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What does bx stand for in bx cable?

Bronx


Write a program to subtract two 16 bit numbers in microprocessor 8086?

.code main proc mov ax,@data mov ds,ax lea dx,msg ;printing msg mov ah,09h int 21h mov ax,x ;ax=x mov bx,y ;bx=y cmp ax,0 ;jump to l3 if ax is negtive jb l3 cmp bx,0 ;jump to l6 if bx is negative jb l6 cmp ax,bx ;if ax<bx,then jump to l1 jl l1 sub ax,bx ;else normal sub mov diff,ax ;diff=result is stored jmp l2 l1: ;iff (+)ax<(+)bx neg bx ;bx=-bx clc add ax,bx neg ax ;-ans=ans mov diff,ax mov dx,2dh ;print '-' mov ah,02h int 21h jmp l2 l3: ;iff (-)ax neg ax ;-ax=ax cmp bx,0 ;jump to l4 if bx is negative jb l4 clc add ax,bx ;ax=(+)ax+(+)bx mov ax,diff mov dx,2dh ;print '-' mov ah,02h int 21h jmp l2 l4: ;if (-)ax & (-)bx neg bx ;-bx=bx cmp ax,bx ;if ax>bx then jump to l5 jg l5 sub ax,bx ;else ax-bx mov diff,ax mov dx,2dh ;print '-' mov ah,02h int 21h jmp l3 l5: ;if(-)ax>(-)bx xchg ax,bx ;exchange ax and bx sub ax,bx ;ax-bx mov diff,ax ;ans is positive jmp l2 l6: ;iff (-)bx neg bx ;-bx=bx add ax,bx ;ax-(-)bx mov diff,ax ;ans will be positive mov ah,4ch int 21h main endp


Program to subtract two 8 bit numbers using 8086 microprocessor?

I have a code for 16 bit subtraction.. just replace ax by al,bx by bl etc... .code main proc mov ax,@data mov ds,ax lea dx,msg ;printing msg mov ah,09h int 21h mov ax,x ;ax=x(any number) mov bx,y ;bx=y( " ") cmp ax,0 ;jump to l3 if ax is negtive jb l3 cmp bx,0 ;jump to l6 if bx is negative jb l6 cmp ax,bx ;if ax<bx,then jump to l1 jl l1 sub ax,bx ;else normal sub mov diff,ax ;diff=result is stored jmp l2 l1: ;iff (+)ax<(+)bx neg bx ;bx=-bx clc add ax,bx neg ax ;-ans=ans mov diff,ax mov dx,2dh ;print '-' mov ah,02h int 21h jmp l2 l3: ;iff (-)ax neg ax ;-ax=ax cmp bx,0 ;jump to l4 if bx is negative jb l4 clc add ax,bx ;ax=(+)ax+(+)bx mov ax,diff mov dx,2dh ;print '-' mov ah,02h int 21h jmp l2 l4: ;if (-)ax & (-)bx neg bx ;-bx=bx cmp ax,bx ;if ax>bx then jump to l5 jg l5 sub ax,bx ;else ax-bx mov diff,ax mov dx,2dh ;print '-' mov ah,02h int 21h jmp l3 l5: ;if(-)ax>(-)bx xchg ax,bx ;exchange ax and bx sub ax,bx ;ax-bx mov diff,ax ;ans is positive jmp l2 l6: ;iff (-)bx neg bx ;-bx=bx add ax,bx ;ax-(-)bx mov diff,ax ;ans will be positive mov ah,4ch int 21h main endp


When was Toyota BX created?

Toyota BX was created in 1951.


When was Citroën BX created?

Citroën BX was created in 1982.


When was BX Warrior created?

BX Warrior was created in 2006.


Where did BX get its name?

The term BX is a trademark owned by GE. The term "BX" is a short form for Bronx because GE had a facility to make the cable.


Was silver ever used in bx wire?

No, silver was never used in BX cable.


How do you install games on china mobile A300 BX?

how to install china mobile BX a300


What is Ax plus bx plus c?

Ax + Bx + C is called an algebraic expression.


Solve bx plus cy equals d for x?

bx+cy=d bx=d-cy x=d-cy ----- b answer: d-cy ----- b


How do you write a code to complement the corresponding bit in bx if AX contains 6complement the 6th bit of BX in assembly?

To complement the 6th bit of the BX register in assembly language, you can use the XOR instruction. The 6th bit corresponds to the bit mask 0x20 (binary 0010 0000). The code would look like this: MOV AX, 6 ; Load AX with 6 (not directly relevant to complementing BX) XOR BX, 0x20 ; Complement the 6th bit of BX This will toggle the 6th bit of BX, effectively complementing it.