. Assuming that DI=100 , DS= 140016, SI-FF00, BX= AE001, AX-100614 and the memory contents explained in the Tablel. Hand trace the execution of the following Table 1 Memory Memory program MOV SI, (BX+33H) Address Content MUL SI IEE30 92 ADD AL, DH IEE31 EA CMP AL, 4E14 IEE32 F5 Then answer the following questions: IEE33 62 1- What are the values of registers in each statement? 1EE34 22 2- What is the effect to the flag registers after execution ADD instruction in 1EE35 8B the third statement of the program?
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- Memory address Data According to the memory view given below, if RO = Ox20008002 then LDRSB r1, [r0, #-4] is executed as a result of r1 = ?(data overlay big endian)? Øx20008002 ØXA1 Øx20008001 ØXB2 Øx20008000 Øx73 ØX20007FFE ØXD4 ØX20007FFE Lütfen birini seçin: O A. R1 = 0X7F O B. R1 = Oxffffffd4 O C. R1 = Oxffffff7F O D. R1=0XD4000000 O E. R1 = 0XD4The memory location at address of 0X003FB01 contains 1-byte memory variable J (0010_0001), and the memory location at the address of 0X003FB02 contains 1-byte memory variable K (0001 0010), see figure below. There is a 2-byte variable M which hold binary information M (1110 0101 0000 1i11). What is the address in hexadecimal format for 2-byte memory variable M, following little Endian computer? 7 Address in Data in Hex. Format Hex. Format 0X003FBF04 1110 0101 M OX003FBF03 0000 1111 0X003FBF02 0001 0010 0X003FBF01 0010 0001 J Its address in hexadecimal is 0X003FBF02. а. Its address in hexadecimal is 0×003FBF03. O b. Its address in hexadecimal is 0X003FBF04. Its address in hexadecimal is 0×003FBF01. d.Write a code in sim8085 for the following problem: The pressure of two boilers is monitored and controlled by a microcomputer works based on microprocessor programming. A set of 6 readings of first boiler, recorded by six pressure sensors, which are stored in the memory location starting from 2050H. A corresponding set of 6 reading from the second boiler is stored at the memory location starting from 2060H. Each reading from the first set is expected to be higher than the corresponding position in the second set of readings. Write an 8085 sequence to check whether the first set of reading is higher than the second one or not. If all the readings of first set is higher than the second set, store 00 in the ‘D’ register. If any one of the readings is lower than the corresponding reading of second set, stop the process and store FF in the register ‘D’. Data (H): First set: 78, 89, 6A, 80, 90, 85 Second Set:71, 78, 65, 89, 56, 75
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- Computer organization and assembly language Please help me with this. I have to write line by line what each line of codes does. CODE IS BELOW: .model small .386 .stack 100h .data msg1 db 13, 10, "Enter any number --> ", "$" msg2 db "Enter an operation +,- * or / --> ",13, 10, "$" msg3 db "The Operation is --> ", "$" msg4 db "The result is --> ", "$" By_base dd 21 by_10 dd 10 ; 32 bits variable with initial value = 10 sp_counter db 0 ; 8 bits variable with initial value of zero disp_number dd 0 ; 32 bits variable with initial value = 0 disp_number2 dd 0 disp_number3 dd 0 op_type db 0 last_key dd 0 ; 32 bits variable with initial value of zero remainder db 0 .code main proc mov ax,@data;set up datasegment movds,ax mov dx,offset msg1 call display_message callm_keyin calloperation mov dx,offset msg1 calldisplay_message callm_keyin cmpop_type, "+" jnz short skip_plus callop_plus skiP_plus: cmp op_type, "-" jnz short skip_minus callop_minus…Select the correct Value of CL and DL registers after execution of below given program. MOV AL, B2 MOV BX, 0000 MOV CL, [0001] MOV AX, BX MOV DL,[BX+1] INT 7 0090:0000 DB AC, 7E, 3B a. CL = 7E and DL = 7E O b. CL = AC and DL = AC O c. CL = 3B and DL = 3B O d. CL = AC and DL = 7EQuiz 5: In this problem we want to set the control signals of the datapath shown below (also in in slide # 1 of "chapter3_single_cycle_datapaths.pptx") so that it supports execution of a new instruction called swi. Single Cycle Datapath: PC Read Instru- address ction [31-0] Instruction memory Sns Add Ins 1 [25-21] 1 [20-16] [15-11]. 1[10-0] RegWrite Read register 1 Read register 2 Write register Write data Read data 1 Read data 2 Read Ins Write 3ns Sign extend 2ns MemWrite Read Read address data Write address Read Gns Write data Write 10ns ins ALUSTO1 MemRead ALU Result 2ns ALUOP1 -XEWO) ins ALUSrc2 ALUSrc3 x=3 ins ALU Result 2ns ALUOP2 swi rd, rs, rt, imm # Memory [R[rs]]= R[rt], R[rd] =R [rs]+R [rt]+Imm #this instruction copies contents of "rt" register into the main memory addressed by the "rs" register. In the same cycle it add "rs" and "rt" register contents along with the "imm" field of the instruction and writes the final result into the "rd" register. You are NOT allowed to…
- Create a program in C++ which simulates a direct cache. The memory array that contains the data to be cached is byte addressable and can contain 256 single byte entries or lines. The cache has only 8 entries or lines. The Data field in each line of the cache is 8 bits. Since the data stored in each line of the cache is only 8 bits, there is no need for a line field. Only a tag field is needed which is log2(256) = 8 bits. The memory array can be filled with any values of your choice. The program should work by taking user input of a memory address (index). This input represents the memory data that should be cached. Your program will check the cache to see if the item is already cached. If it is not, your program should count a cache miss, and then replace the item currently in the cache with the data from the inputted address. Allow the user to input addresses (in a loop), until they so choose to end the program. The program should output the number of cache misses upon ending.Let's say that p is a pointer to memory and the next six bytes in memory (in hex) beginning at p's address are: aa bb cc dd ee ff. What value would be in x if the following code is run on a little- endian computer? uint16_t *q uint16_t x = (uint16_t *)p; q[0]; aa aabb bbaa aabbccdd ddccbbaaDescribe Describe addl moub subg suppox as the difference moug $0x8000, % rax movg 0x8000, rax the following addressing modes. 12 (%rbp), % ecx. (%rax, rcx), % de %rdx, (%rcx, %/orax, 8) OXA (%rcx, 8) array in C is declared! valiable. incw write 18 global long array [34]; @ C use between: assembly code that: sets rsi to the address of array Sets rbx to the Constant g loads array [9] into register so scales index Mc Mory no de!