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LISTING 10.5 L10-5.ASM
; Finds and returns the greatest common divisor of two integers.
; Uses Euclids algorithm: divides the larger integer by the
; smaller; if the remainder is 0, the smaller integer is the GCD,
; otherwise the smaller integer becomes the larger integer, the
; remainder becomes the smaller integer, and the process is
; repeated. Avoids code recursion.
;
;
;
; C near-callable as:
; unsigned int gcd(unsigned int int1, unsigned int int2);
; Parameter structure:
parms struc
dw ? ;pushed BP
dw ? ;pushed return address
int1 dw ? ;integers for which to find
int2 dw ? ; the GCD
parms ends
.model small
.code
public _gcd
align 2
_gcd proc near
push bp ;preserve callers stack frame
mov bp,sp ;set up our stack frame
push si ;preserve callers register variables
push di
;Swap if necessary to make sure that int1 >= int2
mov ax,int1[bp]
mov bx,int2[bp]
cmp ax,bx ;is int1 >= int2?
jnb IntsSet ;yes, so were all set
xchg ax,bx ;no, so swap int1 and int2
IntsSet:
; Now loop, dividing int1 by int2 and checking the remainder, until
; the remainder is 0. At each step, if the remainder isnt 0, assign
; int2 to int1, and the remainder to int2, then repeat.
GCDLoop:
;if the remainder of int1 divided by
; int2 is 0, then int2 is the gcd
sub dx,dx ;prepare int1 in DX:AX for division
div bx ;int1/int2; remainder is in DX
and dx,dx ;is the remainder zero?
jz Done ;yes, so int2 (BX) is the gcd
;no, so move int2 to int1 and the
; remainder to int2, and repeat the
; process
mov ax,bx ;int1 = int2;
mov bx,dx ;int2 = remainder from DIV
;start of loop unrolling; the above is repeated three times
sub dx,dx ;prepare int1 in DX:AX for division
div bx ;int1/int2; remainder is in DX
and dx,dx ;is the remainder zero?
jz Done ;yes, so int2 (BX) is the gcd
mov ax,bx ;int1 = int2;
mov bx,dx ;int2 = remainder from DIV
;
sub dx,dx ;prepare int1 in DX:AX for division
div bx ;int1/int2; remainder is in DX
and dx,dx ;is the remainder zero?
jz Done ;yes, so int2 (BX) is the gcd
mov ax,bx ;int1 = int2;
mov bx,dx ;int2 = remainder from DIV
;
sub dx,dx ;prepare int1 in DX:AX for division
div bx ;int1/int2; remainder is in DX
and dx,dx ;is the remainder zero?
jz Done ;yes, so int2 (BX) is the gcd
mov ax,bx ;int1 = int2;
mov bx,dx ;int2 = remainder from DIV
;end of loop unrolling
jmp GCDLoop
align2
Done:
mov ax,bx ;return the GCD
pop di ;restore callers register variables
pop si
pop bp ;restore callers stack frame
ret
_gcd endp
end
Assembly language optimization is pattern matching on a local scale. Frankly, its also the sort of boring, brute-force work that people are lousy at; compilers could out-optimize you at this level with one pass tied behind their back if they knew as much about the code youre writing as you do, which they dont.
![]() | Design optimizationconceptual breakthroughs in understanding the relationships between the needs of an application, the nature of the data the application works with, and what the computer can dois global pattern matching. |
Computers are much worse at that sort of pattern matching than humans; computers have no way to integrate vast amounts of disparate information, much of it only vaguely defined or subject to change. People, oddly enough, are better at global optimization than at local optimization. For one thing, its more interesting. For another, its complex and imprecise enough to allow intuition and inspiration, two vastly underrated programming tools, to come to the fore. And, as I pointed out earlier, people tend to perform instantaneous solutions to even the most complex problems, while computers bog down in geometrically or exponentially increasing execution times. Oh, it may take days or weeks for a person to absorb enough information to be able to reach a solution, and the solution may only be near-optimalbut the solution itself (or, at least, each of the pieces of the solution) arrives in a flash.
Those flashes are your programming pattern matcher doing its job. Your job is to give your pattern matcher the opportunity to get to know each problem and run through it two or three times, from different angles, to see what unexpected solutions it can come up with.
Pull back the reins a little. Dont measure progress by lines of code written today; measure it instead by overall progress and by quality. Relax and listen to that quiet inner voice that provides the real breakthroughs. Stop, look, listenand think. Not only will you find that its a more productive and creative way to programbut youll also find that its more fun.
And think what you could do with all those extra computer years!
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