Write the assembly code for this C statement as a MIPS procedure: for ( i=0; i<100; i++) A[i] = 0; Assumptions: • A: .space 400 • la $s0, A #Array A is defined # $$0 = Address of A
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- Convert the following to Pep/9 assembly: #include <iostream>using namespace std;int main(){ int number; cin >> number; if (number % 2 == 0) { cout << “Even\n”; } else { cout << “Odd\n”; } return 0;} show the screen shot to of pep/9 code workingTake 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 commentsPROBLEM 20 - 0494: Given a two-dimensional array AMATR which contains 10 rows and 10 columns and a one- dimensional array called DIAG which contains 10 elements, write a program segment in FORTRAN to compute the elements of D1AG from DIAG(I) = AMATR(I, I) for I =1,2,. .,10. Then compute the trace of AMATR, which is defined as trace(AMATR) = 1°£i = 1 %3D AMATR(I, I) = 10Ei = 1 DIAG(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…Language:C Write the function that produces the largest and smallest elements of an integer array sent to it, in accordance with the main function below. (Access to array elements should be done according to offset-address increment notation.)Sample Run:Biggest : 23Smallest : 0Sample Main................................/*prototip*/int main(){int array[5]={23, 4, 2, 0, 8};int biggest;int smallest;bigSmallFind(.............................)printf("Biggest : %d \n", biggest);printf("Smallest : %d \n", smallest);return 0; }MIPS Assembly The program: Write a function in MIPS assembly that takes an array of integers and finds local minimum points. i.e., points that if the input entry is smaller than both adjacent entries. The output is an array of the same size of the input array. The output point is 1 if the corresponding input entry is a relative minimum, otherwise 0. (You should ignore the output array's boundary items, set to 0.) My code: # (Note: The first/last entry of the output array is always 0# since it's ignored, never be a local minimum.)# $a0: The base address of the input array# $a1: The base address of the output array with local minimum points# $a2: Size of arrayfind_local_minima:############################ Part 2: your code begins here ###la $t1, ($t2)la $t1, ($t2)move $a1, $s0 li $a2, 4jal find_local_minima print:ble $a2, 0, exitlw $a0, ($s0)li $v0, 1syscall addi $s0, $s0, 4addi $a2, $a2, -1 ############################ Part 2: your code ends here ###jr $ra I am not getting the correct…
- Conver the program into assembly language#include <iostream>using namespace std;int sum(int n){ if (n <= 0) return 0; else return n + sum(n-1);}int main(){ cout << "Range num? "; int num; cin >> num; cout << sum(num) << endl; return 0;}Debug my C code and translate the small code into assembly language. Thank you! #include <stdio.h> #include <stdlib.h> int Length = 4; int Seq[] = {17, 34, 51, 68}; int main() { int Next = 0; //Sets value for Next. int i = 1; // sets initial value for i. while (i != 4){ //condition for while loop. Next = Next + 17; //adds integer value to Next. i++; //post increment for @param i. return Next; //returns value for Next. } // your program should use this print statement printf("The next integer in the sequence: %d\n", Next); return 0; }In c++ write VM translator in which it will read a program written in HACK vm from an external file and ultimatley translate each line of code into Hack asm (assembly) , so, Higher level Hack Virtual machine language to Hack level assembly language. For example make it possible to translate the following: One Arithmetic (SUB), one logic (AND), and then Memory access command POP.
- F Convert the following C++ program into assembly language. int table[10]; void get(int &x ) { cout > X; } int main() { int total = 0; int i; for (i = 0; i <10; i++) { get( table[i]); total = total + table[i]; } cout << total << endl; }Objectives: -write assembly language programs to: -define a recursive procedure/function and call it. -use syscall operations to display integers and strings on the console window -use syscall operations to read integers from the keyboard. Assignment Description: Implement a MIPS assembly language program that defines "main", and "function1" procedures. The function1 is recursive and should be defined as: function1 (n) = (n mod 7) - 11 if n <= 4 = function1(n-2)*n - n*function1(n-4) - 5*n otherwise# //Write the assembly for the following loop # //Use indexed array access (no pointers) 15; const int AMAX # # # # . data A: int A[] = {0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15}; for (int i=AMAX; i > 0; i--) { A[i-1] = A[i] * 2; } # DO NOT CHANGE THIS TEST DATA .word 1, 2, 3, 5, 6, 7, 4, 9, 10, 11, 12, 13, 14, 15 .word AMAX: . word 15 .text .globl main $t0, AMAX $s0, A #TODO: Write the loop code: A[i+1] = A[i] * 2 main: #highest index #&A[0] lw la loоp: exit: li $v0, 10 syscall