int x = 0; co x = x + 3; || x = x + 1; oc Command A = B + C is implemented with machine instructions; it is mapped to the atomic instruction TEMP = B + C; A = TEMP; where TEMP is the local process variable (so each process has its own). List all possible program traces and give the final value for each trace.
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co x = x + 3;
|| x = x + 1;
oc
Command A = B + C is implemented with machine instructions; it is mapped to the atomic instruction TEMP = B + C; A = TEMP; where TEMP is the local process variable (so each process has its own). List all possible
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- MIPS Simulator QtSpim: You are to have a complete program in MIPS assembly language that behaves exactly as the included C program. This program contains four functions in addition to the main() one. Your solution must contain all five C routines as they have been coded in the example. Make sure to run the program in MIPS and show the same output on MIPS as well to make sure there are no errors. Below is the five C routines and attached is the image of what the output must print out on QtSpim. #include <stdio.h> int getMax(int arr[], int n){int mx = arr[0];for (int i = 1; i < n; i++)if (arr[i] > mx)mx = arr[i];return mx;}void countSort(int arr[], int n, int exp){int output[n];int i, count[10] = { 0 };for (i = 0; i < n; i++)count[(arr[i] / exp) % 10]++;for (i = 1; i < 10; i++)count[i] += count[i - 1];for (i = n - 1; i >= 0; i--) {output[count[(arr[i] / exp) % 10] - 1] = arr[i];count[(arr[i] / exp) % 10]--;}for (i = 0; i < n; i++)arr[i] = output[i];}void…E In the following code block(Reference:Q11), you will a set of assembly instructions with corresponding line numbers (line numbers are for informational purpose only and they are not part of the source code). 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 mov edx, 5 dec ecx jmp LABEL1 mov eax, 1 LABEL1: mul edx jmp ecx mov edx, 0678h sub edx, eax jmp DWORD PTR [edx] neg ebx add ecx, ebx mov eax, 0 For each of the conditions/scenario listed below, indicate the corresponding line number (that cause or is associated with the condition/scenario). Enter 0 (Zero) if the condition is not caused by the block of code. 1) Memory indirect jump: type your answer... type your answer... type your answer... type your answer... 2) Register indirect jump: 3) Relative short jump: type your answer... 5) Two's complement type your answer... 4) Relative near jump: 6) Unreachable code8. Consider the following C code snippet. // c code void setArray(int num) { int i=5; compare (4, i); int compare (int a, int b) { if (sub (a, b) >= 0) return 1; else return 0; int sub (int a, int b) { return a - b; Implement the C code snippet in ARM assembly language. Use R4 to hold the variable i.
- Example: The Problem Input File Using C programming language write a program that simulates a variant of the Tiny Machine Architecture. In this implementation memory (RAM) is split into Instruction Memory (IM) and Data Memory (DM). Your code must implement the basic instruction set architecture (ISA) of the Tiny Machine Architecture: //IN 5 //OUT 7 //STORE O //IN 5 //OUT 7 //STORE 1 //LOAD O //SUB 1 55 67 30 55 67 1 LOAD 2- ADD 3> STORE 4> SUB 5> IN 6> OUT 7> END 8> JMP 9> SKIPZ 31 10 41 30 //STORE O 67 //OUT 7 11 /LOAD 1 //OUT 7 //END 67 70 Output Specifications Each piece of the architecture must be accurately represented in your code (Instruction Register, Program Counter, Memory Address Registers, Instruction Memory, Data Memory, Memory Data Registers, and Accumulator). Data Memory will be represented by an integer array. Your Program Counter will begin pointing to the first instruction of the program. Your simulator should provide output according to the input file. Along with…do { flag[i] = true; turn = i; while (flag[j] && turn == j); Critical section flag[i] = FALSE; Remainder section } Three processes P1, P2, and P3 are sharing a resource R1 in a mutually exclusive manner. Using compare_and_swap instruction, show code for P1, P2, and P3 that shows the entry section and exit section (similar to the code shown in Question 1). Using this code, show a scenario if all three processes want to enter their critical sections.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
- Implement Round Robin scheduling algorithm in C programming. NOTE: Please do not compute and print the average waiting time and the average turnaround time. You are required to compute these for each process and print these times for each process separately. Add comments in the code so that the code is understandable. Take input from the user: the number of processes, the time quantum and their CPU burst times for n number of processes. Compute and print the completion Time(CT), Turnaround Time(TAT), and Waiting Time(WT) for each process using C Programming. Assuming all the processes have arrived at time 0, please do not take arrival times as input from the user.int total; int i; total = 0; for (i = 10; i > 0; i--) { total + i; } The following translates this into ARM assembler: /* -- sum-to-ten.s */ .text .global start _start: mov rº, #0 mov r1, #10 again: end: add r0, r0, r1 subs r1, r1, #1 bne again mov r7, #1 swi 0 @r0 := 0 @r1 = 10 @r0 := r0 + r1 @ r1 := r1 - 1 @loop again if we haven't hit zero @setup exit @exit In the above assembly language program, we first use the MOV instruction to initialize RO at O and R1 at 10. The ADD instruction computes the sum of RO and R1 (the second and third arguments) and places the result into RO (the first argument); this corresponds to the total += i; line of the equivalent C program. Note the use of the label at the beginning of the ADD instruction. The label is simply the relative memory address of where that instruction is when the assembler loads it into memory. The subsequent SUBS instruction decreases R1 by 1. To understand the next instruction, we need to understand that in addition to the…2) Translate the function origami into a MIPS assembly code fragment: int origami (int Aŋ, int BØ, int a2) { int to = 0; int vo = 0; int t1; do { t1 = A[t0]; if (t1 >= 0) { B[v0] = A[t0]; vo++; to++; } while (t0The Problem Using C programming language write a program that simulates a variant of the Tiny Machine Architecture. In this implementation memory (RAM) is split into Instruction Memory (IM) and Data Memory (DM). Your code must implement the basic instruction set architecture (ISA) of the Tiny Machine Architecture: 1 LOAD 2→ ADD 3→ STORE 4 → SUB 5> IN 6> OUT 7> END 8 → JMP 9> SKIPZ Each piece of the architecture must be accurately represented in your code (Instruction Register, Program Counter, Memory Address Registers, Instruction Memory, Data Memory, Memory Data Registers, and Accumulator). Data Memory will be represented by an integer array. Your Program Counter will begin pointing to the first instruction of the program. For the sake of simplicity Instruction Memory (IM) and Data Memory (DM) may be implemented as separate arrays. Hint: Implementing a struct for your Instructions and an array of these structs as your Instruction Memory greatly simplifies this program. Example:…Branches and Branches Here is a portion of a program. The initial condition codes have the values N = 0, Z = 1, and P=0. Consider the values for these codes after the LC-3 executes the sequence of instructions starting at address X384F. X384F 0101111111100000 X3050 0000100000000001 x3051 | 0000001000000010 x3052 0001000000111111 x3053 0000101000000001 x3054 0001111111101111 Value in condition cell N: Value in condition cell z: Value in condition cell P:Write the mnemonics of the following program: 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, 75SEE MORE QUESTIONS