1. Given the x86’s register and memory contents below: Perform the following ope
ID: 2072473 • Letter: 1
Question
1. Given the x86’s register and memory contents below: Perform the following operations below independently.
EAX 300000020H DS:230H 1008H DS: 232H 30000H EBX 00000030H DS: 250H 30002H ECX 00000040H DS:252H 0004H EDX 00000050H DS:300H 1111H ESI 00000200H DS:302H 30000H EDI 300000300H DS:310H 36740H DS:312H 30000H DS:200H 1020H DS:330H 0008H DS:202H 0000H DS:332H 4230H DS:220H ECFFH DS:122H DAFFH Perform the following operations below independently. a. ADD ECX, AX b. ADD AX, 01 FEH c. ADC DI, BX d. INC WORD PTR [0122H] e. SUB CL, AL SBB CL, [0250H] g. MUL BX h. NEG WORD PTR [0332 i. DEC WORD PTR [SI+BXI j. DIV WORD PTR [0330H1Explanation / Answer
Given the following register & memory contents of 8086:
EAX =00000020H DS: 230H = 1008H
EBX = 00000030H DS: 232H = 0000H
ECX = 00000040H DS: 250H = 0002H
EDX = 00000050H DS: 252H = 0004H
ESI = 00000200H DS: 300H = 1111H
EDI = 00000300H DS: 302H = 0000H
CF = 1 DS: 310H = 6740H
DS: 200H = 1020H DS: 312H = 0000H
DS: 202H = 0000H DS: 330H = 0008H
DS: 220H = ECFFH DS: 332H = 4230H
DS: 122H = DAFFH
(AX)<= (AX) + 01FEH. 00000020+000001FE = 0000021EH.
so the contents of (AX) = 0000021EH
(DI)= (DI) + (BX) + 1. 00000300 + 00000030 + 00000001=00000331H
so the contents of (DI) = 00000331H
(CL)=(CL)-(AL), 0040 – 0020 =0020H. SO the content of (CL) = 0020H
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