Question

Given an array of words aw, an array of double words adw and a byte variable called vb, write the code that does the following: adw[5]=aw[3]+8-vb. "Assembly Language"(x86).

Answer 1

Answer:

DATA SEGMENT

vb DB 5 ; declare a byte variable 'vb'

aw DW 6, 7, 8, 5, 22, 12, 11. ;. declare a word array

adw DD. 12, 45, 56, 12, 5, 34, 65. ; declare a double word array

DATA ENDS

CODE SEGMENT

ASSUME. CS: CODE, DS: DATA

BEGIN:

MOV AX, DATA

MOV DS , AX

MOV AX, aw[3]

MOV BX, 8

MOV BH, 0. ; Fill the upper byte of BX with zeros

ADD AX, BX . ; AX contains aw[3] + 8

MOV BX, vb

MOV BH, 0. ; Fill the upper byte of BX with zeros

SUB AX, BX . ; AX now contains aw[3] + 8 - vb

MOV adw[5], AX . ; adw[5] now contains aw[3] + 8 - vb

CODE ENDS

END BEGIN

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