Compute Pi using Assembly Language 8086

Monday, November 7, 2011 Posted by Glenn Von Posadas

Source Codes:

compiled using any 8086 Assembler/Emulator
.radix 10
.model tiny

; This program uses almost all of the 64K code segment. There might
; be rare situations where the last 4 digits of the printout are
; wrong, but this is less likely than in the previous version.
; The last 4 digits should be 1224. (9,304 digits of pi)

; Program assumes there is 64K of memory available for the program,
; without checking. In the rare case where there isn't 64K
; available, it may crash the computer.
; i.e. don't LOADHIGH this program.

org 100h
mov cx,2326*14 ;32564 ; this MUST be a multiple of 14
; might be possible to increase
; to 2327*14, but that last 4 digits
; is much more likely to be corrupted
; and/or corrupt the rest of the array
; since that ends the scratch area
; at FFF1h.
mov di,Offset Buffer
mov bx,cx
mov ax,2000 ; initial value of the array
mov bp,10000 ; keeping 10000 constant in extra register
rep stosw ; wipes the buffer used.
; offset 0172h through offset FFD9

; FFDA-FFFE is stackspace for 18 words.
; the workspace shrinks by an additional 14 words
; before calling an INT. This program is smaller
; than previous version by 12 bytes, meaning that
; stack is larger by 6 words, making it less likely
; that the stack will overrun the workspace.

; cx = 0000, which is the original value of the remainder.

; bx is decremented from 32564 by 14's. Lesser value would
; not have correct values toward the low end of the output.
; outer-counter is "i", inner-counter is "j"
; because counter is decremented at the end, array is accessed
; by subtracting 2 from the counter.

xor ax,ax
push bx
cwd ; dx:ax initialized as 0000:0000

push bx
mov si,dx ; save dx thru the multiply
mul bx ; bx * ax -> dx:ax
xchg si,ax ; get the original dx to multiply it
push dx
mul bx
pop dx
add ax,dx
xchg di,ax ; dx:ax * bx -> di:si

mov ax,bp ; ax = 10,000
add bx,bx ; align for 2-byte array elements
mul word ptr [bx+offset Buffer-2]
; because bx is decremented at the end,
; must subract 2 to point at correct item
add ax,si
adc dx,di ; dx:ax = dx:ax + di:si
dec bx ; bx = 2*j -1
xor di,di
xchg di,ax
xchg dx,ax
div bx ; dx = 0, ax = high part of prev di:si
; dx:ax / bx -> ax, dx=remainder
xchg di,ax
mov si,dx
xor dx,dx
div bx ; dx = 0, ax = low part of prev di:si
xchg si,ax
xchg dx,ax
div bx
mov [bx+offset buffer-1],dx ; because bx was decremented
; above, just subtract 1 to
; access correct element.
xor dx,dx
add ax,si ; si:di + ax -> dx:ax
adc dx,di
pop bx
dec bx
jnz InnerLoop

div bp ; div by 10,000 - resulting mod has 4 decimal digits

push dx ; save remainder for use in next loop.
add ax,cx
mov bl,10 ; dividing by 10 yields decimal digits.
mov cx,4

cwd ; "xor dx,dx" not needed, value is always < 7fff
div bx ; divide by 10
push dx
loop Num1

mov cl,4

; pop dx
; add dl,"0"
; mov ah,2
; int 21H

pop ax ; these 3 lines save 3 bytes compared to
add al,"0" ; the previous 4 lines, but the output
int 29H ; can't be redirected to a file

loop Num2

pop cx ; restore remainder pushed in DX above
pop bx ; restore outer-loop's value
sub bx,+14 ; decrement by 14.

; jnz OuterLoop
jnbe OuterLoop ; changed from JNZ in case someone
; didn't make BX a multiple of 14.

ret ; .COM always has 0000 on the stack.
; instruction at offset 0000 is always
; an INT 20H, which ends the program.

Buffer=$ ;+1 ;align even without adding a byte

end start
; This is offset 016Eh. Next 32564 words through FFD5h
; are used as workspace.

glenn von

Author: Glenn Posadas

20 year old 5th year Computer Engineering Student, a blogger, photographer, programmer, and an electronics hobbyist; a Christian who loves God very much..

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