Assume the memory for x and y has been initialized as: x = 10 y = 2 What will be stored in "d" after the following virtual machine code is executed? push 2 push x sub push y push 9 add add pop d
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- Q4: Disk controllers map logical blocks to physical locations on the disk. Suppose that a 10MB file consisting of 1024-byte logical blocks is stored on a disk drive with the following characteristics: Rotational rate: Average seek time: Average sectors/track: Surfaces: Sector size: 12,000 RPM 8ms 1024 8 1024 bytes Suppose that a program reads all the blocks of this file sequentially, and that the time to position the head over the first block is the average seek time plus the average rotational latency. 4a) What is the best case for mapping logical blocks to disk sectors? Estimate the time required to read the file in this best-case scenario. 4b) Suppose that the logical blocks are mapped randomly to disk sectors. Estimate the time required to read the file in this scenarioGenerate a listing file for AddTwoSum.asm(in 3.4.3 of the text) and write a description of the machine code bytes generated for each instruction. You can write your descriptions directly over the .lst file. You might have to guess at some of the meanings of the byte values. ; AddTwoSum.asm - Chapter 3 example. .386.model flat,stdcall.stack 4096ExitProcess proto,dwExitCode:dword .datasum dword 0 .codemain proc mov eax,5 add eax,6 mov sum,eax invoke ExitProcess,0main endpend mainConsider the following assembly program of the basic computer as seen in class. What would be the value of variable X after executing the following instructions? Address Assembly ORG 2FE 2FE CLE 2FF LDA X 300 BSA rot 301 STA X 302 HLT х, НЕХ ЗЕАТ Y, HEX 3FA7 303 304 305 rot, HEX 0 306 CIL 307 CIL 308 CIL 309 CIL 30A BUN rot I END O 3FA7 O FA71 None of these ZF4E O FE9C FD38 O 7D3A
- Consider the following assembly program of the basic computer as seen in class. What would be the value of variable X after executing the following instructions? Address Assembly ORG 2FE 2FE 2FF CLE LDA X BSA rot 300 301 STA X 302 HLT Х, НЕХ ЗFА7 Y, НЕХ ЗҒАТ rot, HEX 0 CIL 303 304 305 306 307 CLE 308 CIL 309 CLE 30A BUN rot I END None of these 7D3A O 7F4E FA71The Subject is Microprocessor Systems.INSTRUCTIONS: 1. Implement the memory placement algorithms in Java Language Assume we have this scenario: A system contains the following list of memory partitions with the specified size (unit measured by Kilobytes/Kb), where partition marked with 'X' assumed is occupied by a process and thus can't be used for allocation/placement, and 'R' is the most recent placement inside the partitions: X R X 100 20 80 50 50 120 100 Based the given scenario, create a single or several Java programs which perform a simulation of memory placement algorithm using Next Fit algorithm To allow user to test the simulation, each program will allow the user to specify as input, a total (N) of new processes to be placed inside the partitions • a list of N integers (each integer, in Kb, represents the size of a new process to be placed) Assume the user entered some data as input (total=5 processes). Then the program must display the resulting list of process placements such as follows: Total of processes to…
- Case Study Assume a computer system has a main memory of 256 Bytes. The following is a memory byte-access trace history of a program run on this system. For example, as it is shown, the program first accesses memory address 0000 0000, and then it accesses memory address 0000 0001 and so on. Note the memory addresses are represented in binary: 00000000, 00000001, 00000010, 00000011, 00001000, 00010000, 00010001, 00000100, 00000101, 00000110, 00000111, 00001001, 00001010, 00001011, 00001100, 00001000, 00001001, 00001010, 00011100, 00011101. Q3. Assume the system has a 16-Byte direct mapped unified L1 cache with a block size of 2 Bytes. The following table shows how the cache looks like after the first access to the memory is finished. Please show; in the provided table, how it looks like after the 20th access is finished. You could ignore the "Data" Column. (Add or remove Rows/Columns in the provided table for your answer, as you see fit). Cache contents after the 1st access: Cache Index…In c++ write an assembler in which it will read a program written in HACK assembly language from an external file and ultimatley translate each line of code into the binary equivalent that can be run on the computer I built so based off the following hdl files Computer below others are in the images. CHIP Computer { IN reset; PARTS: //Read-only memory (ROM) for instruction fetch ROM32K(address=PC,out=instruction); //Central Processing Unit (CPU) for instruction executionCPU(instruction=instruction,reset=reset,inM=outMemo,outM=CPUoutM,writeM=wM,addressM=adM,pc=PC); //Memory for data storage and control logic Memory(in=CPUoutM,load=wM,address=adM,out=outMemo); }Write a Java program that prompts the user for the page size used in a virtual memory system; this will be a power of two between 512 (29) and 16384 (214), inclusive. Your program should check the user input for page size to make sure it is one of the allowable inputs (must be a power of 2 and cannot be smaller than 512 or larger than 16384), and should then prompt the user for a virtual address (assume 32-bit virtual addressing, so your code must be able to accept any input between 0 and 4294967295, which is 232-1). Given this input, the program must output the virtual page number and the offset within the page. Sample output might look like this: Please enter the system page size: 1024 Please enter the virtual address: 10000 This address is in virtual page: 9 At offset: 784
- Return-Oriented Programming (ROP). Return-oriented programming uses sequences of instructions that arealready part of the executable program. These are called gadgets; each gadget ends with the single-byte instruction ret (binary code 0xc3). The exploit input consists of (i) enough padding to reachthe return address from the vulnerable function (can be anything), followed by (ii) a sequence ofaddresses of gadget code in the program. Each gadget executes a few instructions that do somethinguseful (e.g., move a particular value into a particular register), then “return” to the address on thetop of the stack, which starts the next gadget.ROP is important to know about because attackers can use it when countermeasures like non-executable stack make code injection infeasible. In this problem, you will search for gadgets inprograms on your VM.(Nota bene: ROP is a technique for exploiting an existing vulnerability—i.e., buffer overflow—in a program; it does not create any vulnerabilities. So…EX2: « Explain what happens as the following sequences of instructions are executed : . MOV DL.05H . MOV AX,0A00H . MOV DS AX . MOV SI,0H . MOV CX,0FH « Again: INC SI . CMP [SI].DL . LOGCPNE AgainWhen translating conditionals or loops, the generated low-level code contains jumps. For the CPU to be able to execute a jump, the target of a jump must be a valid memory address. However, code generators typically generate jumps to symbolic addresses (labels). For example: beq exit Answer the following questions about symbolic addresses. i. What are the advantages of using symbolic addresses? ii. Which programs are responsible for translating such symbolic addresses to actual memory addresses? iii. How do code generators create symbolic addresses?
