{ int i; float x[5], y[5]; [ lots of code deleted... ] for (i=0;i<5;i++) if (x[i] > y[i]) x[i] = x[i] + y[i]; [ more code deleted... ] Translate to MIPS assembly language:

Help me with this

Consider the following C code: 1: int main() { 2: 3: 4: 5: } int y = 5; // Return y times 4 return y* 4; What is removed during the preprocessing step of compilation? A. Line 1 B. Line 3 C. Line 5 D. Lines 3 and 5 E. Everything, since it will be assembly after the preprocessing step

NAND2TETRIS HARDWARE SIMULATOR HiLoMux - This has one 8-bit input bus, in, and one 4-bit output bus, out. Alsopresent is a sel input, which is used to select what appears on out. Ifsel is false, then out should contain the lower 4-bits of in (i.e. in[0],in[1], in[2], in[3]). If sel is true, then out should contain theupper 4-bits of in (i.e. in[4] mapped to out[0], in[5], mapped toout[1], etc.). In other words, the HiLoMux can be used to select anibble from a byte please use the skeleton program below CHIP HiLoMux{ IN in[8], sel; OUT out[4]; PARTS: }

Translate C program to Pep/9 assembly language. /C code for 6.19a#include <stdio.h>char myChar;char toLower(char ch) { if ((ch >= ’A’) && (ch <= ’Z’)) { return ch + ’a’ - ’A’; } else { return ch; }}int main () { scanf("%c", &myChar); printf("%c\n", toLower(myChar)); return 0;}

Please your code should be in mips assembly language

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...

Take the following program and translate it into PEP/9 assembly language: #include <iostream> using namespace std; int fib(int n) { int temp; if (n <= 0) return 0; else if (n <= 2) return 1; else { temp = fib(n – 1); return temp + fib(n-2); } } int main() { int num; cout << "Which fibonacci number? "; cin >> num; cout << fib(num) << endl; return 0; } You must use equates to access the stack and follow the call to the function as discussed in the book (pass the parameter, return address, return a value and so on). There are NO global variables in the resulting code (except a global message of "Range num? "). It must be able to do sum a range greater than 2. comments would be appreciated

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.

Take the following program and translate it into PEP/9 assembly language: #include <iostream> using namespace std; int theArray[] = { 5, 11, -29, 45, 9, -1}; void sumPos(int ary[], int len, int &sum) { sum = 0; for (int i = 0; i < len; i++) if (ary[i] > 0) sum = sum + ary[i]; } int main() { int total; sumPos(theArray, 6, total); for (int k=0; k < 6; k++) cout << theArray[k] << endl; cout << "Positive sum is " << total << endl; return 0; } You must use equates to access the stack and the index register in accessing the array. Remember, the sumPos array does NOT know about the global "theArray" – the address of the array must be passed via the parameter. make theArray a local array in main. Submit source code add comments

s. Convert the following C code to assembly language using vector operations. (Just do your best, make necessary assumptions, don't worry about syntax too much.) for (i-0;i <300; i++) { c_re[i] = a_re[i] b_re[i]-a_im[i] * b_im[i]: c_im[i] - a_re[i] b_im[i]+a_im[i] * b_re[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...