Lab 02: Arithmetic Operations
OBJECTIVE
To learn the basic arithmetic commands and their use.
Arithmetic Operations
1. ADD: A source operand is added to a destination operand and the sum is stored in the destination. Operands must be the same size.
Syntax: Add destination, source
Instruction Formats:
Add reg,reg Add reg,operand
Add mem,reg Add mem,operand
Add reg,mem
Example: Add al,20h ; al = al + 20h
Add al,bl ; al = al + bl
Add eax,[ebx] ; Add the 4 bytes in memory at the address contained in EBX into EAX
2. Sub: Subtracts the source operand from the destination operand.
Syntax: Sub destination, source
Instruction Formats:
Sub reg,reg Sub reg,operand
Sub mem,reg Sub mem,operand
Sub reg,mem
Example: Sub al,20h ; al = al – 20h
Sub al,bl ; al = al – bl
Sub eax,[ebx] ; Sub the 4 bytes in memory at the address contained in EBX into EAX
3. Inc: Increments or adds 1 to a register or memory operand.
Syntax: inc destination
Instruction Formats:
inc reg inc mem
Example: inc al ; al = al + 1
4. Dec: Decrements or subtracts 1 from an operand. Does not affect carry.
Syntax: dec destination
Instruction Formats:
dec reg dec mem
Example: dec al ; al = al – 1
5. Neg: It reverses the sign of a number by converting the number to its two’s complement and stores the result in the destination.
Syntax: Neg destination
Instruction Formats:
neg reg neg mem
Example: neg al ; al = -5 if it initially contined 5
Code: (Add, Sub, Inc, Dec)
.model small
.stack 100h
.data
.code
main proc
mov al,5
add al,2
;inc al
;sub al,1
;dec al
mov ah,2
mov dl,al
int 21h
main endp
end main
5. Mul: Multiplies AL or AX by a source operand. If the source is 8 bits, it is multiplied by AL and the product (16 bit) is stored in AX. If the source is 16 bits, it is multiplied by AX and the product (32 bit) is stored in DX:AX.
Syntax: Mul source ; the destination is Ax, where AL is mul by source.
Instruction Formats:
mul reg mul mem
Example: mov al,2
mov bl,4
mul bl ; Ax = al*bl = 8
6. Div: Performs either 8 or 16-bit unsigned integer division. If the divisor is 8 bits, the dividend is in AX, the quotient is AL, and the remainder is AH i.e. (AL = AX/operand & AH = remainder (modulus).
If the divisor is 16 bits, the dividend is DX:AX, the quotient is AX, and the remainder is DX.
Syntax: Div source
Instruction Formats:
|
div reg div mem
Example: mov ax,9
mov bl,2
div bl ; AL = 4, AH =1
Code for Division:
.model small
.stack 100h
.data
.code
main proc
mov ax,100h
mov bl,8h
div bl
;we want to display reminder, so divide quotient by 10
mov ch,ah ; remainder moved to ch to display later
mov bh,10h ; to disply values in seperate digit we divide by 10
div bh ; e.g. 20/10 now al=2
mov cl,ah ; move remainder from ah to cl to avoid loss of contents
mov ah,2
mov dl,al ; display quotient 20/10 = 2 will display
add dl,30h ; ascii conversion
int 21h
mov ah,2
mov dl,cl ; now display the remainder contained in cl=2
add dl,30h ; ascii conversion
int 21h
main endp
end main
Lab Code 01: Perform the following arithmetic expression
Result = -5 + (8 - 2)
Hint: Use neg, sub and add commands.
.model small
.stack 100h
.data
.code
main proc
mov al,5
neg al ; al = -5
mov bl,8
sub bl,2 ; (8-2)=6
add al,bl ; -5 + 6 =1
mov ah,2
mov dl,al
int 21h
main endp
end main
Lab Code 02: To input a character at run time, convert its case (from lower to upper) and display it.
Hint: ASCII codes for ‘upper case’ and ‘lower case’ characters differ by 32d or 20h. (a=61h & A=41h).
.model small
.stack 100h
.data
.code
main proc
mov ah,1
int 21h
sub al,20h
mov ah,2
;mov dl,0dh
;int 21h
;mov dl,0ah
;int 21h
mov dl,al
int 21h
mov ah,4ch
int 21h
main endp
end main
Lab Task: To input a character at run time, convert its case (from upper to lower) and display it.
Hint: Now use add command to convert the case.
