Suppose the state of the RAM is as follows and the adjacent assembly codewill execute:RAM031220فيا364565141234567$1A = MA = MA = MD = M@4M = DWhat will be the value of the RAM[4] following the assembly codeexecution?

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Answered: Suppose the state of the RAM is as…

C++ Programming: From Problem Analysis to Program Design

8th Edition

ISBN:9781337102087

Author:D. S. Malik

Publisher:D. S. Malik

Chapter12: Points, Classes, Virtual Functions And Abstract Classes

Section: Chapter Questions

Problem 19SA

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solve fast | microprocessor 8086
There is an application that requires the following hardware: an Intel 8031, a Program ROM of 8Kx8, a Data ROM of 4Kx8 for look-up tables and a Data RAMs of 8Kx8. The memory map of the design: Program ROM should start at address 0000H. Then, the Data ROM should come above the Program ROM. Finally the Data RAM must go to the top of the memory map. There should be no gaps between the memory addresses of the external ROMs. Show the design’s address space on a memory map, starting with 0000H at the bottom and FFFFH at the top.
Please solve this problem in emu8086 microprocessor in Assembly Language. Please solve properly, will upvote after getting proper solve Read 10 inputs (range 0-9) from the user and store them in an array. Find the position (or index) of the maximum and minimum values in the array, and swap them (move the biggest element to the position of the smallest, and move the smallest element to the position of the biggest), and print that again. If there are multiple maximum or minimum numbers, then choose the first occurrence, values. Your program should be able to handle negative values. Explanation: Here from the given input we see that 9 is the maximum number and it has occurred twice, first in the 2nd index and again in the 5th index. Similarly here 3 is the least and it also has appeared twice, once in the 3rd index and again in the 6th index. So we swap 9 and 3 from the 2nd and 3rd indices and then after swapping, print the new array. Input: 7935935874 Output: 7395645874 Comment:Range 0…
There is an application that requires the following hardware: an Intel 8031, a Program ROM of 8Kx8, a Data ROM of 4Kx8 for look-up tables and a Data RAMs of 8Kx8. The memory map of the design: Program ROM should start at address 0000H. Then, the Data ROM should come above the Program ROM. Finally the Data RAM must go to the top of the memory map. There should be no gaps between the memory addresses of the external ROMs. Calculate the address space of the ROMs and RAMs of your design.
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Write a program in HACK assembly, without using symbols, that computes thebitwise exclusive or (XOR) of the values stored in RAM[1] and the value of thememory location with address stored in RAM[2]. The result of the computationshould be stored in RAM[0].You can think of RAM[2] as being a pointer to where the second operand of the XORis stored.
Computer science
Suppose a program segment consists of a purely sequential part which takes 100 cycles to execute, and an iterated loop which takes 400 cycles per iteration. Assume that the loop is dependent on the sequential part, i.e., both parts cannot run in parallel. Also assume that the loop iterations are independent, and cannot be further parallelized. If the loop is to be executed 100 times, what is the maximum speedup possible using an infinite number of processors (as many processors as you could possibly need) compared to a single processor?
Suppose you have an Intel 8086 Microprocessor which is running at a frequency of 50 MHz. Now what is the maximum number of times it can read a byte of data from the memory in 3 seconds? 12500000 37500000 1250000 3750000 Suppose an Intel 8086 Microprocessor is operating at 8 MHz. Which of the following statements are true? ] The clock cycle is 125 ns | In each bus cycle, the clock remains high for 41.25 ns Data is supplied during the 250 ns to 375 ns period in a bus cycle. While reading, ALE stays high during the 125 ns to 250 ns period in a bus cycle. The Status Bus bits becomes available before 250 ns in a bus cycle.
Write a assembly code for 8086/8088 that replace the contents of each cell with the sum of the contents of all the cells in the original array from the left end to the cell in question. Thus, for example, if the array passed to the function looks like this: 52937 then when the function returns, the array will have been changed so that it looks like this: 5 7 16 19 26
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 = 0XD4
Design a Read Only Memory (ROM) to implement the following, A097803, polynomial A=3(2x2 +1), with input (x): 0≤(x)≤7 (a) What is the size of the initial (unsimplified) ROM ? (b) What is the size of the final (simplified) ROM ? (c) Show in detail the final ROM layout, using LogiSim.can you do letter C using Logisim APPthank you!

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