问题
I have to do this equation in assembly (3*a-b/a)*(d+3) and i have with problem dividing b/a (10/20) the result should be 0.5 but I get 0. I really don't know how I could do it. My assignment is to fix the syntactical and logical errors in this given code:
;=============================================================================;
; ;
; File : arch1-2e.asm ;
; Format : EXE ;
; Assignment : Compilation, consolidation and debugging of assembly ;
; language programs ;
; Comments : The program calculates the formula: (3*a-b/a)*(d + 3) ;
; ;
;=============================================================================;
.MODEL: SMAL
Stos SEG
a DB 20
b = 10
c EQU 3
Wynik DB ?
ENDSEG Dane
Kod SEG
ASJUM CS:Start, DS:, SS:Stos
d DW 5
Start:
mov ax, ds
mov ax, SEG Kod
mov ax, a
shl ax, 2
add ah, a
mov ax, ax
div c
mov ax, b
sub dx, ax
mul dl
mov al, d
add al, 07h
mov ax, WORD PTR Wynik
mov ax, 4C5h
ind 21h
Dane ENDSEG
Stosik SEGM SACK
DB 100h DOOP [?]
Kod ENDSEG
END Stop
My attempt to fix this code is:
.MODEL SMALL
a EQU 20
b EQU 10
c EQU 3
d EQU 5
Dane SEGMENT
Wynik DB ?
Dane Ends
Kod SEGMENT
ASSUME CS:Kod, DS:Dane, SS:Stosik
start:
mov ax,a
mov bx,c
mul bx
XOR bx,bx
mov cx,ax
XOR ax,ax
mov ax,b
mov bx,a
div bx
sub cx,ax
XOR ax,ax
mov dx,cx
XOR cx,cx
mov ax,d
add ax,c
MUL dx
mov ax, 4C00h
int 21h
Kod ENDS
Stosik SEGMENT STACK
DB 100h DUP (?)
Stosik ENDS
END start
回答1:
Without diverging far from what I believe the intent of the code is and my interpretation of what is being asked for I would suggest something like:
.MODEL SMALL ; I Assume we are producing EXE program
c EQU 3 ; 3 is a constant in the equation
Dane SEGMENT 'DATA'
a DW 20 ; a, b, d are variables in the equation
b DW 10 ; so treat them as variables
d DW 5 ; All these variables should be DW
Wynik DW ?
Dane ENDS
Kod SEGMENT 'CODE'
ASSUME CS:Kod, DS:Dane, SS:Stosik
Start:
mov ax, SEG Dane ; For EXE we need to set DS
mov ds, ax ; To Dane segment manually
mov ax, a ; Multiplying a by 3 is the same
; as multiplying a by 2 and adding a
shl ax, 1 ; Multiply a*2
add ax, a ; Add a to previous result in a
mov cx, ax ; Copy result of a*3 to CX
mov ax, b ; Do div b/a
xor dx, dx ; We need to ensure DX is zerofor this div
; as Div is result of DX:AX by a
div a
sub cx, ax ; Subtract reslt of b/a from result of a*3
mov ax, d ; ax = d + 3
add ax, c
mul cx ; Multiple d+3 (AX) by a*3-b/a (cx)
mov Wynik, ax ; Save 16-bit result in memory
mov ax, 4C05h ; Exit with value 5
int 21h
Kod ENDS
Stosik SEGMENT STACK
DB 100h DUP (?)
Stosik ENDS
END
The program keeps with the spirit of the original fixing the syntax and logic errors. b/a is still using integer division (you will have to ask your TA or professor about that) which will round result down to nearest whole number (in case of 10/20 that is 0). Main problems in this code are:
- Some of the code was placed out of order
- Your
divis the division of DX:AX by a 16-bit value so we need to zero DX. - In some places the register names were altered.
- In this code 3*a is being represented as a*2+a=3a. Multiplying by 2 is the same as shifting the value left by 1.
If the professor requires a better approximation to the result by still using integer division then Jester's suggestion of rearranging the equation to be 3*a*(d+3)-(b*(d+3))/a is a good one. This defers the division to a point where the rounding down of integer division has less effect on the result, so the final result should only be off by almost 1. Code that uses this revised equation would look like:
mov ax, SEG Dane ; For EXE we need to set DS
mov ds, ax ; To Dane segment manually
mov cx, a
shl cx, 1
add cx, a ; cx = 2*a+a = a*3
mov ax, d
add ax, c ; ax = d+c = d+3
mov bx, ax ; bx = copy of d+3
mul cx
mov si, ax ; si = a*3*(d+3)
mov ax, bx
mul b ; ax = b*(d+3)
xor dx, dx ; Avoid division overflow, set DX=0
div a ; ax = b*(d+3)/a
sub si, ax ; si = a*3*(d+3) - b*(d+3)/a
mov Wynik, si ; Save 16-bit result in memory
A slight improvement can be made with this variation. When integer division produces a result it's rounded down to the nearest whole number. If you divide 99/100 you will get 0 with div and a remainder of 99. The answer is much closer to 1 than 0. Usually you round up when something is >= .5 and round down < .5 . It is possible to use the remainder (DX) from div to adjust the final result up by 1 if need be or to keep the result as is. The amended code could look like:
mov ax, SEG Dane ; For EXE we need to set DS
mov ds, ax ; To Dane segment manually
mov cx, a
shl cx, 1
add cx, a ; cx = a*3
mov ax, d
add ax, c ; ax = d+c = d+3
mov bx, ax ; bx = copy of d+3
mul cx
mov si, ax ; si = a*3*(d+3)
mov ax, bx
mul b ; ax = b*(d+3)
xor dx, dx ; Avoid division overflow, set DX=0
div a ; ax = b*(d+3)/a
shl dx, 1 ; Remainder(DX) = Remainder(DX) * 2
cmp dx, a ; Ajustment of whole nuber needed?
jb .noadjust ; No? Then skip adjust
add ax, 1 ; Else we add 1 to quotient
.noadjust:
sub si, ax ; si = a*3*(d+3) - b*(d+3)/a
mov Wynik, si ; Save 16-bit result in memory
mov ax, 4C05h ; Exit with value 5
int 21h
The adjustment is based on the method in Rounding Half Up. Essentially if the remainder (DX) times 2 is less than the divisor a then no adjustment is needed, otherwise the quotient (AX) needs to be increased by 1
The results of the first version would by 480. The result of the second is 476. The second will be closer to the expected value. In this case the result of 476 happens to be exact. (3*20-10/20)*(5+3) = 59.5*8 = 476.
来源：https://stackoverflow.com/questions/49267371/assembly-equation-divide-to-get-float-value