int total; int i; total = 0; for (i = 10; i > 0; i--) { totali; } The following translates this into ARM assembler: /* -- sum-to-ten.s */ .text .global start _start: end: mov re, #0 mov r1, #10 again: @re: 0 @r1 = 10 add re, re, r1@re: re + r1 r1 - 1 subs r1, r1, #1 @ r1 bne again. @loop again if we haven't hit zero. mov r7, #1 swie @setup exit @exit In the above assembly language program, we first use the MOV instruction to initialize RO at 0 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 registers RO through R15, the ARM processor also incorporates a set of four "flags," labeled the zero flag (Z), the negative flag (N), the carry flag (C), and the overflow flag (V). Whenever an arithmetic instruction has an S at its end, as SUBS does, these flags will be updated based on the result of the

convert this code to java language     /* OPERATING SYSTEMS LAB PROJECT* AKASH JAIN* 179303013* DESIGNING A VIRTUAL MEMORY MANAGER*/ #include<stdio.h>#include<stdlib.h>#include<string.h> const int VM_SIZE=256;const int PAGE_SIZE=256;const int TLB_SIZE=16;const int MM_SIZE=256; int main(int argc, char* argv[]){FILE *fd;if(argc<2){printf("NOT ENOUGH AGRUMENTS\n EXITING\n");return 0;}fd=fopen(argv[1],"r");if(fd==NULL){printf("ERROR OPENING FILE\n FILE FAILED TO OPEN\n");return 0;}char *value=NULL;size_t len=0;ssize_t read;long long page_no,offset,page_table,totalhits=0,fault=0,pages=0;int qp=0; //to maintain the queue positionint physicalad=0,frame,logicalad;int tlb[TLB_SIZE][2];int pagetable[PAGE_SIZE]; memset(tlb,-1,TLB_SIZE*2*sizeof(tlb[0][0]));memset(pagetable,-1,sizeof(pagetable));int mask=255,mask1=62580,i,hit;while((read=getline(&value,&len,fd))!=-1){pages++;//get page number and offset from logical…

Convert the following into Pep/9 Assembler: #include using namespace std; int square(int n){     int i;     int sq;     sq = 0;     for (i = 0; i < n; i++){         sq = sq + n;     }     return sq; } int main (){     int num;     cout << "Enter a number: ";     cin >> num;     cout << num << " squared = "          << square(num)          << endl;     return 0; }

Convert the following into Pep/9 Assembler: #include <iostream>using namespace std;int square(int n){    int i;    int sq;    sq = 0;    for (i = 0; i < n; i++){        sq = sq + n;     }    return sq;}int main (){    int num;    cout << "Enter a number: ";     cin >> num;    cout << num << " squared = "          << square(num)          << endl;    return 0;} Submit: Source file along with screen captures showing the program running in the Pep simulator.

PLEASE DON'T USE STRUCTS Using C programming language write a program that simulates a variant of the Tiny Harvard 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 TinyMachine 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, Instruction Memory (IM), MAR1, MDR-1(MAR-1 and MDR-1 are connected to the IM). Data Memory, MAR-2, MDR2 (MAR-2 and MDR-2 are connected to the DM), and Accumulator. Instruction Memory will be represented by an integer array and each instruction will use 2 elements of the array(one for OP and the other one for address) Data Memory will be represented by an integer array and each data value uses an element of the DM array. Your Program Counter will begin pointing to the first…

In assembly 68k write the: Initialize the supervisor stack pointer to $8000 Initialize the program counter to $1000 Reserve locations at $2000 to hold your first name only (Joshua) as text and call it NAM Reserve locations at $3000 to store 9 longwords, call it MEM, and write your family name (Kimmich) as a comment Reserve locations at $12400 to store 32-bit unsigned integer variable called RES (to be computed)

Answser must be in MIPSzy assembly language. Max of 3 - Branch

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Computer organisation ! Use RARS simulator to write and execute the two codes below using RISC-V Assembly language.

C PROGRAMMING HELP! I tried a threaded reply and bartleby wasn't having it. Soooo: Please create the header file as specified and implement it into my script. Thank you for your help, it is seriously appreciated! The header file deals with the following instruction prompt: Find the roots of the second order nontrivial solution.D2y+a1Dy+a2 = 0 => λ 2+a1 λ +a2 = 0Create a header file for this step. I believe the portion of the script that needs to be turned into a utilized header file is located in the main() function. Thank you so much for your help, pasted is a copy of my script so far: #include <stdio.h>#include <math.h> void compute_coefficients(double lambda1, double lambda2, double *C1, double *C2, double *C_alpha, double *theta);void display_output(double lambda1, double lambda2, double C1, double C2, double C_alpha, double theta); int main(){    int a, b, c;    double lambda1, lambda2, C1, C2, C_alpha, theta;     printf("Enter a, b, and c where a*D2y + b*Dy + c*y =…