Translate the following C program to Pep/9 assembly language. #include int inpNum, sum, count, min; int main() { sum = 0; count = min = 0; 32767; printf("? "); scanf("%d", &inpNum); while (inpNum != 0 ) { sum += inpNum; count++; if (inpNum < min) { min = inpNum; } printf("? "); scanf("%d", &inpNum); } // end while printf("Sum printf("Count = printf("Min } // end of main = %d\n", sum); = %d\n", count); %d\n", min);

How does the concept of 'garbage collection' contribute to memory management in programming languages?

Convert the following c++ code into pep9 assembly language. #include <iostream> using namespace std; void times(int& prod, int mpr, int mcand) { prod = 0; while (mpr != 0) { if (mpr % 2 == 1) prod = prod + mcand; mpr /= 2; mcand *= 2; } } int main(){ int product, n, m; cout << "Enter two numbers: "; cin >> n >> m; times(product, n, m); cout << "Product: " << product << endl; return 0; }

Convert the following c++ code into pep9 assembly language #include <iostream> using namespace std; void times(int& prod, int mpr, int mcand) { prod = 0; while (mpr != 0) { if (mpr % 2 == 1) prod = prod + mcand; mpr /= 2; mcand *= 2; } } int main(){ int product, n, m; cout << "Enter two numbers: "; cin >> n >> m; times(product, n, m); cout << "Product: " << product << endl; return 0; }

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.

Complete the following C++ programs into Pep/9 assembly language: 1) int main(){int cop; int driver; cop = 0; driver = 40; while (cop <= driver) { cop += 25; driver += 20; } cout << cop; return 0;}

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...

translate a c++ program into pep/9 assembly language #include <iostream> using namespace std; int myAge;void ShowVal(int age){ cout << "Age:" << age << endl;} int main() {cout << "Enter age: ";cin >> myAge;ShowVal(myAge);return 0;}

15. Translate the following C program to Pep/9 assembly language. #include int main() { int numitms, j, data, sum; scanf("%d", &numItms); sum = 0; for (j = 1; j <= numitms; j++) { scanf("%d", &data); sum += data; } printf ("Sum: %d\n", sum); return 0; } Sample Input 48-376 Sample Output Sum: 18