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Re: Looking for 32-bit signed integer math routines in 6502 assembly
In article <Pine.SOL.3.96.970824193831.22408C-100000@post.its.mcw.edu> Ronald
Kneusel <rkneusel@post.its.mcw.edu> writes:
>
>Subject says it all. Specifically, I'm looking for 32-bit _signed_
>multiplication and division:
>
>32-bit * 32-bit = 64-bit (or 32-bit ignoring overflow)
>
>32-bit / 32-bit = 32-bit (plus remainder)
>
>I've found 16-bit routines but was wondering if anyone had 32-bit routines
>in 6502 assembly handy.
First, my apologies for posting all this machine language in comp.sys.apple2,
but that *is* where the original question is. I've tried to direct followups
to comp.sys.apple2.programmer.
Attached below is a 32*32=64 multiplication routine. I just banged it out
tonight, but it seems to give the correct answers for a few test problems.
Alas, it's unsigned, but that's not much of a problem. If a 32-bit result
is sufficient, just use MUL4X4 as-is...the low-order 32 bits are the
correct low-order bits of the signed product. If you need a 64-bit result,
you'll have to do a little pre- and post-processing:
LDA A4 ;Compute sign of result
EOR B4
PHP ;Save on stack
LDY #0
BIT A4 ;A<0?
BPL TRYB ;If not, try B
LDX #$FC ;Else negate A
SEC
L1 TYA
SBC A4+1,X ;(this only works if A1..A4 are on page 0)
STA A4+1,X
INX
BNE L1
TRYB BIT B4 ;B<0?
BPL DOMUL ;If not, multiply
LDX #$FC ;Else negate B
SEC
L2 TYA
SBC B4+1,X ;(this only works if B1..B4 are on page 0)
STA B4+1,X
INX
BNE L2
DOMUL JSR MUL4X4 ;Do the (unsigned) multiplication
PLP ;Should result be <0?
BPL DONE ;If not, then done
LDY #0 ;Else negate result
LDX #$F8
SEC
L3 TYA
SBC R8+1,X ;(this only works if R1..R8 are on page 0)
STA R8+1,X
INX
BNE L3
DONE RTS
Here's the multiplication routine. It wants the four-byte multiplicand
in locations A1 (low byte) through A4 (high byte), and the four-byte
multiplier in B1 through B4. The result comes out in R1 through R8.
The four low-order bytes of R must be the same memory locations as the four
bytes of B--i.e. B1 = R1, B2 = R2, etc. Calling MUL4X4 destroys B.
For efficiency, all twelve memory locations should reside on page 0, but
this is not required (unlike the signed routine above, which does require
everything to be on page 0).
This is not the most compact possible routine...some loops have been
unrolled for speed, and some code has been added to optimize for zero bytes
in B.
MUL4X4 LDA #0 ;Init result to 0
STA R5
STA R6
STA R7
STA R8
JSR MUL1 ;Multiply A by 1st byte of B
JSR MUL1 ;Multiply A by 2nd byte of B
JSR MUL1 ;Multiply A by 3rd byte of B
; Multiply A by 4th byte of B, by falling into...
MUL1 LDY B1 ;Get low byte of B
LDA B2 ;Shift remaining bytes down, to make room for result
STA B1
LDA B3
STA B2
LDA B4
STA B3
TYA ;Get low byte of B back
BNE NON0 ;Is it 0?
LDA R5 ;If so, shift result down by a whole byte, and return
STA R4
LDA R6
STA R5
LDA R7
STA R6
LDA R8
STA R7
STY R8 ;(note Y=0 here)
RTS
NON0 LDX #8 ;Otherwise prepare to shift 8 bits
M1 LSR ;Get low bit of multiplier
TAY ;Save remaining multiplier bits
BCC M2 ;Low bit 0?
CLC ;If not, add multiplicand to result...
LDA A1
ADC R5
STA R5
LDA A2
ADC R6
STA R6
LDA A3
ADC R7
STA R7
LDA A4
ADC R8
STA R8
M2 ROR R8 ;...and shift everything down 1 bit
ROR R7
ROR R6
ROR R5
ROR R4
TYA ;Get multiplier bits back
DEX ;Done 8 bits yet?
BNE M1 ;If not, go do more
RTS ;Else done
The 32/32=32+remainder division routine will have to wait for another day--
I don't have one handy at the moment, and besides, this article is already
getting a bit too long.
- Neil Parker
P.S. Concerning the division routine: Do you care which way the quotient
rounds when the result is negative? Should it go to the more negative result,
or toward 0, or something else?
--
Neil Parker | Unsolicited commercial e-mail to my
nparker@cie-2.uoregon.edu | address is not welcome, and will be
nparker@cie.uoregon.edu | discarded unread.
http://cie-2.uoregon.edu/~nparker |