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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Q: Q1) If BX=1000, DS=0200, SS=0100, CS=0300 and AL=EDH, for the following instruction: MOV [BX] +…
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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: Answer: I have given answered in the handwritten format in brief explanation.
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A: Here is the solution to the above problem: -
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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: given, the contents of the register SI=AECDH. convert, AECDH from hexadecimal to decimal. 160 * 13 +…
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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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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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- Q1:/ Show the contents in hexadecimal of registers PC, AR, DR, AC, IR and SC of the basic computer when an instruction at address 021 in the basic computer has I = 1, an operation code of the ADD instruction, and an address part equal to 051. The memory word at address 051 contains 0083. The memory word at address 083 contains B8F2. The memory word at address 038 contains A837 and the content of AC is A937. Give the answer in a table with six columns, one for each register and a row for each timing signal. (All numbers are in hexadecimal) uipors - ea02:- (A) Find the phicycal address if (BP) = 0100H, (SI) = 0200H , (SS) = 2000H and a displacement of 10H, of the instruction MOV AL. (BP+S+10H]. Which the name of Addressing Modes? 02:- (B) Find the result of the following: 1111-iIH+110010111011B 03: Write a piece of code to find the number of negative integers in an array of size 1024 bytes contain signed numbers stored at addresses starting at 21000H, store the result in a location 51000HRegister R1 (used for indexed addressing mode) contains the value: 0x200 Memory contains the values below (memory address -> contents) : 0x100 -> 0x600 ... 0x400 -> 0x300 ... 0x500 -> 0x100 ... 0x600 -> 0x500 ... 0x700 -> 0x800 When the instruction "Load 0x500" is executed, the value loaded into the AC is when using immediate addressing, it is when using direct addressing, it is when using indirect addressing, and when using indexed addressing.
- Memory 12200 12201 12202 12203 12204 Content %D AA EE FF 22 What result is produced in the destination operand by execution the following instruction? a- LEA SI[DI+Bx+5] b- LDS SI.[200]Q: Compute the physical address for the specified operand in each of the following instructions. The register contents and variable are as follows: (CS)=D0A00H, (DS)=OBOOH, (SS)=0DO0H, (SI)=OFFOH, (DI)=00BOH, (BP)=00EAH and (IP)=0000H, LIST=D00FOH, AX=4020H, BX=2500H. 1) Destination operand of the instruction MOV LIST (BP+DI], AX 2) Source operand of the instruction MOV CL, [BX+200H] 3) Destination operand of the instruction MOV [DI+6400H] , DX 4) Source operand of the instruction MOV AL, [BP+SI-400H] 5) Destination operand of the instruction MOV (DI+SP] , AX Source operand of the instruction MOV CL, [SP+200H] 7) Destination operand of the instruction MOV [BX+DI+640O0H], CX 8) Source operand of the instruction MOV AL , [BP- 0200H] 9) Destination operand of the instruction MOV [SI] , AX 10) Destination operand of the instruction MOV [BX][DI]+0400H,AL 11) Source operand of the instruction MOV AX, [BP+200H] 12) Source operand of the instruction MOV AL, [SI-0100H] 13) Destination operand…5. Load the register (CL) from the memory location [0500H] then subtract the content of this register from the accumulator (AL). Correct the result as a (BCD) numbers . Let [0500H] = 12H & AL = 3FH %3D
- Complete the following table: MIPS Instruction op code rs rt rd shamt funct imm. /address Hexadecimal Representation add $t4, $s2, $s1 addi $s0, $t0, 123 lw $s6, -88($t7) Note: In MIPS register file, temporary registers $t0-$t7 have indices 8-15 (respec- tively). Also, the saved registers $s0-$s7 have indices 16-23 (respectively).Given a memory load instruction, "mov R0; [R1+1000]," please give the input that should be selectedat each multiplexer. You can write "none" for the multiplexers that are not used for this instruction.(a) MUX1:(b) MUX2:(c) MUX3:(d) MUX4:Q: Compute the physical address for the specified operand in each of the following instructions. The register contents and variable are as follows: (CS)=0A00H, (DS)=0B00H, (SS)=0D00H, (SI)=OFFOH, (DI)=00BOH, (BP)=00EAH and (IP)=00O0H, LIST=00FOH, AX=4020H, BX=2500H. 1) Destination operand of the instruction MOV LIST [BP+DI] , AX 2) Source operand of the instruction MOV CL, [BX+200H] 3) Destination operand of the instruction MOV [DI+6400H] , DX 4) Source operand of the instruction MOV AL, [BP+SI-400H] 5) Destination operand of the instruction MOV [DI+SP] , AX 6) Source operand of the instruction MOV CL, [SP+200H] 7) Destination operand of the instruction MOV [BX+DI+6400H] , CX 8) Source operand of the instruction MOV AL , [BP- 0200H] 9) Destination operand of the instruction MOV [SI] , AX 10) Destination operand of the instruction MOV [BX][DI]+0400H,AL 11) Source operand of the instruction MOV AX, [BP+200H] 12) Source operand of the instruction MOV AL, [SI-0100H] 13) Destination operand…
- if BX=1000, DS=0400, and AL=EDH, for the following instruction: MOV [BX] + 1234H, AL. the physical address is O 6243H O 4234H O 6234H O 6324H O 4244HYou are given the following instruction: ADD AX, [10h] You are provided the following data: DS = AB12h ; SS = 2567h ; CS= 29C1h Find the effective address location for the given instruction.25: . Find the time delay in the following program if the crystal frequency is 1 MHz. Do not ignore the time delay due to the first and last instruction. DELAY: LDI R16, 30 AGAIN: LDI R17, 35 HERE: NOP NOP DEC R17 BRNE HERE DEC R16 BRNE AGAIN RET 26: Write a program to display 2 on 7 segment. A 7 segment is connected to PortD.