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kc5tja
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PostPosted: Mon Nov 23, 2009 1:11 am 
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Bogax's solution is 4 cycles faster too, despite the use of the stack.

But, in either case, that's damn slick, and represents a style of coding I'd generally not think of. I rarely think of the carry bit as a single-bit storage location, because it's so often overwritten by other CPU instructions.

This is a good example of keeping multiple views of data in your head at once -- a register isn't "just a number."
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BigDumbDinosaur
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PostPosted: Mon Nov 23, 2009 2:54 am 
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ElEctric_EyE wrote:
Forgive my ignorance,but what is the point of a "CMP #$10" without a branch instruction afterwards?

Only to condition flags to facilitate or affect a later operation. CoMParisons don't necessarily have to be followed by branches. In the above case, the effect on carry is what is wanted to possibly modify the result of the following ADC #$F0 instruction.

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bogax
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PostPosted: Mon Nov 23, 2009 7:37 am 
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ElEctric_EyE wrote:
Forgive my ignorance,but what is the point of a "CMP #$10" without a branch instruction afterwards?


CMP does a subtract (without borrow) and sets the flags without
affecting the accumulator.

Since in this case all bits to the left of the fourth bit are zero
and carry will be set (no borrow) if the accumulator is greater than
or equal to $10 ie if bit four of the accumulator is set, the effect
is to copy bit four (p) to the carry


edit: borrowed the castrated swine :oops: , revised clumsy wording


Last edited by bogax on Sun Dec 20, 2009 10:53 am, edited 1 time in total.

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debounce
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PostPosted: Mon Nov 23, 2009 2:55 pm 
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Location: London, UK
bogax wrote:
ElEctric_EyE wrote:
Forgive my ignorance,but what is the point of a "CMP #$10" without a branch instruction afterwards?


CMP does a subtract (without barrow) and sets the flags without
affecting the accumulator.

Since in this case all bits to the left of the fourth bit are zero
carry will be set (no barrow) if the accumulator is greater than
or equal to $10 ie if bit four of the accumulator is set, in effect
copying bit four (p) to the carry
BigDumbDinosaur wrote:
Only to condition flags to facilitate or affect a later operation. CoMParisons don't necessarily have to be followed by branches. In the above case, the effect on carry is what is wanted to possibly modify the result of the following ADC #$F0 instruction.
At that point I'm trying to add p to the bit count twice. Shifting it out with ADC leaves inverted copies in A, which would take another instruction to rectify anyway. So I copy p with the CMP then take the second copy destructively with ADC #$F0.
The first copy is taken from C and added into the bottom of A (stu), while simultaneously the second copy is put back into C - so the carry flag doesn't change during the ADC.
--Greg
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bogax
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PostPosted: Sun Sep 05, 2010 7:29 am 
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I think I've managed to do it without the stack (haven't tested it yet)
No longer runs in constant time but it is a little faster

Code:

            hgfedcba ?
 asl        gfedcba0 h
 bcc SKIP   gfedcba0 h
 adc #$7F   Sfedcba0 S  ;SS is the Sum h+g
SKIP
 rol        fedcba0S S
 rol        edcba0SS f
 adc #$80   ?dcba0SS e
 and #$7F   0dcba0SS e
 adc #$C0   ??cbaSSS d
 and #$3F   00cbaSSS d
 adc #$E0   ???baSSS c
 and #$1F   000baSSS c
 adc #$F0   ????aSSS b 
 and #$0F   0000aSSS b
 adc #$F8   ?????SSS a 
 and #$07   00000SSS a
 adc #$00   0000SSSS 0


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debounce
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PostPosted: Thu Apr 14, 2011 5:50 pm 
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Joined: Tue Nov 23, 2004 2:11 pm
Posts: 27
Location: London, UK
I don't think I can beat that, bogax... :)
Code:
                        ; C ---A----
                        ; ? hgfedcba
        asl             ; h gfedcba.
        rol             ; g fedcba.h
        adc     #$80    ; f Fedcbapq   pq = h+g
        rol             ; F edcbapqf
        asl             ; e dcbapqf.
        rol             ; d cbapqf.e
        adc     #$84    ; c CbapqFtu   pq = h+g+f  tu = e+d
        and     #$7b    ; c .bapq.tu
        adc     #$c0    ; b BBapq.tu   tu = e+d+c
        and     #$3b    ; b ..apq.tu
        adc     #$e0    ; a AAApqstu  stu = e+d+c+b
        and     #$1f    ; a ...pqstu
        adc     #$f0    ; p PPPPqstu  stu = e+d+c+b+a
        adc     #$18    ; P ..mnQstu  stu = e+d+c+b+a+p  mn = p+q
        cmp     #$10    ; k ..mnQstu    k = m|n = p|q = h|g|f
        adc     #$e0    ; m MMMnQstu  stu = e+d+c+b+a+p+p|q
        and     #$07    ; m .....stu
        adc     #$00    ; . ....rstu rstu = e+d+c+b+a+p+p|q+p&q
;31b 36c
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