offset storage for forming physical address
computer has different registers each of which has different functions. ax - accumulator register bx - base register cx - counter register computer has different registers each of which has different functions. ax - accumulator register bx - base register cx - counter register
EBX --> Extended Base Register. Extended means, it stores more data than 'Base Register' OR BX. BX is a 16 bit register, where EBX is 32 bit. Base register stores the base address, ie the starting address with respect to a segment.
•The mov instruction carries out assignment in 8086 assembly language. •It which allows us place a number in a register or in a memory location (a variable) i.e. it assigns a value to a register or variable. Example: Store the ASCII code for the letter A in register bx.•A has ASCII code 65D (01000001B, 41H) •The following mov instruction carries out the task:•mov bx, 65d•We could also write it as:•mov bx, 41h•or mov bx, 01000001b•or mov bx, 'A'
The BX register is a general purpose register. It is a 16 bit register that represents the contatenation of the two 8 bit registers BH (B High) and BL (B Low)
8086 has four multipurpose registers. 1. AX (Accumulator Register) 2. BX (Base Register) 3. CX (Count Register) 4. DX (Data Register) By Aneeta Arshad
used as implicit operand
asymptote
x-axis
Bronx
You just have to follow the rule of quadratic functions. Example y = mx+b is the rule for linear functions. ax^2+bx+c is the rule of quadratic equation.
x axis
.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