问题
I'm working on a micro-controller without hardware multiply and divide. I need to cook up software algorithms for these basic operations that are a nice balance of compact size and efficiency. My C compiler port will employ these algos, not the the C developers themselves.
My google-fu is so far turning up mostly noise on this topic.
Can anyone point me to something informative? I can use add/sub and shift instructions. Table lookup based algos might also work for me, but I'm a bit worried about cramming so much into the compiler's back-end...um, so to speak.
回答1:
Here's a simple multiplication algorithm:
Start with rightmost bit of multiplier.
If bit in multiplier is 1, add multiplicand
Shift multiplicand by 1
Move to next bit in multiplier and go back to step 2.
And here's a division algorithm:
If divisor is larger than dividend, stop.
While divisor register is less than dividend register, shift left.
Shift divisor register right by 1.
Subtract divisor register from dividend register and change the bit to 1 in the result register at the bit that corresponds with the total number of shifts done to the divisor register.
Start over at step 1 with divisor register in original state.
Of course you'll need to put in a check for dividing by 0, but it should work.
These algorithms, of course, are only for integers.
回答2:
My favorite reference for things like this, available in book form:
http://www.hackersdelight.org/
Also you can't go wrong with TAoCP: http://www-cs-faculty.stanford.edu/~uno/taocp.html
回答3:
Here's a division algorithm: http://www.prasannatech.net/2009/01/division-without-division-operator_24.html
I assume we're talking about ints?
If there's no hardware support, you'll have to implement your own divide-by-zero exception.
(I didn't have much luck quickly finding a multiplication algorithm, but I'll keep looking if someone else doesn't find one).
回答4:
One simple and fairly performant multiplication algorithm for integers is Russian Peasant Multiplication.
For rationals, you could try a binary quote notation, for which division is easier than usual.
回答5:
It turns out that I still have some old 68000 assembler code for long multiplication and long division. 68000 code is pretty clean and simple, so should be easy to translate for your chip.
The 68000 had multiply and divide instructions IIRC - I think these were written as a learning exercise.
Decided to just put the code here. Added comments and, in the process, fixed a problem.
;
; Purpose : division of longword by longword to give longword
; : all values signed.
; Requires : d0.L == Value to divide
; : d1.L == Value to divide by
; Changes : d0.L == Remainder
; : d2.L == Result
; : corrupts d1, d3, d4
;
section text
ldiv: move #0,d3 ; Convert d0 -ve to +ve - d3 records original sign
tst.l d0
bpl.s lib5a
neg.l d0
not d3
lib5a: tst.l d1 ; Convert d1 -ve to +ve - d3 records result sign
bpl.s lib5b
neg.l d1
not d3
lib5b: tst.l d1 ; Detect division by zero (not really handled well)
bne.s lib3a
rts
lib3a: moveq.l #0,d2 ; Init working result d2
moveq.l #1,d4 ; Init d4
lib3b: cmp.l d0,d1 ; while d0 < d1 {
bhi.s lib3c
asl.l #1,d1 ; double d1 and d4
asl.l #1,d4
bra.s lib3b ; }
lib3c: asr.l #1,d1 ; halve d1 and d4
asr.l #1,d4
bcs.s lib3d ; stop when d4 reaches zero
cmp.l d0,d1 ; do subtraction if appropriate
bhi.s lib3c
or.l d4,d2 ; update result
sub.l d1,d0
bne.s lib3c
lib3d: ; fix the result and remainder signs
; and.l #$7fffffff,d2 ; don't know why this is here
tst d3
beq.s lib3e
neg.l d2
neg.l d0
lib3e: rts
;
; Purpose : Multiply long by long to give long
; Requires : D0.L == Input 1
; : D1.L == Input 2
; Changes : D2.L == Result
; : D3.L is corrupted
;
lmul: move #0,d3 ; d0 -ve to +ve, original sign in d3
tst.l d0
bpl.s lib4c
neg.l d0
not d3
lib4c: tst.l d1 ; d1 -ve to +ve, result sign in d3
bpl.s lib4d
neg.l d1
not d3
lib4d: moveq.l #0,d2 ; init d2 as working result
lib4a: asr.l #1,d0 ; shift d0 right
bcs.s lib4b ; if a bit fell off, update result
asl.l #1,d1 ; either way, shift left d1
tst.l d0
bne.s lib4a ; if d0 non-zero, continue
tst.l d3 ; basically done - apply sign?
beq.s lib4e ; was broken! now fixed
neg.l d2
lib4e: rts
lib4b: add.l d1,d2 ; main loop body - update result
asl.l #1,d1
bra.s lib4a
By the way - I never did figure out whether it was necessary to convert everything to positive at the start. If you're careful with the shift operations, that may be avoidable overhead.
回答6:
To multiply, add partial products from the shifted multiplicand to an accumulator iff the corresponding bit in the multiplier is set. Shift multiplicand and multiplier at end of loop, testing multiplier & 1 to see if addition should be done.
回答7:
The Microchip PICmicro 16Fxxx series chips do not have a multiply or divide instruction. Perhaps some of the soft multiply and soft divide routines for it can be ported to your MCU.
PIC Microcontroller Basic Math Multiplication Methods
PIC Microcontroller Basic Math Division Methods
Also check out "Newton's method" for division. I think that method gives the smallest executable size of any division algorithm I've ever seen, although the explanation makes it sound more complicated than it really is. I hear that some early Cray supercomputers used Newton's method for division.
来源：https://stackoverflow.com/questions/2390354/where-can-i-find-soft-multiply-and-divide-algorithms