Given the following memory addresses and opcodes to be executed, if the current value of the Instruction Register (IR) is EBOA, what is the value of the Instruction Pointer Register (RIP) after the instruction in IR is executed? 00000000 EBOA 00000002 B805000000 00000007 BF01000000 0000000C BAODO00000
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- Compute the physical address for the specified operand in each of the following instructions from previousproblem. The register contents and variables are as follows : (CS) = 0A0016, (DS) = 0B0016, (SI) =010016, (DI) = 020016 and (BX) = 030016.a) Destination operand of the instruction in (c)b) Source operand of the instruction in (d)c) Destination operand of the instruction in (e)d) Destination operand of the instruction in (f)e) Destination operand of the instruction in (g)In a computer instruction format, the instruction length is 16 bits, and the size of an address field is 4 Is it possible to have: 15 instructions with 3 addresses, 14 instructions with 2 addresses, 31 instructions with one address, and 16 instructions with zero addresses, using this format? Justify your answer.1. T/F - if (B)=006000 (PC)=003600 (X)=000090, for the machine instruction 0x032026, the target address is 003000.2. T/F – PC register stores the return address for subroutine jump.3. T/F – S register contains a variety of information such as condition code.4. T/F – INPUT WORD 1034 – This means Operating system should reserve 1034 bytes in memory5. T/F - In a two pass assembler, adding literals to literal table and address resolution of local symbol are done using first pass and second pass respectively.
- Given following code: 00 9000 01 1150 02 1251 03 5300 04 6420 05 7541 06 8414 07 D20A 08 2523 09 0000 0A 3331 0B B200 0C D20A 0D 2523 0E 0000 50 000A 51 0005 52 0000 Explain what each instruction does in the program, stored in the memory locatons 00 through 0E. When the program stops, what are values in registers R0, R1, R2, R3, R4, R5, and in memory location 52? What does the program do?Assume that the state of the 8088’s registers and memory just prior to the executionof each instruction in problem 15 is as follows: * in photos*What result is produced in the destination operand by executing instructions (a)through (k)? *only h through k* (h) MUL DX(i) IMUL BYTE PTR [BX+SI](j) DIV BYTE PTR [SI]+0030H(k) IDIV BYTE PTR [BX][SI]+0030HWhat are the values of ECX, EBP, and ESP registers after executing the following instruction? push ECX Before ECX: 0xBABE EBP: 0x0012FF24 ESP: 0x0012FF24 After ECX: EBP: ESP:
- Assume that the state of the 8088’s registers and memory just prior to the executionof each instruction in problem 15 is as follows: * in photos*What result is produced in the destination operand by executing instructions (a)through (k)? *only b,c,e,g* b) ADC SI, AX(c) INC BYTE PTR [0100H] (e) SBB DL, [0200H] (g) NEG BYTE PTR [DI]+0010HIf R0 = 0x20008000, after STMDA r0!, {r3, r9, r7, r1, r2} instruction is executed, register r7 will be stored in memory starting from which memory base address. A. R0 = 0x20007ff0 B. R0=0x20007fec C. R0 = 0x20007fff D. R0= 0x20007ff4 E. R0 = 0x20007ffefWill 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
- 333: E8B3 8B3: 0A12 A12: 2153 Since the contents of the relevant addresses are 333 in the basic computer given above, the initial value of the PC is 333; When fetching and executing an indirectly addressing ISZ instruction, write the contents of the following registers in hexadecimal (hex) format for the T5 time. AR ........... DR......... IR ........... PC ........... SC ...........Question 30 Assume that EBX and ECX have the following values EBX: FF FF FF 75 ECX: 00 00 01 A2 After the execution of the instruction INC EBX, ECX The Value in EBX is _________ Group of answer choices FFFFFF76 FFFFFF75 000001A2 none of them Question 31 The Hexadecimal Representation for each of the following Binary number 11101111 Group of answer choices 239 EF FE none of them Question 32 Assume that EBX and ECX have the following values EBX: FF FF FF 75 ECX: 00 00 01 A2 After the execution of the instruction MOV EBX, ECX The Value in EBX is _________ Group of answer choices 00000117 FFFFFF75 000001A2 none of themQuestion 13 Assume that EBX and ECX have the following values EBX: FF FF FF 75 ECX: 00 00 01 A2 After the execution of the instruction ADD EBX, ECX The Value in EBX is _________ Group of answer choices 000001A2 00000117 00000117 none of them Question 14 The MOV instruction changes the flags Group of answer choices True False Question 15 The Binary Representation for each of the following Hexadecimal number 7C Group of answer choices 1111000 11110100 1111100 none of them