Assume X address is 29D, find the value of AX for each of the following instruction .Data X byte 12, 3.5 , 2 dup (2, 2 dup(2,1)), 65 , 1 15, 88.29 , 506 '1', "23", '32', '0', '1',0, '2', '13' W Real 4 Z Word "198", 0 Qword Y 1936 , 7610 , 3.5 • Mov AL, Offset AX=. • Mov AH, Length Of AX=. • Mov AX , Size of Y AX= . Mov AH , TYPE W AX=.. Z .......... .... .... .....
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A: your answer is given below:
Q: Consider the following expression: M = U/(V*W + X*Y - Z). a) List a sequence of instructions to…
A: Actually, given expression : M = U/(V*W + X*Y - Z)..
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A: Here is the answer with an explanation:-
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A: Here, I have to provide a solution to the above question.
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A: Answer: I have given answered in the handwritten format in brief explanation.
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A: the option c is correct
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A: Here is the solution to the above problem: -
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A: According to the information given:- We have to find the register values of after execution.
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A: Actually, binary numbers are nothing but a 0's and 1's.
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A: The answer as follows
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Q: 2- Find the physical destination address of last instruction below MOV BX,0AAH MOV AX,1BBH MOV DS,AX…
A: Please give positive ratings for my efforts. Thanks. ANSWER BX = 0AA H AX = 1BB H DS = AX = 1BB…
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A:
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A: ff
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Q: B2.JPG MOV SP,[BP+DI+ACBAH] ВJPG PHY. ADD| С6079H АСH C607AH BDH С607вн | СЕН C607CH| F1H C607DH 02H…
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A: The Answer is in Below Steps
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Q: Q1) If BX=1000, DS=0200, SS=0100, CS=0300 and AL=EDH, for the following instruction: MOV [BX] +…
A: Given, BX =1000 DS =0200 SS =0100 CS =0300 AL =EDH Instruction = MOV [BX]+1234H,AL Physical…
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- A(n) ________________ instruction always alters the instruction execution sequence. A(n) ______________ instruction alters the instruction execution sequence only if a specified Condition is true.6. Assume that two numbers: dividend and divisor are saved in memory address M1 and M2 respectively. Quotient and remainder should be saved in R1 and R2 respectively. Write assembly language instructions and then list microoperations for each instruction and list the control signals required to be activated for each microoperation. MBR is used as buffer for any register to register transfer operation. Signal Description: Control signals operation Comments C0 MAR to RAM (through address bus) C1 PC to MBR C2 PC to MAR C3 MBR to PC C4 MBR to IR C5 RAM to MBR C6 MBR to ALU C7 Accumulator to ALU C8 IR to MAR C9 ALU to Accumulator C10 MBR to Accumulator C11 Accumulator to MBR C12 MBR to RAM (through data bus) C13 IR to Control Unit C14 MBR to R1 C15 MBR to R2 C16 MBR to R3 C17 MBR to R4…Assume the following register contents: $t0 = 0x01234567, $t1 = 0x56781234. Set back to these values back after answering each question. For the register values shown above, what is the value of registers ($t0, $t1, $t2, $t3) after executing each instruction? sll $t2, $t0, 2 and $t3, $t2, $t1
- Assume the register ($s1) contains (0x12345678). Write at most two instructions to move ONLY the fourth byte value in the register ($s1) into the data memory at address stored in ($s0). Hint: In this problem, the fourth byte value in the register ($s1) = "0x12"Consider the following store instruction: SW R1, 0x000F(R0). Assume that the registers R0 and R1 are initialized with 0x00000001 and 0x53A78BC Frespectively. A section of the MIPS byte addressable data memory is shown. Give the memory word of the following memory locations after the SW operation: (a). 0x00000015. (b). 0x00000014. (c). 0x00000013. (d) 0x00000012.(e). 0x00000011. (f). 0x00000010.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]
- For the MIPS assembly instructions below, what is thecorresponding C statement? Assume that the variables f, g, h, i, and j areassigned to registers $s0, $s1, $s2, $s3, and $s4, respectively. Assume thatthe base address of the arrays A and B are in registers $s6 and $s7,respectively. Note: for each line of MIPS code below, write the respective Ccode. After that, write the corresponding C code for the MIPS.sll $t0, $s0, 2add $t0, $s6sll $t1, $s1, 2 add $t1, $s7, $t1lw $s0, 0($t0)addi $t2, $t0, 4lw $t0, 0($t2)add $t0, $t0, $s0sw $t0, 0($t1)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. IndexedAssume the register ($s1) contains (0x12345678). Write at most two instructions to move ONLY the third byte value in the register ($s1) into the data memory at address stored in ($s0). Hint: In this problem, the third byte value in the register ($s1) = "0x34"
- If r0 = 0x20000000 and r1 = 0xC3B2A, after STR r1, [r0], #4 instruction is executed, which of the following is the data in memory address 0x20000002 (data layout big endian)? A. 0xB2 B. 0x2A C. 0x3B D. 0xC3 E. 0xC01. Name all of the general purpose registers and some of their special functions. 2. How are the segment registers used to form a 20-bit address? 3. (a) If CS contains 03E0H and IP contains 1F20H, from what address is the next instruction fetched? (b) If SS contains 0400H and SP contains 3FFEH, where is the top of the stack located? (c) If a data segment deigns at address 24000H, what is the address of the last location in the segment? 4. Explain what the instruction array and data caches are used for. 5. What is the EU and BIU, and what purpose in the microcomputer? 6. Two memory locations, beginning at address 3000H, contain the bytes 34H and 12H. What is the word stored at location 3000H? See Figure 2.26 for details. Address 3000 Data 34 3001 12 Figure 2.26 For question 6 7. What is a physical address? What are the differences between the 8086 logical and physical memory maps? 8. May memory segments overlap? If so, what is the minimum number of overlapped bytes…For the MIPS assembly instructions below, what is the corresponding C statement?Assume that the variables f, g, h, i, and j are assigned to registers $s0, $s1, $s2, $s3, and$s4, respectively. Assume that the base address of the arrays A and B are in registers $s6 and $s7, respectively. sll $t1, $s1, 2add $t1, $t1, $s6lw $t1, 0($t1)sub $t0, $s3, $s4sll $t0, $t0, 2add $t0, $t0, $s7lw $t0, 0($t0)add $t1, $t1, $t0sll $t0, $s0, 2add $t0, $t0, $s7sw $t1, 0($t0)