CS 206D Computer OrganizationLab8

CS 111

CS 111

Exercise 1

- Write an assembly program that ask the user to enter a positive hex (2 digits) number and multiply the input by 2. Then print the result in Binary form.

.MODEL SMALL .STACK 100H

.DATA MSG1 DB 'ENTER A POSITIVE HEX NUMBER (2 digit) : $' MSG2 DB 0AH,0DH,'THE RESULT IN BINARY: ','$‘

.CODE MAIN PROC MOV AX, @DATA ; initialize DS MOV DS, AX LEA DX, MSG1 ; load and display the string MOV AH, 9 INT 21H

XOR BX,BX ; BX WILL HOLD THE INPUT MOV AH,1 INT 21H WHILE: CMP AL, 0DH ; CR? JE ENDWHILE CMP AL,'9' ; AL>9? JG LETTER ; Al is letter ;ELSe AL is digit AND AL , 0FH JMP SHIFT LETTER: SUB AL , 37H SHIFT: SHL BX,4 OR BL,AL ;Insert value to bx INT 21H ; Read again JMP WHILE

ENDWHILE: LEA DX, MSG2 MOV AH, 9 INT 21H MOV AL,BL MOV CL , 2 ; Multiply by 2 MUL CL ; The result will be stored in AX MOV BX,AX ;Print ax in binary form MOV CX,16 ; AX is 16 bits MOV AH,2 PRINT: ROL BX,1 JC ONE ;IF (CF == 1) ;Else cf = 0 MOV DL, 30H ; To print 0 JMP DISPLAY ONE: MOV DL, 31H ;To print 1 DISPLAY: INT 21H LOOP PRINT

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CS 111

Addressing mode

CS 111

Register Indirect Mode

• The offset address of the operand is contained in a register. I.e. The register acts as a pointer to the memory location.

• The register must be one of the following: BX, SI, DI. BP.• Format: [register]

Ex. suppose that SI contains 0110h, and the word at 0110h contains 1884h.•MOV AX, [SI]•MOV AX, SI

CS 111

Based and Indexed Addressing Modes

• The operand’s offset address is obtained by adding a number called a displacement to the contents of a register.

• Syntax:[register + displacement][displacement + register][register] + displacementdisplacement + [register]displacement [register]

CS 111

Based and Indexed Addressing Modes

Ex. Suppose that ALPHA is declared as:

ALPHA DW 0123H, 0116H, 0789H, 0ABCDH

In segment address by DS. Suppose also that: • BX contains 2 offset 0002 contains 1022 h• SI contains 4 offset 0004contains 2DDCh• DI contains 1

instruction NUMBER MOVED

MOV AX,[ALPHA+BX]MOV BX,[BX+2]MOV CX,ALPHA[SI]MOV AX,-2[SI]MOV BX,[ALPHA+3+DI]

MOV AX,[BX]2

• What will be the result of following instructions:

CS 111

Based and Indexed Addressing Modes

instruction NUMBER MOVED

MOV AX,[ALPHA+BX] 0116H

MOV BX,[BX+2] 2DDCh

MOV CX,ALPHA[SI] 0789H

MOV AX,-2[SI] 1022H

MOV BX,[ALPHA+3+DI] 0789H

MOV AX,[BX]2 illegal

CS 111

Homework 6

Question 1: Write a program that Sum the values of even index of array A. Suppose that Array A contains:

CS 111

Homework 6.DATA MSG DB 'THE SUM OF ARRAY ELEMENTS: $' A DB 2,3,5,8,1,4.CODEMOV AX, @DATA ; initialize DSMOV DS, AXXOR SI,SI ;clear SIXOR AX,AX ;AX holds sumMOV CX,3 ; no. of indexes in ATOP:ADD AL,A[SI] ;Sum= Sum + elementADD SI,2; to go the next indexLOOP TOP ;loop until doneMOV AH,9; display the string MSGLEA DX, MSGINT 21hMOV DL,ALMOV AH,2INT 21H