in 8086 Identify the addressing modes used for the source and destination operands, and then compute the physical address for the specified operand in each of the following instructions: (Assume all missing data.) 1. MOV CX, [DI] (Source operand) 2. MOV [BP]+BETA, AX (Destination operand) 3. XCHG BX, [FF02] (Source operand)

in 8086 Identify the addressing modes used for the source and destination operands, and then compute the physical address for the specified operand in each of the following instructions: (Assume all missing data.) 1. MOV CX, [DI] (Source operand) 2. MOV [BP]+BETA, AX (Destination operand) 3. XCHG BX, [FF02] (Source operand)

Systems Architecture

7th Edition

ISBN:9781305080195

Author:Stephen D. Burd

Publisher:Stephen D. Burd

Chapter6: System Integration And Performance

Section: Chapter Questions

Problem 19VE

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Identify the addressing modes used for the source and destination operands, and then compute the physical address for the specified operand in each of the following instructions: a) MOV [DI]+XYZ,AH (Destination operand)b) MOV [BX][DI]+XYZ,AL (Destination operand)c) MOV DI,[SI] (Source operand)d) MOV CL,XYZ[BX+SI] (Source operand)e) MOV DX, [BP] (Source operand)f) MOV AL, XYZ[BP+DI] (Source operand)g) MOV AL, [BP+SI-20016] (Source operand)
in 80886 microprocessor Suppose that. DS = 0200H, BX = 0300H, and DI-400H Determine the memory address accessed by each of the following instructions, assuming real mode operation: (1) MOV AL,[1234H] (2) MOV AX,[BX] (3) MOV [DI]AL
liomework: Identify the addressing modes used for the source and destination operands, and then compute the physical address for the specified orx•rand in each of the following instructions: a) Mov +XYZ,AH (Destination operand) b) Mov tBX) (Destination operand) Suppose that BX=030016, 010016.
Suppose we have the instruction Load 0000. Given memory and register R1 contain thevalues below:R130Memory Address Content0000 40...0010 30...0020 78...0030 55...0040 77...0050 84 Assuming R1 is implied in the indexed addressing mode, determine the actual value loaded into the accumulator using the following addressing modes: a. Immediateb. Directc. Indirectd. Indexed
Will upvote! Suppose that ES= 6DF2, DS=83AC, SS=EBD2, AX=B75, DI=DC7, BP=51A, SI=FB3, REN=75D and YEN=A8E. Determine the address accessed by each of the following instructions and state what addressing mode is used:   a. MOV REN[DI][AX][9F], BX b. MOV BL, YEN[BP+SI-72A]
A) Identify the addressing modes used for the source and destination operands and determine the physical address accessed by each of the following instructions (Assume all missing data.) (Choose Two Only): 1. MOV CX, [DI] (Source operand) 2. MOV [BP]+BETA, AX (Destination operand) 3. XCHG BX, [FF02] (Source operand)
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. directB - physical address/es e.g. 19000-19001 C - content of the destination after executione.g. if register: AX=1234 if memory (lower address first) : 12000=34;12001=12 GIVENDS = ACF7HSS = BAC9HDI = ECABHSI = ABAEHBP = BAD2HAX = 6FCBHBX = 7BCFHCX = 5CADHDX =D1BCHSP = 4FABH
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. directB - physical address/es e.g. 19000-19001 C - content of the destination after executione.g. if register: AX=1234 if memory (lower address first) : 12000=34;12001=12 DS = 8EFDHSS = AFDEHDI = B8F9HSI = D3CDHBP = A5D8HAX = 5FFEHBX = B3DFHCX = CADEHDX = B2BCHSP = A4FBH
Will upvote! Find the memory address of the next instruction executed by the microprocessor, when operated in the real mode, for the following CS:IP and 80286 register combinations: a. DS=2F2E & DX=9D64 b. CS=9F7A & IP=AB27 c. ES=DE21 & DI=D75F d. SS=FF5C & BP=92B8 e. DS=DC67 & CX=2FE8
The x86 CMP (compare) instruction updates the flags (C, P, A, Z, S, O) in the status register by subtracting the source operand from the destination operand without saving the result. Note 1: In case of x86 processors, C flag is interpreted as borrow for subtraction. Note 2: In two’s compliment arithmetic, C flag is not used. Indicate how to determine the followings: 2.1 destination < source (two operands are unsigned) 2.2 destination > source (two operands are signed) 2.3 destination = source (two operands are unsigned)
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 DS = 9CEFH SS = CB8DH DI = BACEH SI = 9EBCH BP = BAD2H AX = 6FCBH BX = 7BCFH CX = 5CADH DX = D1BCH SP = 4FABH MOV CX, [BP+DI+ACEH} Find the addressing mode: Physical address/es: Content of the destination:
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