ASSIGNMENT NO 2

section .data
welmsg db 10,"Welcome to count positive and negative numbers in an array",10
welmsg_len equ
−
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−welmsgpmsgdb10,"Countof+venumbers::"pmsg
l
​
enequ-pmsg
nmsg db 10,"Count of -ve numbers::"
nmsg_len equ $-nmsg
nwline db 10
array dw 8505h,90ffh,87h,88h,8a9fh,0adh,02h,8507h
arrcnt equ 8
pcnt db 0
ncnt db 0
section .bss
dispbuff resb 2
%macro print 2 ;defining print function
mov eax, 4 ; this 4 commands signifies the print sequence
mov ebx, 1
mov ecx, %1 ; first parameter
mov edx, %2 ;second parameter
int 80h ;interrupt command
%endmacro
section .text ;code segment
global _start ;must be declared for linker
_start: ;tells linker the entry point ;i.e start of code
print welmsg,welmsg_len ;print title
mov esi,array
mov ecx,arrcnt ;store array count in extended counter reg

up1: ;label
bt word[esi],15
;bit test the array number (15 th byte) pointed by esi.
;It sets the carray flag as the bit tested

jnc pnxt ;jump if no carry to label pskip
inc byte[ncnt] ;if the 15 th bit is 1 it signifies it is a ;negative no and so we ;use this
command to increment ncnt counter.
jmp pskip ;unconditional jump to label skip
pnxt: inc byte[pcnt] ;label pnxt if there no carry then it is ;positive no
;and so pcnt is incremented
pskip: inc esi ;increment the source index but this ;instruction only increments it by
8 bit but the no’s in array ;are 16 bit word and hence it needs to be incremented twice.
inc esi
loop up1 ;loop it ends as soon as the array end “count” or
;ecx=0 loop automatically assums ecx has the counter
print pmsg,pmsg_len ;prints pmsg
mov bl,[pcnt] ;move the positive no count to lower 8 bit of B reg
call disp8num ;call disp8num subroutine
print nmsg,nmsg_len ;prints nmsg
mov bl,[ncnt] ;move the negative no count to lower 8 bits of b reg
call disp8num ;call disp8num subroutine

print nwline,1 ;New line char
exit:
mov eax,01
mov ebx,0
int 80h
disp8num:
mov ecx,2 ;move 2 in ecx ;Number digits to display
mov edi,dispbuff ;Temp buffer
dup1: ;this command sequence which converts hex to bcd
rol bl,4 ;Rotate number from bl to get MS digit to LS digit
mov al,bl ;Move bl i.e. rotated number to AL
and al,0fh ;Mask upper digit (logical AND the contents ;of lower8 bits of
accumulator with 0fh )
cmp al,09 ;Compare al with 9

jbe dskip ;If number below or equal to 9 go to add only 30h
;add al,07h ;Else first add 07h to accumulator
dskip:
add al,30h ;Add 30h to accumulator

mov [edi],al ;Store ASCII code in temp buff (move contents ;of accumulator to the
location pointed by edi)
inc edi
;Increment destination index i.e. pointer to ;next location in temp buff
loop dup1 ;repeat till ecx becomes zero
print dispbuff,2 ;display the value from temp buff
ret ;return to calling program