iorlw    0xFC
>    movwf    Out1
> #endasm
> }
>
>
> Fast (enough), clean (very), and (I think) isochronous to boot...

You're not likely to find faster.
It is very simple to follow and 'clean'.
It isn't too long.
And yes, it is isochronous.

Sometimes the obvious is what makes the most sense.

Bob Ammerman
RAm Systems
(contract development of high performance, high function, low-level
software)

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