B2.JPG MOV SP,[BP+DI+ACBAH] ВJPG PHY. ADD| С6079H АСH C607AH BDH С607вн | СЕН C607CH| F1H C607DH 02H C607EH 13H С607FH 24H PHY. ADD PHY. ADD PHY. ADD PHY. ADD F6960H FEH F695FH| DCH F695EH | ВАН F695DH 98H DBD4EH 13Н DBD4DH 24H DBD4CH 35H DBD4BH 46H E47A7H ACH E47A6H BDH E47A5H CEH E47A4H F1H Е47АЗН 02H E47A2H 13H E47A1H 24H F5C8FH 11H F5C90H 84H F5C91H 52H F5C92H A1H F695CH 76H F695BH 54H F695AH 32H DBD4AH 57H DBD49H 68H DBD48H 79H F5C93H | 27H F5C94H 05H F5C95H 87H ADDRESSING MODE: Blank 1 PHYSICAL ADDRESS/ES: Blank 2 CONTENT OF THE DESTINATION: Blank 3 Blank 1 Add your answer Blank 2 Add your answer Blank 3 Add your answer
B2.JPG MOV SP,[BP+DI+ACBAH] ВJPG PHY. ADD| С6079H АСH C607AH BDH С607вн | СЕН C607CH| F1H C607DH 02H C607EH 13H С607FH 24H PHY. ADD PHY. ADD PHY. ADD PHY. ADD F6960H FEH F695FH| DCH F695EH | ВАН F695DH 98H DBD4EH 13Н DBD4DH 24H DBD4CH 35H DBD4BH 46H E47A7H ACH E47A6H BDH E47A5H CEH E47A4H F1H Е47АЗН 02H E47A2H 13H E47A1H 24H F5C8FH 11H F5C90H 84H F5C91H 52H F5C92H A1H F695CH 76H F695BH 54H F695AH 32H DBD4AH 57H DBD49H 68H DBD48H 79H F5C93H | 27H F5C94H 05H F5C95H 87H ADDRESSING MODE: Blank 1 PHYSICAL ADDRESS/ES: Blank 2 CONTENT OF THE DESTINATION: Blank 3 Blank 1 Add your answer Blank 2 Add your answer Blank 3 Add your answer
Database System Concepts
7th Edition
ISBN:9780078022159
Author:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan
Publisher:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan
Chapter1: Introduction
Section: Chapter Questions
Problem 1PE
Related questions
Question
Determine the specific type of addressing mode (SMALL LETTERS only) and compute for the address/es. If applicable, determine the content of the destination after the execution of the instruction. Otherwise, NA. For physical address and content of the destination, use CAPITAL LETTERS,NO SPACE/S in between and NO need to include "H" or the unit.
A - is for addressing mode type, e.g. direct
B - physical address/es e.g. 19000-19001
C - content of the destination after execution
e.g. if register: AX=1234
if memory (lower address first): 12000=34,12001=12
GIVEN
AX =AD6EH
DX = 9ECFH
SS = BEDAH
SI = AECDH
BX =AFECH
CS = EBDCH
ES = CFEDH
BP = EFBAH
CX =FADEH
DS = AF0EH
DI = BD8FH
IP = AB8FH
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